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    NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

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    NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

    Edited By Ramraj Saini | Updated on Sep 23, 2023 06:43 PM IST

    Straight Lines Class 11 Questions And Answers

    NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines are provided here. In earlier classes, you have studied 2D coordinate geometry. This NCERT Book chapter is a continuation of the coordinate geometry to study the simplest geometric figure – a straight line. A straight line is a line which is not bent or curved. In this article, you get straight lines class 11 NCERT solutions. These NCERT Solutions are prepared by experts keeping in mind CBSE latest syllabus 2023. The Important topics included in the chapter 10 class 11 maths are definition of the straight line, the slope of the line, collinearity between two points, the angle between two points, horizontal lines, vertical lines, general equation of a line, conditions for being parallel or perpendicular lines, the distance of a point from a line. Also, students can practice NCERT solutions for class 11 to command class 11th concepts which are foundation for class 12th concepts.

    In this NCERT Book ch 10 maths class 11, there are 3 exercises with 52 questions. All these questions are explained in comprehensively by Careers360 experts. This chapter is very important for CBSE class 11 final examination as well as in various competitive exams like JEEmains, JEEAdvanced, BITSAT etc. There are 24 questions are given in a miscellaneous exercise. You should solve all the NCERT problems including examples and miscellaneous exercise to get command on this chapter.

    Also read:

    Straight Lines Class 11 Questions And Answers PDF Free Download

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    Straight Lines Class 11 Solutions - Important Formulae

    Distance Formula:

    The distance between two points A(x1, y1) and B (x2, y2) is given by:

    • AB = √((x2-x1)2 + (y2-y1)2)

    Distance of a Point from the Origin:

    The distance of a point A(x, y) from the origin O(0, 0) is given by:

    • OA = √(x2 + y2)

    Section Formula for Internal Division:

    The coordinates of the point which divides the line segment joining (x1, y1) and (x2, y2) in the ratio m:n internally is:

    • ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))

    Section Formula for External Division:

    The coordinates of the point which divides the line segment joining (x1, y1) and (x2, y2) in the ratio m:n externally is:

    • ((mx2 - nx1)/(m - n), (my2 - ny1)/(m - n))

    Midpoint Formula:

    The midpoint of the line segment joining (x1, y1) and (x2, y2) is:

    • ((x1 + x2)/2, (y1 + y2)/2)

    Coordinates of Centroid of a Triangle:

    For a triangle with vertices (x1, y1), (x2, y2), and (x3, y3), the coordinates of the centroid are:

    • ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3)

    Area of a Triangle:

    The area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is given by:

    • (1/2) |x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|

    Collinearity of Points:

    If the points (x1, y1), (x2, y2), and (x3, y3) are collinear, then:

    • x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0

    Slope or Gradient of a Line:

    The slope (m) of a line passing through points P(x1, y1) and Q(x2, y2) is given by:

    • m = (y2 - y1)/(x2 - x1)

    Angle between Two Lines:

    The angle (θ) between two lines with slopes m1 and m2 is given by:

    • tan θ = |(m2 - m1)/(1 + m1m2)|

    Point of Intersection of Two Lines:

    Let the equations of lines be a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. The point of intersection is:

    • ((b1c2 - b2c1)/(a1b2 - a2b1), (a2c1 - a1c2)/(a1b2 - a2b1))

    Distance of a Point from a Line:

    The perpendicular distance (d) of a point P(x1, y1) from the line Ax + By + C = 0 is given by:

    • d = |(Ax1 + By1 + C)/√(A2 + B2)|

    Distance Between Two Parallel Lines:

    The distance (d) between two parallel lines y = mx + c1 and y = mx + c2 is given by:

    • d = |c1 - c2|/√(1 + m2)

    Different Forms of Equation of a Line:

    Various forms of the equation of a line include:

    • Normal form: x cos α + y sin α = p

    • Intercept form: x/a + y/b = 1

    • Slope-intercept form: y = mx + c

    • One-point slope form: y - y1 = m(x - x1)

    Two-point form: y - y1 = ((y2 - y1)/(x2 - x1))(x - x1)

    Free download NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines for CBSE Exam.

    Straight Lines Class 11 NCERT Solutions (Intext Questions and Exercise)

    Straight lines class 11 solutions - Exercise: 10.1

    Question:1 Draw a quadrilateral in the Cartesian plane, whose vertices are (-4,5),(0,7),(5,-5) and (-4,-2) . Also, find its area.

    Answer:

    1646804957415

    Area of ABCD = Area of ABC + Area of ACDNow, we know that the area of a triangle with vertices (x_1,y_1),(x_2,y_2) \ and \ (x_3,y_3) is given byA = \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|Therefore,Area of triangle ABC = \frac{1}{2}|-4(7+5)+0(-5-5)+5(5-7)|= \frac{1}{2}|-48-10|= \frac{58}{2}=29Similarly,Area of triangle ACD = \frac{1}{2}|-4(-5+2)+5(-2-5)+-4(5+5)|= \frac{1}{2}|12-35-40|= \frac{63}{2}Now,Area of ABCD = Area of ABC + Area of ACD
    =\frac{121}{2} \ units

    Question:2 The base of an equilateral triangle with side 2a lies along the y -axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

    Answer:

    1646804999680 it is given that it is an equilateral triangle and length of all sides is 2a
    The base of the triangle lies on y-axis such origin is the midpoint
    Therefore,
    Coordinates of point A and B are (0,a) \ \ and \ \ (0,-a) respectively
    Now,
    Apply Pythagoras theorem in triangle AOC
    AC^2=OA^2+OC^2
    (2a)^2=a^2+OC^2
    OC^2= 4a^2-a^2=3a^2
    OC=\pm \sqrt3 a
    Therefore, coordinates of vertices of the triangle are
    (0,a),(0,-a) \and \ (\sqrt3a,0) \ \ or \ \ (0,a),(0,-a) \and \ (-\sqrt3a,0)

    Question:3(i) Find the distance between P(x_1,y_1) and Q(x_2,y_2) when :

    PQ is parallel to the y -axis.

    Answer:

    When PQ is parallel to the y-axis
    then, x coordinates are equal i.e. x_2 = x_1
    Now, we know that the distance between two points is given by
    D = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|
    Now, in this case x_2 = x_1
    Therefore,
    D = |\sqrt{(x_2-x_2)^2+(y_2-y_1)^2}| = |\sqrt{(y_2-y_1)^2}|= |(y_2-y_1)|
    Therefore, the distance between P(x_1,y_1) and Q(x_2,y_2) when PQ is parallel to y-axis is |(y_2-y_1)|

    Question:3(ii) Find the distance between P(x_1,y_1) and Q(x_2,y_2) when :

    PQ is parallel to the x -axis.

    Answer:

    When PQ is parallel to the x-axis
    then, x coordinates are equal i.e. y_2 = y_1
    Now, we know that the distance between two points is given by
    D = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|
    Now, in this case y_2 = y_1
    Therefore,
    D = |\sqrt{(x_2-x_1)^2+(y_2-y_2)^2}| = |\sqrt{(x_2-x_1)^2}|= |x_2-x_1|
    Therefore, the distance between P(x_1,y_1) and Q(x_2,y_2) when PQ is parallel to the x-axis is |x_2-x_1|

    Question:4 Find a point on the x-axis, which is equidistant from the points (7,6) and (3,4) .

    Answer:

    Point is on the x-axis, therefore, y coordinate is 0
    Let's assume the point is (x, 0)
    Now, it is given that the given point (x, 0) is equidistance from point (7, 6) and (3, 4)
    We know that
    Distance between two points is given by
    D = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|
    Now,
    D_1 = |\sqrt{(x-7)^2+(0-6)^2}|= |\sqrt{x^2+49-14x+36}|= |\sqrt{x^2-14x+85}|
    and
    D_2 = |\sqrt{(x-3)^2+(0-4)^2}|= |\sqrt{x^2+9-6x+16}|= |\sqrt{x^2-6x+25}|
    Now, according to the given condition
    D_1=D_2
    |\sqrt{x^2-14x+85}|= |\sqrt{x^2-6x+25}|
    Squaring both the sides
    x^2-14x+85= x^2-6x+25\\ 8x = 60\\ x=\frac{60}{8}= \frac{15}{2}
    Therefore, the point is ( \frac{15}{2},0)

    Question:5 Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P(0,-4) and B(8,0) .

    Answer:

    Mid-point of the line joining the points P(0,-4) and B(8,0) . is
    l = \left ( \frac{8}{2},\frac{-4}{2} \right ) = (4,-2)
    It is given that line also passes through origin which means passes through the point (0, 0)
    Now, we have two points on the line so we can now find the slope of a line by using formula
    m = \frac{y_2-y_1}{x_2-x_1}
    m = \frac{-2-0}{4-0} = \frac{-2}{4}= \frac{-1}{2}
    Therefore, the slope of the line is \frac{-1}{2}

    Question:6 Without using the Pythagoras theorem, show that the points (4,4),(3,5) and (-1,-1), are the vertices of a right angled triangle.

    Answer:

    It is given that point A(4,4) , B(3,5) and C(-1,-1) are the vertices of a right-angled triangle
    Now,
    We know that the distance between two points is given by
    D = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|
    Length of AB = |\sqrt{(4-3)^2+(4-5)^2}|= |\sqrt{1+1}|= \sqrt2
    Length of BC = |\sqrt{(3+1)^2+(5+1)^2}|= |\sqrt{16+36}|= \sqrt{52}
    Length of AC = |\sqrt{(4+1)^2+(4+1)^2}|= |\sqrt{25+25}|= \sqrt{50}
    Now, we know that Pythagoras theorem is
    H^2= B^2+L^2
    Is clear that
    (\sqrt{52})^2=(\sqrt{50})^2+(\sqrt 2)^2\\ 52 = 52\\ i.e\\ BC^2= AB^2+AC^2
    Hence proved

    Question:7 Find the slope of the line, which makes an angle of 30^{\circ} with the positive direction of y -axis measured anticlockwise.

    Answer:

    It is given that the line makes an angle of 30^{\circ} with the positive direction of y -axis measured anticlockwise
    Now, we know that
    m = \tan \theta
    line makes an angle of 30^{\circ} with the positive direction of y -axis
    Therefore, the angle made by line with the positive x-axis is = 90^{\degree}+30^{\degree}= 120\degree
    Now,
    m = \tan 120\degree = -\tan 60\degree = -\sqrt3
    Therefore, the slope of the line is -\sqrt3

    Question:8 Find the value of x for which the points (x,-1),(2,1) and (4,5) are collinear.

    Answer:

    Point is collinear which means they lie on the same line by this we can say that their slopes are equal
    Given points are A(x,-1) , B(2,1) and C(4,5)
    Slope = m = \frac{y_2-y_1}{x_2-x_1}
    Now,
    The slope of AB = Slope of BC
    \frac{1+1}{2-x}= \frac{5-1}{4-2}
    \frac{2}{2-x}= \frac{4}{2}\\ \\ \frac{2}{2-x} = 2\\ \\ 2=2(2-x)\\ 2=4-2x\\ -2x = -2\\ x = 1
    Therefore, the value of x is 1

    Question:9 Without using the distance formula, show that points (-2,-1),(4,0),(3,3) and (-3,2) are the vertices of a parallelogram.

    Answer:

    Given points are A(-2,-1),B(4,0),C(3,3) and D(-3,2)
    We know the pair of the opposite side are parallel to each other in a parallelogram
    Which means their slopes are also equal
    Slope = m = \frac{y_2-y_1}{x_2-x_1}
    The slope of AB =

    \frac{0+1}{4+2} = \frac{1}{6}

    The slope of BC =

    \frac{3-0}{3-4} = \frac{3}{-1} = -3

    The slope of CD =

    \frac{2-3}{-3-3} = \frac{-1}{-6} = \frac{1}{6}

    The slope of AD

    = \frac{2+1}{-3+2} = \frac{3}{-1} = -3
    We can clearly see that
    The slope of AB = Slope of CD (which means they are parallel)
    and
    The slope of BC = Slope of AD (which means they are parallel)
    Hence pair of opposite sides are parallel to each other
    Therefore, we can say that points (-2,-1),(4,0),(3,3) and (-3,2) are the vertices of a parallelogram

    Question:10 Find the angle between the x-axis and the line joining the points (3,-1) and (4,-2) .

    Answer:

    We know that
    m = \tan \theta
    So, we need to find the slope of line joining points (3,-1) and (4,-2)
    Now,
    m = \frac{y_2-y_1}{x_2-x_1}= \frac{-2+1}{4-3} = -1
    \tan \theta = -1
    \tan \theta = \tan \frac{3\pi}{4} = \tan 135\degree
    \theta = \frac{3\pi}{4} = 135\degree
    Therefore, angle made by line with positive x-axis when measure in anti-clockwise direction is 135\degree

    Question:11 The slope of a line is double of the slope of another line. If tangent of the angle between them is \frac{1}{3}, find the slopes of the lines

    Answer:

    Let m_1 \ and \ m_2 are the slopes of lines and \theta is the angle between them
    Then, we know that
    \tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |
    It is given that m_2 = 2m_1 and

    \tan \theta = \frac{1}{3}
    Now,
    \frac{1}{3}= \left | \frac{2m_1-m_1}{1+m_1.2m_1} \right |
    \frac{1}{3}= \left | \frac{m_1}{1+2m^2_1} \right |
    Now,
    3|m_1|= 1+2|m^2_1|\\ 2|m^2_1|-3|m_1|+ 1 = 0\\ 2|m^2_1|-2|m_1|-|m_1|+1=0\\ (2|m_1|-1)(|m_1|-1)= 0\\ |m_1|= \frac{1}{2} \ \ \ \ \ or \ \ \ \ \ \ |m_1| = 1
    Now,
    m_1 = \frac{1}{2} \ or \ \frac{-1}{2} \ or \ 1 \ or \ -1
    According to which value of m_2 = 1 \ or \ -1 \ or \ 2 \ or \ -2
    Therefore, m_1,m_2 = \frac{1}{2},1 \ or \ \frac{-1}{2},-1 \ or \ 1,2 \ or \ -1,-2

    Question:12 A line passes through (x_1,y_1) and (h,k) . If slope of the line is m , show that k-y_1=m(h-x_1) .

    Answer:

    Given that A line passes through (x_1,y_1) and (h,k) and slope of the line is m
    Now,
    m = \frac{y_2-y_1}{x_2-x_1}
    \Rightarrow m = \frac{k-y_1}{h-x_1}
    \Rightarrow (k-y_1)= m(h-x_1)
    Hence proved

    Question:13 If three points (h,0),(a,b) and (0,k) lie on a line, show that \frac{a}{h}+\frac{b}{k}=1.

    Answer:

    Points A(h,0),B(a,b) and C(0,k) lie on a line so by this we can say that their slopes are also equal
    We know that
    Slope = m = \frac{y_2-y_1}{x_2-x_1}

    Slope of AB = \frac{b-0}{a-h} = \frac{b}{a-h}

    Slope of AC = \frac{k-b}{0-a} = \frac{k-b}{-a}
    Now,
    Slope of AB = slope of AC
    \frac{b}{a-h} = \frac{k-b}{-a}
    -ab= (a-h)(k-b)
    -ab= ak -ab-hk+hb\\ ak +hb = hk
    Now divide both the sides by hk
    \frac{ak}{hk}+\frac{hb}{hk}= \frac{hk}{hk}\\ \\ \frac{a}{h}+\frac{b}{k} = 1
    Hence proved

    Question:14 Consider the following population and year graph, find the slope of the line AB and using it, find what will be the population in the year 2010?

    1646805059743

    Answer:

    Given point A(1985,92) and B(1995,97)
    Now, we know that
    Slope = m = \frac{y_2-y_1}{x_2-x_1}
    m = \frac{97-92}{1995-1985} = \frac{5}{10}= \frac{1}{2}
    Therefore, the slope of line AB is \frac{1}{2}
    Now, the equation of the line passing through the point (1985,92) and with slope = \frac{1}{2} is given by
    (y-92) = \frac{1}{2}(x-1985)\\ \\ 2y-184 = x-1985\\ x-2y = 1801
    Now, in the year 2010 the population is
    2010-2y = 1801\\ -2y = -209\\ y = 104.5
    Therefore, the population in the year 2010 is 104.5 crore

    Straight lines class 11 solutions - Exercise: 10.2

    Question:1 Find the equation of the line which satisfy the given conditions:

    Write the equations for the x -and y -axes.

    Answer:

    Equation of x-axis is y = 0
    and
    Equation of y-axis is x = 0

    Question:2 Find the equation of the line which satisfy the given conditions:

    Passing through the point (-4,3) with slope \frac{1}{2} .

    Answer:

    We know that , equation of line passing through point (x_1,y_1) and with slope m is given by
    (y-y_1)=m(x-x_1)
    Now, equation of line passing through point (-4,3) and with slope \frac{1}{2} is
    (y-3)=\frac{1}{2}(x-(-4))\\ 2y-6=x+4\\ x-2y+10 = 0
    Therefore, equation of the line is x-2y+10 = 0

    Question:3 Find the equation of the line which satisfy the given conditions:

    Passing through (0,0) with slope m .

    Answer:

    We know that the equation of the line passing through the point (x_1,y_1) and with slope m is given by
    (y-y_1)=m(x-x_1)
    Now, the equation of the line passing through the point (0,0) and with slope m is
    (y-0)=m(x-0)\\ y = mx
    Therefore, the equation of the line is y = mx

    Question:4 Find the equation of the line which satisfy the given conditions:

    Passing through (2,2\sqrt{3}) and inclined with the x-axis at an angle of 75^{\circ} .

    Answer:

    We know that the equation of the line passing through the point (x_1,y_1) and with slope m is given by
    (y-y_1)=m(x-x_1)
    we know that
    m = \tan \theta
    where \theta is angle made by line with positive x-axis measure in the anti-clockwise direction
    m = \tan75\degree \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \theta=75\degree \ given)
    m = \frac{\sqrt3+1}{\sqrt3-1}
    Now, the equation of the line passing through the point (2,2\sqrt3) and with slope m = \frac{\sqrt3+1}{\sqrt3-1} is
    (y-2\sqrt3)=\frac{\sqrt3+1}{\sqrt3-1}(x-2)\\ \\ (\sqrt3-1)(y-2\sqrt3)=(\sqrt3+1)(x-2)\\ (\sqrt3-1)y-6+2\sqrt3= (\sqrt3+1)x-2\sqrt3-2\\ (\sqrt3+1)x-(\sqrt3-1)y = 4(\sqrt3-1)
    Therefore, the equation of the line is (\sqrt3+1)x-(\sqrt3-1)y = 4(\sqrt3-1)

    Question:5 Find the equation of the line which satisfy the given conditions:

    Intersecting the x -axis at a distance of 3 units to the left of origin with slope -2 .

    Answer:

    We know that the equation of the line passing through the point (x_1,y_1) and with slope m is given by
    (y-y_1)=m(x-x_1)
    Line Intersecting the x -axis at a distance of 3 units to the left of origin which means the point is (-3,0)
    Now, the equation of the line passing through the point (-3,0) and with slope -2 is
    (y-0)= -2(x-(-3))\\ y = -2x-6\\ 2x+y+6=0
    Therefore, the equation of the line is 2x+y+6=0

    Question:6 Find the equation of the line which satisfy the given conditions:

    Intersecting the y -axis at a distance of 2 units above the origin and making an angle of 30^{\circ} with positive direction of the x-axis.

    Answer:

    We know that , equation of line passing through point (x_1,y_1) and with slope m is given by
    (y-y_1)=m(x-x_1)
    Line Intersecting the y-axis at a distance of 2 units above the origin which means point is (0,2)
    we know that
    m = \tan \theta\\ m = \tan 30\degree \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \theta = 30 \degree \ given)\\ m = \frac{1}{\sqrt3}
    Now, the equation of the line passing through the point (0,2) and with slope \frac{1}{\sqrt3} is
    (y-2)= \frac{1}{\sqrt3}(x-0)\\ \sqrt3(y-2)= x\\ x-\sqrt3y+2\sqrt3=0
    Therefore, the equation of the line is x-\sqrt3y+2\sqrt3=0

    Question:7 Find the equation of the line which satisfy the given conditions:

    Passing through the points (-1,1) and (2,-4) .

    Answer:

    We know that , equation of line passing through point (x_1,y_1) and with slope m is given by
    (y-y_1)=m(x-x_1)
    Now, it is given that line passes throught point (-1 ,1) and (2 , -4)
    m = \frac{y_2-y_1}{x_2-x_1}\\ \\ m = \frac{-4-1}{2+1}= \frac{-5}{3}
    Now, equation of line passing through point (-1,1) and with slope \frac{-5}{3} is
    (y-1)= \frac{-5}{3}(x-(-1))\\ \\3(y-1)=-5(x+1)\\ 3y-3=-5x-5\\ 5x+3y+2=0

    Question:8 Find the equation of the line which satisfy the given conditions:

    Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x -axis is 30^{\circ} .

    Answer:

    It is given that length of perpendicular is 5 units and angle made by the perpendicular with the positive x -axis is 30^{\circ}
    Therefore, equation of line is
    x\cos \theta + y \sin \theta = p
    In this case p = 5 and \theta = 30\degree
    x\cos 30\degree + y \sin 30\degree = 5\\ x.\frac{\sqrt3}{2}+\frac{y}{2}= 5\\ \sqrt3x+y =10
    Therefore, equation of the line is \sqrt3x+y =10

    Question:9 The vertices of \Delta \hspace{1mm}PQR are P(2,1),Q(-2,3) and R(4,5) . Find equation of the median through the vertex R .

    Answer:

    1646805124218 The vertices of \Delta \hspace{1mm}PQR are P(2,1),Q(-2,3) and R(4,5)
    Let m be RM b the median through vertex R
    Coordinates of M (x, y ) = \left ( \frac{2-2}{2},\frac{1+3}{2} \right )= (0,2)
    Now, slope of line RM
    m = \frac{y_2-y_1}{x_2-x_1} = \frac{5-2}{4-0}= \frac{3}{4}
    Now, equation of line passing through point (x_1,y_1) and with slope m is
    (y-y_1)= m(x-x_1)
    equation of line passing through point (0 , 2) and with slope \frac{3}{4} is
    (y-2)= \frac{3}{4}(x-0)\\ \\ 4(y-2)=3x\\ 4y-8=3x\\ 3x-4y+8=0
    Therefore, equation of median is 3x-4y+8=0

    Question:10 Find the equation of the line passing through (-3,5) and perpendicular to the line through the points (2,5) and (-3,6) .

    Answer:

    It is given that the line passing through (-3,5) and perpendicular to the line through the points (2,5) and (-3,6)
    Let the slope of the line passing through the point (-3,5) is m and
    Slope of line passing through points (2,5) and (-3,6)
    m' = \frac{6-5}{-3-2}= \frac{1}{-5}
    Now this line is perpendicular to line passing through point (-3,5)
    Therefore,
    m= -\frac{1}{m'} = -\frac{1}{\frac{1}{-5}}= 5

    Now, equation of line passing through point (x_1,y_1) and with slope m is
    (y-y_1)= m(x-x_1)
    equation of line passing through point (-3 , 5) and with slope 5 is
    (y-5)= 5(x-(-3))\\ \\ (y-5)=5(x+3)\\ y-5=5x+15\\ 5x-y+20=0
    Therefore, equation of line is 5x-y+20=0

    Question:11 A line perpendicular to the line segment joining the points (1,0) and (2,3) divides it in the ratio 1:n . Find the equation of the line.

    Answer:

    Co-ordinates of point which divide line segment joining the points (1,0) and (2,3) in the ratio 1:n is
    \left ( \frac{n(1)+1(2)}{1+n},\frac{n.(0)+1.(3)}{1+n} \right )= \left ( \frac{n+2}{1+n},\frac{3}{1+n} \right )
    Let the slope of the perpendicular line is m
    And Slope of line segment joining the points (1,0) and (2,3) is
    m'= \frac{3-0}{2-1}= 3
    Now, slope of perpendicular line is
    m = -\frac{1}{m'}= -\frac{1}{3}
    Now, equation of line passing through point (x_1,y_1) and with slope m is
    (y-y_1)= m(x-x_1)
    equation of line passing through point \left ( \frac{n+2}{1+n},\frac{3}{1+n} \right ) and with slope -\frac{1}{3} is
    (y- \frac{3}{1+n})= -\frac{1}{3}(x- (\frac{n+2}{1+n}))\\ 3y(1+n)-9=-x(1+n)+n+2\\ x(1+n)+3y(1+n)=n+11
    Therefore, equation of line is x(1+n)+3y(1+n)=n+11

    Question:12 Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2,3) .

    Answer:

    Let (a, b) are the intercept on x and y-axis respectively
    Then, the equation of the line is given by
    \frac{x}{a}+\frac{y}{b}= 1
    Intercepts are equal which means a = b
    \frac{x}{a}+\frac{y}{a}= 1\\ \\ x+y = a
    Now, it is given that line passes through the point (2,3)
    Therefore,
    a = 2+ 3 = 5
    therefore, equation of the line is x+ y = 5

    Question:13 Find equation of the line passing through the point (2,2) and cutting off intercepts on the axes whose sum is 9 .

    Answer:

    Let (a, b) are the intercept on x and y axis respectively
    Then, the equation of line is given by
    \frac{x}{a}+\frac{y}{b}= 1
    It is given that
    a + b = 9
    b = 9 - a
    Now,
    \frac{x}{a}+\frac{y}{9-a } = 1\\ \\ x(9-a)+ay= a(9-a)\\ 9x-ax+ay=9a-a^2
    It is given that line passes through point (2 ,2)
    So,
    9(2)-2a+2a=9a-a^2\\ a^2-9a+18=0\\ a^2-6a-3a+18=0\\ (a-6)(a-3)= 0\\ a=6 \ \ \ \ \ \ or \ \ \ \ \ \ a = 3

    case (i) a = 6 b = 3
    \frac{x}{6}+\frac{y}{3}= 1\\ \\ x+2y = 6

    case (ii) a = 3 , b = 6
    \frac{x}{3}+\frac{y}{6}= 1\\ \\ 2x+y = 6
    Therefore, equation of line is 2x + y = 6 , x + 2y = 6

    Question:14 Find equation of the line through the point (0,2) making an angle \frac{2\pi }{3} with the positive x -axis. Also, find the equation of line parallel to it and crossing the y -axis at a distance of 2 units below the origin .

    Answer:

    We know that
    m = \tan \theta \\ m = \tan \frac{2\pi}{3} = -\sqrt3
    Now, equation of line passing through point (0 , 2) and with slope -\sqrt3 is
    (y-2)= -\sqrt3(x-0)\\ \sqrt3x+y-2=0
    Therefore, equation of line is \sqrt3x+y-2=0 -(i)

    Now, It is given that line crossing the y -axis at a distance of 2 units below the origin which means coordinates are (0 ,-2)
    This line is parallel to above line which means slope of both the lines are equal
    Now, equation of line passing through point (0 , -2) and with slope -\sqrt3 is
    (y-(-2))= -\sqrt3(x-0)\\ \sqrt3x+y+2=0
    Therefore, equation of line is \sqrt3x+y+2=0

    Question:15 The perpendicular from the origin to a line meets it at the point (-2,9) , find the equation of the line.

    Answer:

    Let the slope of the line is m
    and slope of a perpendicular line is which passes through the origin (0, 0) and (-2, 9) is
    m' = \frac{9-0}{-2-0}= \frac{9}{-2}
    Now, the slope of the line is
    m = -\frac{1}{m'}= \frac{2}{9}
    Now, the equation of line passes through the point (-2, 9) and with slope \frac{2}{9} is
    (y-9)=\frac{2}{9}(x-(-2))\\ \\ 9(y-9)=2(x+2)\\ 2x-9y+85 = 0
    Therefore, the equation of the line is 2x-9y+85 = 0

    Question:16 The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C . In an experiment, if L=124.942 when C=20 and L=125.134 when C=110 , express L in terms of C

    Answer:

    It is given that
    If C=20 then L=124.942
    and If C=110 then L=125.134
    Now, if assume C along x-axis and L along y-axis
    Then, we will get coordinates of two points (20 , 124.942) and (110 , 125.134)
    Now, the relation between C and L is given by equation
    (L-124.942)= \frac{125.134-124.942}{110-20}(C-20)
    (L-124.942)= \frac{0.192}{90}(C-20)
    L= \frac{0.192}{90}(C-20)+124.942
    Which is the required relation

    Question:17 The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs\hspace{1mm}14/litre and 1220 litres of milk each week at Rs\hspace{1mm}16/litre . Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs\hspace{1mm}17/litre

    Answer:

    It is given that the owner of a milk store sell
    980 litres milk each week at Rs\hspace{1mm}14/litre
    and 1220 litres of milk each week at Rs\hspace{1mm}16/litre
    Now, if we assume the rate of milk as x-axis and Litres of milk as y-axis
    Then, we will get coordinates of two points i.e. (14, 980) and (16, 1220)
    Now, the relation between litres of milk and Rs/litres is given by equation
    (L-980)= \frac{1220-980}{16-14}(R-14)
    (L-980)= \frac{240}{2}(R-14)
    L-980= 120R-1680
    L= 120R-700
    Now, at Rs\hspace{1mm}17/litre he could sell
    L= 120\times 17-700= 2040-700= 1340
    He could sell 1340 litres of milk each week at Rs\hspace{1mm}17/litre

    Question:18 P(a,b) is the mid-point of a line segment between axes. Show that equation of the line is \frac{x}{a}+\frac{y}{b}=2 .

    Answer:

    1646805177906 Now, let coordinates of point A is (0 , y) and of point B is (x , 0)
    The,
    \frac{x+0}{2}= a \ and \ \frac{0+y}{2}= b
    x= 2a \ and \ y = 2b
    Therefore, the coordinates of point A is (0 , 2b) and of point B is (2a , 0)
    Now, slope of line passing through points (0,2b) and (2a,0) is
    m = \frac{0-2b}{2a-0} = \frac{-2b}{2a}= \frac{-b}{a}
    Now, equation of line passing through point (2a,0) and with slope \frac{-b}{a} is
    (y-0)= \frac{-b}{a}(x-2a)
    \frac{y}{b}= - \frac{x}{a}+2
    \frac{x}{a}+\frac{y}{b}= 2
    Hence proved

    Question:19 Point R(h,k) divides a line segment between the axes in the ratio 1:2 . Find equation of the line.

    Answer:

    1646805218702 Let the coordinates of Point A is (x,0) and of point B is (0,y)
    It is given that point R(h , k) divides the line segment between the axes in the ratio 1:2
    Therefore,
    R(h , k) =\left ( \frac{1\times 0+2\times x}{1+2},\frac{1\times y+2\times 0}{1+2} \right )=\left ( \frac{2x}{3},\frac{y}{3} \right )
    h = \frac{2x}{3} \ \ and \ \ k = \frac{y}{3}
    x = \frac{3h}{2} \ \ and \ \ y = 3k
    Therefore, coordinates of point A is \left ( \frac{3h}{2},0 \right ) and of point B is (0,3k)
    Now, slope of line passing through points \left ( \frac{3h}{2},0 \right ) and (0,3k) is
    m = \frac{3k-0}{0-\frac{3h}{2}}= \frac{2k}{-h}
    Now, equation of line passing through point (0,3k) and with slope -\frac{2k}{h} is
    (y-3k)=-\frac{2k}{h}(x-0)
    h(y-3k)=-2k(x)
    yh-3kh=-2kx
    2kx+yh=3kh
    Therefore, the equation of line is 2kx+yh=3kh

    Question:20 By using the concept of equation of a line, prove that the three points (3,0),(-2,-2) and (8,2) are collinear.

    Answer:

    Points are collinear means they lies on same line
    Now, given points are A(3,0),B(-2,-2) and C(8,2)
    Equation of line passing through point A and B is
    (y-0)=\frac{0+2}{3+2}(x-3)
    y=\frac{2}{5}(x-3)\Rightarrow 5y= 2(x-3)
    2x-5y=6
    Therefore, the equation of line passing through A and B is 2x-5y=6

    Now, Equation of line passing through point B and C is
    (y-2)=\frac{2+2}{8+2}(x-8)
    (y-2)=\frac{4}{10}(x-8)
    (y-2)=\frac{2}{5}(x-8) \Rightarrow 5(y-2)=2(x-8)
    5y-10=2x-16
    2x-5y=6
    Therefore, Equation of line passing through point B and C is 2x-5y=6
    When can clearly see that Equation of line passing through point A nd B and through B and C is the same
    By this we can say that points A(3,0),B(-2,-2) and C(8,2) are collinear points

    Straight lines class 11 NCERT solutions - Exercise: 10.3

    Question:1(i) Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts.

    x+7y=0

    Answer:

    Given equation is
    x+7y=0
    we can rewrite it as
    y= -\frac{1}{7}x -(i)
    Now, we know that the Slope-intercept form of the line is
    y = mx+C -(ii)
    Where m is the slope and C is some constant
    On comparing equation (i) with equation (ii)
    we will get
    m =- \frac{1}{7} and C = 0
    Therefore, slope and y-intercept are -\frac{1}{7} \ and \ 0 respectively

    Question:1(ii) Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts.

    6x+3y-5=0

    Answer:

    Given equation is
    6x+3y-5=0
    we can rewrite it as
    y= -\frac{6}{3}x+\frac{5}{3}\Rightarrow y = -2x+\frac{5}{3} -(i)
    Now, we know that the Slope-intercept form of line is
    y = mx+C -(ii)
    Where m is the slope and C is some constant
    On comparing equation (i) with equation (ii)
    we will get
    m =- 2 and C = \frac{5}{3}
    Therefore, slope and y-intercept are -2 \ and \ \frac{5}{3} respectively

    Question:1(iii) Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts.

    y=0.

    Answer:

    Given equation is
    y=0 -(i)
    Now, we know that the Slope-intercept form of the line is
    y = mx+C -(ii)
    Where m is the slope and C is some constant
    On comparing equation (i) with equation (ii)
    we will get
    m =0 and C = 0
    Therefore, slope and y-intercept are 0 \ and \ 0 respectively

    Question:2(i) Reduce the following equations into intercept form and find their intercepts on the axes.

    3x+2y-12=0

    Answer:

    Given equation is
    3x+2y-12=0
    we can rewrite it as
    \frac{3x}{12}+\frac{2y}{12} = 1
    \frac{x}{4}+\frac{y}{6} = 1 -(i)
    Now, we know that the intercept form of line is
    \frac{x}{a}+\frac{y}{b} = 1 -(ii)
    Where a and b are intercepts on x and y axis respectively
    On comparing equation (i) and (ii)
    we will get
    a = 4 and b = 6
    Therefore, intercepts on x and y axis are 4 and 6 respectively

    Question:2(ii) Reduce the following equations into intercept form and find their intercepts on the axes.

    4x-3y=6

    Answer:

    Given equation is
    4x-3y=6
    we can rewrite it as
    \frac{4x}{6}-\frac{3y}{6} = 1
    \frac{x}{\frac{3}{2}}-\frac{y}{2} = 1 -(i)
    Now, we know that the intercept form of line is
    \frac{x}{a}+\frac{y}{b} = 1 -(ii)
    Where a and b are intercepts on x and y axis respectively
    On comparing equation (i) and (ii)
    we will get
    a = \frac{3}{2} and b = -2
    Therefore, intercepts on x and y axis are \frac{3}{2} and -2 respectively

    Question:2(iii) Reduce the following equations into intercept form and find their intercepts on the axes.

    3y+2=0

    Answer:

    Given equation is
    3y+2=0
    we can rewrite it as
    y = \frac{-2}{3}
    Therefore, intercepts on y-axis are \frac{-2}{3}
    and there is no intercept on x-axis

    Question:3(i) Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

    x-\sqrt{3}y+8=0

    Answer:

    Given equation is
    x-\sqrt{3}y+8=0
    we can rewrite it as
    -x+\sqrt3y=8
    Coefficient of x is -1 and y is \sqrt3
    Therefore, \sqrt{(-1)^2+(\sqrt3)^2}= \sqrt{1+3}=\sqrt4=2
    Now, Divide both the sides by 2
    we will get
    -\frac{x}{2}+\frac{\sqrt3y}{2}= 4
    we can rewrite it as
    x\cos 120\degree + y\sin 120\degree= 4 \ \ \ \ \ \ \ \ \ \ \ -(i)
    Now, we know that the normal form of the line is
    x\cos \theta + y\sin \theta= p \ \ \ \ \ \ \ \ \ \ \ -(ii)
    Where \theta is the angle between perpendicular and the positive x-axis and p is the perpendicular distance from the origin
    On comparing equation (i) and (ii)
    we wiil get
    \theta = 120\degree \ \ and \ \ p = 4
    Therefore, the angle between perpendicular and the positive x-axis and perpendicular distance from the origin is 120\degree \ and \ 4 respectively

    Question:3(ii) Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

    y-2=0

    Answer:

    Given equation is
    y-2=0
    we can rewrite it as
    0.x+y = 2
    Coefficient of x is 0 and y is 1
    Therefore, \sqrt{(0)^2+(1)^2}= \sqrt{0+1}=\sqrt1=1
    Now, Divide both the sides by 1
    we will get
    y=2
    we can rewrite it as
    x\cos 90\degree + y\sin 90\degree= 2 \ \ \ \ \ \ \ \ \ \ \ -(i)
    Now, we know that normal form of line is
    x\cos \theta + y\sin \theta= p \ \ \ \ \ \ \ \ \ \ \ -(ii)
    Where \theta is the angle between perpendicular and the positive x-axis and p is the perpendicular distance from the origin
    On comparing equation (i) and (ii)
    we wiil get
    \theta = 90\degree \ \ and \ \ p = 2
    Therefore, the angle between perpendicular and the positive x-axis and perpendicular distance from the origin is 90\degree \ and \ 2 respectively

    Question:3(iii) Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

    x-y=4

    Answer:

    Given equation is
    x-y=4

    Coefficient of x is 1 and y is -1
    Therefore, \sqrt{(1)^2+(-1)^2}= \sqrt{1+1}=\sqrt2
    Now, Divide both the sides by \sqrt2
    we wiil get
    \frac{x}{\sqrt2}-\frac{y}{\sqrt2}= \frac{4}{\sqrt2}
    we can rewrite it as
    x\cos 315\degree + y\sin 315\degree= 2\sqrt2 \ \ \ \ \ \ \ \ \ \ \ -(i)
    Now, we know that normal form of line is
    x\cos \theta + y\sin \theta= p \ \ \ \ \ \ \ \ \ \ \ -(ii)
    Where \theta is the angle between perpendicular and the positive x-axis and p is the perpendicular distance from the origin
    On compairing equation (i) and (ii)
    we wiil get
    \theta = 315\degree \ \ and \ \ p = 2\sqrt2
    Therefore, the angle between perpendicular and the positive x-axis and perpendicular distance from the origin is 315\degree \ and \ 2\sqrt2 respectively

    Question:4 Find the distance of the point (-1,1) from the line 12(x+6)=5(y-2) .

    Answer:

    Given the equation of the line is
    12(x+6)=5(y-2)
    we can rewrite it as
    12x+72=5y-10
    12x-5y+82=0
    Now, we know that
    d= \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}} where A and B are the coefficients of x and y and C is some constant and (x_1,y_1) is point from which we need to find the distance
    In this problem A = 12 , B = -5 , c = 82 and (x_1,y_1) = (-1 , 1)
    Therefore,
    d = \frac{|12.(-1)+(-5).1+82|}{\sqrt{12^2+(-5)^2}} = \frac{|-12-5+82|}{\sqrt{144+25}}=\frac{|65|}{\sqrt{169}}=\frac{65}{13}= 5
    Therefore, the distance of the point (-1,1) from the line 12(x+6)=5(y-2) is 5 units

    Question:5 Find the points on the x-axis, whose distances from the line \frac{x}{3}+\frac{y}{4}=1 are 4 units.

    Answer:

    Given equation of line is
    \frac{x}{3}+\frac{y}{4}=1
    we can rewrite it as
    4x+3y-12=0
    Now, we know that
    d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}
    In this problem A = 4 , B = 3 C = -12 and d = 4
    point is on x-axis therefore (x_1,y_1) = (x ,0)
    Now,
    4= \frac{|4.x+3.0-12|}{\sqrt{4^2+3^2}}= \frac{|4x-12|}{\sqrt{16+9}}= \frac{|4x-12|}{\sqrt{25}}= \frac{|4x-12|}{5}
    20=|4x-12|\\ 4|x-3|=20\\ |x-3|=5
    Now if x > 3
    Then,
    |x-3|=x-3\\ x-3=5\\ x = 8
    Therefore, point is (8,0)
    and if x < 3
    Then,
    |x-3|=-(x-3)\\ -x+3=5\\ x = -2
    Therefore, point is (-2,0)
    Therefore, the points on the x-axis, whose distances from the line \frac{x}{3}+\frac{y}{4}=1 are 4 units are (8 , 0) and (-2 , 0)

    Question:6(i) Find the distance between parallel lines 15x+8y-34=0 and 15x+8y+31=0 .

    Answer:

    Given equations of lines are
    15x+8y-34=0 and 15x+8y+31=0
    it is given that these lines are parallel
    Therefore,
    d = \frac{ |C_2-C_1|}{\sqrt{A^2+B^2}}
    A = 15 , B = 8 , C_1= -34 \ and \ C_2 = 31
    Now,
    d = \frac{|31-(-34)|}{\sqrt{15^2+8^2}}= \frac{|31+34|}{\sqrt{225+64}}= \frac{|65|}{\sqrt{289}} = \frac{65}{17}
    Therefore, the distance between two lines is \frac{65}{17} \ units

    Question:6(ii) Find the distance between parallel lines l(x+y)+p=0 and l(x+y)-r = 0

    Answer:

    Given equations of lines are
    l(x+y)+p=0 and l(x+y)-r = 0
    it is given that these lines are parallel
    Therefore,
    d = \frac{ |C_2-C_1|}{\sqrt{A^2+B^2}}
    A = l , B = l , C_1= -r \ and \ C_2 = p
    Now,
    d = \frac{|p-(-r)|}{\sqrt{l^2+l^2}}= \frac{|p+r|}{\sqrt{2l^2}}= \frac{|p+r|}{\sqrt{2}|l|}
    Therefore, the distance between two lines is \frac{1}{\sqrt2}\left | \frac{p+r}{l} \right |

    Question:7 Find equation of the line parallel to the line 3x-4y+2=0 and passing through the point (-2,3) .

    Answer:

    It is given that line is parallel to line 3x-4y+2=0 which implies that the slopes of both the lines are equal
    we can rewrite it as
    y = \frac{3x}{4}+\frac{1}{2}
    The slope of line 3x-4y+2=0 = \frac{3}{4}
    Now, the equation of the line passing through the point (-2,3) and with slope \frac{3}{4} is
    (y-3)=\frac{3}{4}(x-(-2))
    4(y-3)=3(x+2)
    4y-12=3x+6
    3x-4y+18= 0
    Therefore, the equation of the line is 3x-4y+18= 0

    Question:8 Find equation of the line perpendicular to the line x-7y+5=0 and having x intercept 3 .

    Answer:

    It is given that line is perpendicular to the line x-7y+5=0
    we can rewrite it as
    y = \frac{x}{7}+\frac{5}{7}
    Slope of line x-7y+5=0 ( m' ) = \frac{1}{7}
    Now,
    The slope of the line is m = \frac{-1}{m'} = -7 \ \ \ \ \ \ \ \ \ \ \ (\because lines \ are \ perpendicular)
    Now, the equation of the line with x intercept 3 i.e. (3, 0) and with slope -7 is
    (y-0)=-7(x-3)
    y = -7x+21
    7x+y-21=0

    Question:9 Find angles between the lines \sqrt{3}x+y=1 and x+\sqrt{3}y=1 .

    Answer:

    Given equation of lines are
    \sqrt{3}x+y=1 and x+\sqrt{3}y=1

    Slope of line \sqrt{3}x+y=1 is, m_1 = -\sqrt3

    And
    Slope of line x+\sqrt{3}y=1 is , m_2 = -\frac{1}{\sqrt3}

    Now, if \theta is the angle between the lines
    Then,

    \tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |

    \tan \theta = \left | \frac{-\frac{1}{\sqrt3}-(-\sqrt3)}{1+(-\sqrt3).\left ( -\frac{1}{\sqrt3} \right )} \right | = \left | \frac{\frac{-1+3}{\sqrt3}}{1+1} \right |=| \frac{1}{\sqrt3}|

    \tan \theta = \frac{1}{\sqrt3} \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \tan \theta = -\frac{1}{\sqrt3}

    \theta = \frac{\pi}{6}=30\degree \ \ \ \ \ \ \ or \ \ \ \ \ \ \theta =\frac{5\pi}{6}=150\degree

    Therefore, the angle between the lines is 30\degree \ and \ 150\degree

    Question:10 The line through the points (h,3) and (4,1) intersects the line 7x-9y-19=0 at right angle. Find the value of h .

    Answer:

    Line passing through points ( h ,3) and (4 ,1)

    Therefore,Slope of the line is

    m =\frac{y_2-y_1}{x_2-x_1}

    m =\frac{3-1}{h-4}

    This line intersects the line 7x-9y-19=0 at right angle
    Therefore, the Slope of both the lines are negative times inverse of each other
    Slope of line 7x-9y-19=0 , m'=\frac{7}{9}
    Now,
    m=-\frac{1}{m'}
    \frac{2}{h-4}= -\frac{9}{7}
    14=-9(h-4)
    14=-9h+36
    -9h= -22
    h=\frac{22}{9}
    Therefore, the value of h is \frac{22}{9}

    Question:11 Prove that the line through the point (x_1,y_1) and parallel to the line Ax+By+C=0 is A(x-x_1)+B(y-y_1)=0.

    Answer:

    It is given that line is parallel to the line Ax+By+C=0
    Therefore, their slopes are equal
    The slope of line Ax+By+C=0 , m'= \frac{-A}{B}
    Let the slope of other line be m
    Then,
    m =m'= \frac{-A}{B}
    Now, the equation of the line passing through the point (x_1,y_1) and with slope -\frac{A}{B} is
    (y-y_1)= -\frac{A}{B}(x-x_1)
    B(y-y_1)= -A(x-x_1)
    A(x-x_1)+B(y-y_1)= 0
    Hence proved

    Question:12 Two lines passing through the point (2,3) intersects each other at an angle of 60^{\circ} . If slope of one line is 2 , find equation of the other line .

    Answer:

    Let the slope of two lines are m_1 \ and \ m_2 respectively
    It is given the lines intersects each other at an angle of 60^{\circ} and slope of the line is 2
    Now,
    m_1 = m\ and \ m_2= 2 \ and \ \theta = 60\degree
    \tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |
    \tan 60\degree = \left | \frac{2-m}{1+2m} \right |
    \sqrt3 = \left | \frac{2-m}{1+2m} \right |
    \sqrt3 = \frac{2-m}{1+2m} \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \sqrt 3 = -\left ( \frac{2-m}{1+2m} \right )
    m = \frac{2-\sqrt3}{2\sqrt3+1} \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ m = \frac{-(2+\sqrt3)}{2\sqrt3-1}
    Now, the equation of line passing through point (2 ,3) and with slope \frac{2-\sqrt3}{2\sqrt3+1} is
    (y-3)= \frac{2-\sqrt3}{2\sqrt3+1}(x-2)
    (2\sqrt3+1)(y-3)=(2-\sqrt3)(x-2)
    x(\sqrt3-2)+y(2\sqrt3+1)=-1+8\sqrt3 -(i)

    Similarly,
    Now , equation of line passing through point (2 ,3) and with slope \frac{-(2+\sqrt3)}{2\sqrt3-1} is
    (y-3)=\frac{-(2+\sqrt3)}{2\sqrt3-1}(x-2)
    (2\sqrt3-1)(y-3)= -(2+\sqrt3)(x-2)
    x(2+\sqrt3)+y(2\sqrt3-1)=1+8\sqrt3 -(ii)

    Therefore, equation of line is x(\sqrt3-2)+y(2\sqrt3+1)=-1+8\sqrt3 or x(2+\sqrt3)+y(2\sqrt3-1)=1+8\sqrt3

    Question:13 Find the equation of the right bisector of the line segment joining the points (3,4) and (-1,2) .

    Answer:

    Right bisector means perpendicular line which divides the line segment into two equal parts
    Now, lines are perpendicular which means their slopes are negative times inverse of each other
    Slope of line passing through points (3,4) and (-1,2) is
    m'= \frac{4-2}{3+1}= \frac{2}{4}=\frac{1}{2}
    Therefore, Slope of bisector line is
    m = - \frac{1}{m'}= -2
    Now, let (h , k) be the point of intersection of two lines
    It is given that point (h,k) divides the line segment joining point (3,4) and (-1,2) into two equal part which means it is the mid point
    Therefore,
    h = \frac{3-1}{2} = 1\ \ \ and \ \ \ k = \frac{4+2}{2} = 3
    (h,k) = (1,3)
    Now, equation of line passing through point (1,3) and with slope -2 is
    (y-3)=-2(x-1)\\ y-3=-2x+2\\ 2x+y=5
    Therefore, equation of line is 2x+y=5

    Question:14 Find the coordinates of the foot of perpendicular from the point (-1,3) to the line 3x-4y-16=0 .

    Answer:

    Let suppose the foot of perpendicular is (x_1,y_1)
    We can say that line passing through the point (x_1,y_1) \ and \ (-1,3) is perpendicular to the line 3x-4y-16=0
    Now,
    The slope of the line 3x-4y-16=0 is , m' = \frac{3}{4}
    And
    The slope of the line passing through the point (x_1,y_1) \ and \ (-1,3) is, m = \frac{y-3}{x+1}
    lines are perpendicular
    Therefore,
    m = -\frac{1}{m'}\\ \frac{y_1-3}{_1+1} = -\frac{4}{3}\\ 3(y_1-3)=-4(x_1+1)\\ 4x_1+3y_1=5 \ \ \ \ \ \ \ \ \ -(i)
    Now, the point (x_1,y_1) also lies on the line 3x-4y-16=0
    Therefore,
    3x_1-4y_1=16 \ \ \ \ \ \ \ \ \ \ \ -(ii)
    On solving equation (i) and (ii)
    we will get
    x_1 = \frac{68}{25} \ and \ y_1 =-\frac{49}{25}
    Therefore, (x_1,y_1) = \left ( \frac{68}{25},-\frac{49}{25} \right )

    Question:15 The perpendicular from the origin to the line y=mx+c meets it at the point (-1,2) . Find the values of m and c .

    Answer:

    We can say that line passing through point (0,0) \ and \ (-1,2) is perpendicular to line y=mx+c
    Now,
    The slope of the line passing through the point (0,0) \ and \ (-1,2) is , m = \frac{2-0}{-1-0}= -2
    lines are perpendicular
    Therefore,
    m = -\frac{1}{m'} = \frac{1}{2} - (i)
    Now, the point (-1,2) also lies on the line y=mx+c
    Therefore,
    2=\frac{1}{2}.(-1)+C\\ C = \frac{5}{2} \ \ \ \ \ \ \ \ \ \ \ -(ii)
    Therefore, the value of m and C is \frac{1}{2} \ and \ \frac{5}{2} respectively

    Question:16 If p and q are the lengths of perpendiculars from the origin to the lines x\cos \theta -y\sin \theta =k\cos 2\theta and x\sec \theta +y\hspace{1mm}cosec\hspace{1mm}\theta =k , respectively, prove that p^2+4q^2=k^2 .

    Answer:

    Given equations of lines are x\cos \theta -y\sin \theta =k\cos 2\theta and x\sec \theta +y\hspace{1mm}cosec\hspace{1mm}\theta =k

    We can rewrite the equation x\sec \theta +y\hspace{1mm}cosec\hspace{1mm}\theta =k as

    x\sin \theta +y\cos \theta = k\sin\theta\cos\theta
    Now, we know that

    d = \left | \frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}} \right |

    In equation x\cos \theta -y\sin \theta =k\cos 2\theta

    A= \cos \theta , B = -\sin \theta , C = - k\cos2\theta \ and \ (x_1,y_1)= (0,0)

    p= \left | \frac{\cos\theta .0-\sin\theta.0-k\cos2\theta }{\sqrt{\cos^2\theta+(-\sin\theta)^2}} \right | = |-k\cos2\theta|
    Similarly,
    in the equation x\sin \theta +y\cos \theta = k\sin\theta\cos\theta

    A= \sin \theta , B = \cos \theta , C = -k\sin\theta\cos\theta \ and \ (x_1,y_1)= (0,0)

    q= \left | \frac{\sin\theta .0+\cos\theta.0-k\sin\theta\cos\theta }{\sqrt{\sin^2\theta+\cos^2\theta}} \right | = |-k\sin\theta\cos\theta|= \left | -\frac{k\sin2\theta}{2} \right |
    Now,

    p^2+4q^2=(|-k\cos2\theta|)^2+4.(|-\frac{k\sin2\theta}{2})^2= k^2\cos^22\theta+4.\frac{k^2\sin^22\theta}{4}
    =k^2(\cos^22\theta+\sin^22\theta)
    =k^2
    Hence proved

    Question:17 In the triangle ABC with vertices A(2,3) , B(4,-1) and C(1,2) , find the equation and length of altitude from the vertex A .

    Answer:

    1646805291744 Let suppose foot of perpendicular is (x_1,y_1)
    We can say that line passing through point (x_1,y_1) \ and \ A(2,3) is perpendicular to line passing through point B(4,-1) \ and \ C(1,2)
    Now,
    Slope of line passing through point B(4,-1) \ and \ C(1,2) is , m' = \frac{2+1}{1-4}= \frac{3}{-3}=-1
    And
    Slope of line passing through point (x_1,y_1) \ and \ (2,3) is , m
    lines are perpendicular
    Therefore,
    m = -\frac{1}{m'}= 1
    Now, equation of line passing through point (2 ,3) and slope with 1
    (y-3)=1(x-2)
    x-y+1=0 -(i)
    Now, equation line passing through point B(4,-1) \ and \ C(1,2) is
    (y-2)=-1(x-1)
    x+y-3=0
    Now, perpendicular distance of (2,3) from the x+y-3=0 is
    d= \left | \frac{1\times2+1\times3-3}{\sqrt{1^2+1^2}} \right |= \left | \frac{2+3-3}{\sqrt{1+1}} \right |= \frac{2}{\sqrt{2}}=\sqrt2 -(ii)

    Therefore, equation and length of the line is x-y+1=0 and \sqrt2 respectively

    Question:18 If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b , then show that \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2} .

    Answer:

    we know that intercept form of line is
    \frac{x}{a}+\frac{y}{b} = 1
    we know that
    d = \left | \frac{Ax_1+bx_2+C}{\sqrt{A^2+B^2}} \right |
    In this problem
    A = \frac{1}{a},B = \frac{1}{b}, C =-1 \ and \ (x_1,y_1)= (0,0)
    p= \left | \frac{\frac{1}{a}\times 0+\frac{1}{b}\times 0-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} \right | = \left | \frac{-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} \right |
    On squaring both the sides
    we will get
    \frac{1}{p^2}= \frac{1}{a^2}+\frac{1}{b^2}
    Hence proved

    Straight lines NCERT solutions - Miscellaneous Exercise

    Question:1(a) Find the values of k for which the line (k-3)x-(4-k^2)y+k^2-7k+6=0 is

    Parallel to the x-axis.

    Answer:

    Given equation of line is
    (k-3)x-(4-k^2)y+k^2-7k+6=0
    and equation of x-axis is y=0
    it is given that these two lines are parallel to each other
    Therefore, their slopes are equal
    Slope of y=0 is , m' = 0
    and
    Slope of line (k-3)x-(4-k^2)y+k^2-7k+6=0 is , m = \frac{k-3}{4-k^2}
    Now,
    m=m'
    \frac{k-3}{4-k^2}=0
    k-3=0
    k=3
    Therefore, value of k is 3

    Question:1(b) Find the values of k for which the line (k-3)x-(4-k^2)y+k^2-7k+6=0 is

    Parallel to the y-axis.

    Answer:

    Given equation of line is
    (k-3)x-(4-k^2)y+k^2-7k+6=0
    and equation of y-axis is x = 0
    it is given that these two lines are parallel to each other
    Therefore, their slopes are equal
    Slope of y=0 is , m' = \infty = \frac{1}{0}
    and
    Slope of line (k-3)x-(4-k^2)y+k^2-7k+6=0 is , m = \frac{k-3}{4-k^2}
    Now,
    m=m'
    \frac{k-3}{4-k^2}=\frac{1}{0}
    4-k^2=0
    k=\pm2
    Therefore, value of k is \pm2

    Question:1(c) Find the values of k for which the line (k-3)x-(4-k^2)y+k^2-7k+6=0 is Passing through the origin.

    Answer:

    Given equation of line is
    (k-3)x-(4-k^2)y+k^2-7k+6=0
    It is given that it passes through origin (0,0)
    Therefore,
    (k-3).0-(4-k^2).0+k^2-7k+6=0
    k^2-7k+6=0
    k^2-6k-k+6=0
    (k-6)(k-1)=0
    k = 6 \ or \ 1
    Therefore, value of k is 6 \ or \ 1

    Question:2 Find the values of \small \theta and \small p , if the equation \small x\cos \theta +y\sin \theta =p is the normal form of the line \small \sqrt{3}x+y+2=0 .

    Answer:

    The normal form of the line is \small x\cos \theta +y\sin \theta =p
    Given the equation of lines is
    \small \sqrt{3}x+y+2=0
    First, we need to convert it into normal form. So, divide both the sides by \small \sqrt{(\sqrt3)^2+1^2}= \sqrt{3+1}= \sqrt4=2
    \small -\frac{\sqrt3\cos \theta}{2}-\frac{y}{2}= 1
    On comparing both
    we will get
    \small \cos \theta = -\frac{\sqrt3}{2}, \sin \theta = -\frac{1}{2} \ and \ p = 1
    \small \theta = \frac{7\pi}{6} \ and \ p =1
    Therefore, the answer is \small \theta = \frac{7\pi}{6} \ and \ p =1

    Question:3 Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are \small 1 and \small -6 , respectively.

    Answer:

    Let the intercepts on x and y-axis are a and b respectively
    It is given that
    a+b = 1 \ \ and \ \ a.b = -6
    a= 1-b
    \Rightarrow b.(1-b)=-6
    \Rightarrow b-b^2=-6
    \Rightarrow b^2-b-6=0
    \Rightarrow b^2-3b+2b-6=0
    \Rightarrow (b+2)(b-3)=0
    \Rightarrow b = -2 \ and \ 3
    Now, when b=-2\Rightarrow a=3
    and when b=3\Rightarrow a=-2
    We know that the intercept form of the line is
    \frac{x}{a}+\frac{y}{b}=1

    Case (i) when a = 3 and b = -2
    \frac{x}{3}+\frac{y}{-2}=1
    \Rightarrow 2x-3y=6

    Case (ii) when a = -2 and b = 3
    \frac{x}{-2}+\frac{y}{3}=1
    \Rightarrow -3x+2y=6
    Therefore, equations of lines are 2x-3y=6 \ and \ -3x+2y=6

    Question:4 What are the points on the \small y -axis whose distance from the line \small \frac{x}{3}+\frac{y}{4}=1 is \small 4 units.

    Answer:

    Given the equation of the line is
    \small \frac{x}{3}+\frac{y}{4}=1
    we can rewrite it as
    4x+3y=12
    Let's take point on y-axis is (0,y)
    It is given that the distance of the point (0,y) from line 4x+3y=12 is 4 units
    Now,
    d= \left | \frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}} \right |
    In this problem A = 4 , B=3 , C =-12 ,d=4\ \ and \ \ (x_1,y_1) = (0,y)
    4 = \left | \frac{4\times 0+3\times y-12}{\sqrt{4^2+3^2}} \right |=\left | \frac{3y-12}{\sqrt{16+9}} \right |=\left | \frac{3y-12}{5} \right |

    Case (i)

    4 = \frac{3y-12}{5}
    20=3y-12
    y = \frac{32}{3}
    Therefore, the point is \left ( 0,\frac{32}{3} \right ) -(i)

    Case (ii)

    4=-\left ( \frac{3y-12}{5} \right )
    20=-3y+12
    y = -\frac{8}{3}
    Therefore, the point is \left ( 0,-\frac{8}{3} \right ) -(ii)

    Therefore, points on the \small y -axis whose distance from the line \small \frac{x}{3}+\frac{y}{4}=1 is \small 4 units are \left ( 0,\frac{32}{3} \right ) and \left ( 0,-\frac{8}{3} \right )

    Question:5 Find perpendicular distance from the origin to the line joining the points (\cos \theta ,\sin \theta ) and (\cos \phi ,\sin \phi ). .

    Answer:

    Equation of line passing through the points (\cos \theta ,\sin \theta ) and (\cos \phi ,\sin \phi ) is
    (y-\sin \theta )= \frac{\sin \phi -\sin \theta}{\cos \phi -\cos \theta}(x-\cos\theta)
    \Rightarrow (\cos \phi -\cos \theta)(y-\sin \theta )= (\sin \phi -\sin \theta)(x-\cos\theta)
    \Rightarrow y(\cos \phi -\cos \theta)-\sin \theta(\cos \phi -\cos \theta)=x (\sin \phi -\sin \theta)-\cos\theta(\sin \phi -\sin \theta)
    \Rightarrow x (\sin \phi -\sin \theta)-y(\cos \phi -\cos \theta)=\cos\theta(\sin \phi -\sin \theta)-\sin \theta(\cos \phi -\cos \theta) \Rightarrow x (\sin \phi -\sin \theta)-y(\cos \phi -\cos \theta)=\sin(\theta-\phi)
    (\because \cos a\sin b -\sin a\cos b = \sin(a-b) )
    Now, distance from origin(0,0) is
    d = \left | \frac{(\sin\phi -\sin\theta).0-(\cos\phi-\cos\theta).0-\sin(\theta-\phi)}{\sqrt{(\sin\phi-\sin\theta)^2+(\cos\phi-\cos\theta)^2}} \right |
    d = \left | \frac{-\sin(\theta-\phi)}{\sqrt{(\sin^2\phi+\cos^2\phi)+(\sin^2\theta+\cos^2\theta)-2(\cos\theta\cos\phi+\sin\theta\sin\phi)}} \right |
    d = \left | \frac{-\sin(\theta-\phi)}{1+1-2\cos(\theta-\phi)} \right | (\because \cos a\cos b +\sin a\sin b = \cos(a-b) \ \ and \ \ \sin^2a+\cos^2a=1)
    d = \left |\frac{ - \sin(\theta-\phi)}{2(1-\cos(\theta-\phi))} \right |
    d = \left | \frac{-2\sin\frac{\theta-\phi}{2}\cos \frac{\theta-\phi}{2}}{\sqrt{2.(2\\sin^2\frac{\theta-\phi}{2})}}\right |
    d = \left | \frac{-2\sin\frac{\theta-\phi}{2}\cos \frac{\theta-\phi}{2}}{2\sin\frac{\theta-\phi}{2}}\right |
    d = \left | \cos\frac{\theta-\phi}{2} \right |

    Question:6 Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines \small x-7y+5=0 and \small 3x+y=0 .

    Answer:

    Point of intersection of the lines \small x-7y+5=0 and \small 3x+y=0
    \left ( -\frac{5}{22},\frac{15}{22} \right )
    It is given that this line is parallel to y - axis i.e. x=0 which means their slopes are equal
    Slope of x=0 is , m' = \infty = \frac{1}{0}
    Let the Slope of line passing through point \left ( -\frac{5}{22},\frac{15}{22} \right ) is m
    Then,
    m=m'= \frac{1}{0}
    Now, equation of line passing through point \left ( -\frac{5}{22},\frac{15}{22} \right ) and with slope \frac{1}{0} is
    (y-\frac{15}{22})= \frac{1}{0}(x+\frac{5}{22})
    x = -\frac{5}{22}
    Therefore, equation of line is x = -\frac{5}{22}

    Question:7 Find the equation of a line drawn perpendicular to the line \small \frac{x}{4}+\frac{y}{6}=1 through the point, where it meets the \small y -axis.

    Answer:

    given equation of line is
    \small \frac{x}{4}+\frac{y}{6}=1
    we can rewrite it as
    3x+2y=12
    Slope of line 3x+2y=12 , m' = -\frac{3}{2}
    Let the Slope of perpendicular line is m
    m = -\frac{1}{m'}= \frac{2}{3}
    Now, the ponit of intersection of 3x+2y=12 and x =0 is (0,6)
    Equation of line passing through point (0,6) and with slope \frac{2}{3} is
    (y-6)= \frac{2}{3}(x-0)
    3(y-6)= 2x
    2x-3y+18=0
    Therefore, equation of line is 2x-3y+18=0

    Question:8 Find the area of the triangle formed by the lines \small y-x=0,x+y=0 and \small x-k=0 .

    Answer:

    Given equations of lines are
    y-x=0 \ \ \ \ \ \ \ \ \ \ \ -(i)
    x+y=0 \ \ \ \ \ \ \ \ \ \ \ -(ii)
    x-k=0 \ \ \ \ \ \ \ \ \ \ \ -(iii)
    The point if intersection of (i) and (ii) is (0,0)
    The point if intersection of (ii) and (iii) is (k,-k)
    The point if intersection of (i) and (iii) is (k,k)
    Therefore, the vertices of triangle formed by three lines are (0,0), (k,-k) \ and \ (k,k)
    Now, we know that area of triangle whose vertices are (x_1,y_1),(x_2,y_2) \ and \ (x_3,y_3) is
    A = \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|
    A= \frac{1}{2}|0(-k-k)+k(k-0)+k(0+k)|
    A= \frac{1}{2}|k^2+k^2|
    A= \frac{1}{2}|2k^2|
    A= k^2
    Therefore, area of triangle is k^2 \ square \ units

    Question:9 Find the value of \small p so that the three lines \small 3x+y-2=0,px+2y-3=0 and \small 2x-y-3=0 may intersect at one point.

    Answer:

    Point of intersection of lines \small 3x+y-2=0 and \small 2x-y-3=0 is (1,-1)
    Now, (1,-1) must satisfy equation px+2y-3=0
    Therefore,
    p(1)+2(-1)-3=0
    p-2-3=0
    p=5
    Therefore, the value of p is 5

    Question:10 If three lines whose equations are y=m_1x+c_1,y=m_2x+c_2 and y=m_3x+c_3 are concurrent, then show that m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0 .

    Answer:

    Concurrent lines means they all intersect at the same point
    Now, given equation of lines are
    y=m_1x+c_1 \ \ \ \ \ \ \ \ \ \ \ -(i)
    y=m_2x+c_2 \ \ \ \ \ \ \ \ \ \ \ -(ii)
    y=m_3x+c_3 \ \ \ \ \ \ \ \ \ \ \ -(iii)
    Point of intersection of equation (i) and (ii) \left ( \frac{c_2-c_1}{m_1-m_2},\frac{m_1c_2-m_2c_1}{m_1-m_2} \right )

    Now, lines are concurrent which means point \left ( \frac{c_2-c_1}{m_1-m_2},\frac{m_1c_2-m_2c_1}{m_1-m_2} \right ) also satisfy equation (iii)
    Therefore,

    \frac{m_1c_2-m_2c_1}{m_1-m_2}=m_3.\left ( \frac{c_2-c_1}{m_1-m_2} \right )+c_3

    m_1c_2-m_2c_1= m_3(c_2-c_1)+c_3(m_1-m_2)

    m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0

    Hence proved

    Question:11 Find the equation of the lines through the point \small (3,2) which make an angle of \small 45^{\circ} with the line \small x-2y=3 .

    Answer:

    Given the equation of the line is
    \small x-2y=3
    The slope of line \small x-2y=3 , m_2= \frac{1}{2}
    Let the slope of the other line is, m_1=m
    Now, it is given that both the lines make an angle \small 45^{\circ} with each other
    Therefore,
    \tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |
    \tan 45\degree = \left | \frac{\frac{1}{2}-m}{1+\frac{m}{2}} \right |
    1= \left | \frac{1-2m}{2+m} \right |
    Now,

    Case (i)
    1=\frac{1-2m}{2+m}
    2+m=1-2m
    m = -\frac{1}{3}
    Equation of line passing through the point \small (3,2) and with slope -\frac{1}{3}
    (y-2)=-\frac{1}{3}(x-3)
    3(y-2)=-1(x-3)
    x+3y=9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)

    Case (ii)
    1=-\left ( \frac{1-2m}{2+m} \right )
    2+m=-(1-2m)
    m= 3
    Equation of line passing through the point \small (3,2) and with slope 3 is
    (y-2)=3(x-3)
    3x-y=7 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)

    Therefore, equations of lines are 3x-y=7 and x+3y=9

    Question:12 Find the equation of the line passing through the point of intersection of the lines 4x+7y-3=0 and 2x-3y+1=0 that has equal intercepts on the axes.

    Answer:

    Point of intersection of the lines 4x+7y-3=0 and 2x-3y+1=0 is \left ( \frac{1}{13},\frac{5}{13} \right )
    We know that the intercept form of the line is
    \frac{x}{a}+\frac{y}{b}= 1
    It is given that line make equal intercepts on x and y axis
    Therefore,
    a = b
    Now, the equation reduces to
    x+y = a -(i)
    It passes through point \left ( \frac{1}{13},\frac{5}{13} \right )
    Therefore,
    a = \frac{1}{13}+\frac{5}{13}= \frac{6}{13}
    Put the value of a in equation (i)
    we will get
    13x+13y=6
    Therefore, equation of line is 13x+13y=6

    Question:13 Show that the equation of the line passing through the origin and making an angle \small \theta with the line \small y=mx+c is \small \frac{y}{x}=\frac{m\pm \tan \theta }{1\mp m\tan \theta } .

    Answer:

    Slope of line \small y=mx+c is m
    Let the slope of other line is m'
    It is given that both the line makes an angle \small \theta with each other
    Therefore,
    \tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |
    \tan \theta = \left | \frac{m-m'}{1+mm'} \right |
    \mp(1+mm')\tan \theta =(m-m')
    \mp\tan \theta +m'(\mp m\tan\theta+1)= m
    m'= \frac{m\pm \tan \theta}{1\mp m\tan \theta}
    Now, equation of line passing through origin (0,0) and with slope \frac{m\pm \tan \theta}{1\mp m\tan \theta} is
    (y-0)=\frac{m\pm \tan \theta}{1\mp m\tan \theta}(x-0)
    \frac{y}{x}=\frac{m\pm \tan \theta}{1\mp m\tan \theta}
    Hence proved

    Question:14 In what ratio, the line joining \small (-1,1) and \small (5,7) is divided by the line x+y=4 ?

    Answer:

    Equation of line joining \small (-1,1) and \small (5,7) is
    (y-1)= \frac{7-1}{5+1}(x+1)
    \Rightarrow (y-1)= \frac{6}{6}(x+1)
    \Rightarrow (y-1)= 1(x+1)
    \Rightarrow x-y+2=0
    Now, point of intersection of lines x+y=4 and x-y+2=0 is (1,3)
    Now, let's suppose point (1,3) divides the line segment joining \small (-1,1) and \small (5,7) in 1:k
    Then,
    (1,3)= \left ( \frac{k(-1)+1(5)}{k+1},\frac{k(1)+1(7)}{k+1} \right )
    1=\frac{-k+5}{k+1} \ \ and \ \ 3 = \frac{k+7}{k+1}
    \Rightarrow k =2
    Therefore, the line joining \small (-1,1) and \small (5,7) is divided by the line x+y=4 in ratio 1:2

    Question:15 Find the distance of the line \small 4x+7y+5=0 from the point \small (1,2) along the line \small 2x-y=0 .

    Answer:

    1646805351289 point \small (1,2) lies on line 2x-y =0
    Now, point of intersection of lines 2x-y =0 and \small 4x+7y+5=0 is \left ( -\frac{5}{18},-\frac{5}{9} \right )
    Now, we know that the distance between two point is given by
    d = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|
    d = |\sqrt{(1+\frac{5}{18})^2+(2+\frac{5}{9})^2}|
    d = |\sqrt{(\frac{23}{18})^2+(\frac{23}{9})^2}|
    d = \left | \sqrt{\frac{529}{324}+\frac{529}{81}} \right |
    d = \left | \sqrt{\frac{529+2116}{324}} \right | = \left | \sqrt\frac{2645}{324} \right | =\frac{23\sqrt5}{18}
    Therefore, the distance of the line \small 4x+7y+5=0 from the point \small (1,2) along the line \small 2x-y=0 is \frac{23\sqrt5}{18} \ units

    Question:16 Find the direction in which a straight line must be drawn through the point \small (-1,2) so that its point of intersection with the line \small x+y=4 may be at a distance of 3 units from this point.

    Answer:

    Let (x_1,y_1) be the point of intersection
    it lies on line \small x+y=4
    Therefore,
    x_1+y_1=4 \\ x_1=4-y_1\ \ \ \ \ \ \ \ \ \ \ -(i)
    Distance of point (x_1,y_1) from \small (-1,2) is 3
    Therefore,
    3= |\sqrt{(x_1+1)^2+(y_1-2)^2}|
    Square both the sides and put value from equation (i)
    9= (5-y_1)^2+(y_1-2)^2\\ 9=y_1^2+25-10y_1+y_1^2+4-4y_1\\ 2y_1^2-14y_1+20=0\\ y_1^2-7y_1+10=0\\ y_1^2-5y_1-2y_1+10=0\\ (y_1-2)(y_1-5)=0\\ y_1=2 \ or \ y_1 = 5
    When y_1 = 2 \Rightarrow x_1 = 2 point is (2,2)
    and
    When y_1 = 5 \Rightarrow x_1 = -1 point is (-1,5)
    Now, slope of line joining point (2,2) and \small (-1,2) is
    m = \frac{2-2}{-1-2}=0
    Therefore, line is parallel to x-axis -(i)

    or
    slope of line joining point (-1,5) and \small (-1,2)
    m = \frac{5-2}{-1+2}=\infty
    Therefore, line is parallel to y-axis -(ii)

    Therefore, line is parallel to x -axis or parallel to y-axis

    Question:17 The hypotenuse of a right angled triangle has its ends at the points \small (1,3) and \small (-4,1) Find an equation of the legs (perpendicular sides) of the triangle.

    Answer:

    1646805393159 Slope of line OA and OB are negative times inverse of each other
    Slope of line OA is , m=\frac{3-y}{1-x}\Rightarrow (3-y)=m(1-x)
    Slope of line OB is , -\frac{1}{m}= \frac{1-y}{-4-x}\Rightarrow (x+4)=m(1-y)
    Now,

    Now, for a given value of m we get these equations
    If m = \infty
    1-x=0 \ \ \ \ and \ \ \ \ \ 1-y =0
    x=1 \ \ \ \ and \ \ \ \ \ y =1

    Question:18 Find the image of the point \small (3,8) with respect to the line x+3y=7 assuming the line to be a plane mirror.

    Answer:

    1646805433376 Let point (a,b) is the image of point \small (3,8) w.r.t. to line x+3y=7
    line x+3y=7 is perpendicular bisector of line joining points \small (3,8) and (a,b)
    Slope of line x+3y=7 , m' = -\frac{1}{3}
    Slope of line joining points \small (3,8) and (a,b) is , m = \frac{8-b}{3-a}
    Now,
    m = -\frac{1}{m'} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because lines \ are \ perpendicular)
    \frac{8-b}{3-a}= 3
    8-b=9-3a
    3a-b=1 \ \ \ \ \ \ \ \ \ \ \ -(i)
    Point of intersection is the midpoint of line joining points \small (3,8) and (a,b)
    Therefore,
    Point of intersection is \left ( \frac{3+a}{2},\frac{b+8}{2} \right )
    Point \left ( \frac{3+a}{2},\frac{b+8}{2} \right ) also satisfy the line x+3y=7
    Therefore,
    \frac{3+a}{2}+3.\frac{b+8}{2}=7
    a+3b=-13 \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
    On solving equation (i) and (ii) we will get
    (a,b) = (-1,-4)
    Therefore, the image of the point \small (3,8) with respect to the line x+3y=7 is (-1,-4)

    Question:19 If the lines \small y=3x+1 and \small 2y=x+3 are equally inclined to the line \small y=mx+4 , find the value of m .

    Answer:

    Given equation of lines are
    \small y=3x+1 \ \ \ \ \ \ \ \ \ \ -(i)
    \small 2y=x+3 \ \ \ \ \ \ \ \ \ \ -(ii)
    \small y=mx+4 \ \ \ \ \ \ \ \ \ \ -(iii)
    Now, it is given that line (i) and (ii) are equally inclined to the line (iii)
    Slope of line \small y=3x+1 is , \small m_1=3
    Slope of line \small 2y=x+3 is , \small m_2= \frac{1}{2}
    Slope of line \small y=mx+4 is , \small m_3=m
    Now, we know that
    \tan \theta = \left | \frac{m_1-m_2}{1+m_1m_2} \right |
    Now,
    \tan \theta_1 = \left | \frac{3-m}{1+3m} \right | and \tan \theta_2 = \left | \frac{\frac{1}{2}-m}{1+\frac{m}{2}} \right |

    It is given that \tan \theta_1=\tan \theta_2
    Therefore,
    \left | \frac{3-m}{1+3m} \right |= \left | \frac{1-2m}{2+m} \right |
    \frac{3-m}{1+3m}= \pm\left ( \frac{1-2m}{2+m} \right )
    Now, if \frac{3-m}{1+3m}= \left ( \frac{1-2m}{2+m} \right )
    Then,
    (2+m)(3-m)=(1-2m)(1+3m)
    6+m-m^2=1+m-6m^2
    5m^2=-5
    m= \sqrt{-1}
    Which is not possible
    Now, if \frac{3-m}{1+3m}= -\left ( \frac{1-2m}{2+m} \right )
    Then,
    (2+m)(3-m)=-(1-2m)(1+3m)
    6+m-m^2=-1-m+6m^2
    7m^2-2m-7=0
    m = \frac{-(-2)\pm \sqrt{(-2)^2-4\times 7\times (-7)}}{2\times 7}= \frac{2\pm \sqrt{200}}{14}= \frac{1\pm5\sqrt2}{7}

    Therefore, the value of m is \frac{1\pm5\sqrt2}{7}

    Question:20 If the sum of the perpendicular distances of a variable point \small P(x,y) from the lines \small x+y-5=0 and \small 3x-2y+7=0 is always \small 10 . Show that \small P must move on a line.

    Answer:

    Given the equation of line are
    x+y-5=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)
    3x-2y+7=0 \ \ \ \ \ \ \ \ \ \ \ -(ii)
    Now, perpendicular distances of a variable point \small P(x,y) from the lines are

    d_1=\left | \frac{1.x+1.y-5}{\sqrt{1^2+1^2}} \right | d_2=\left | \frac{3.x-2.y+7}{\sqrt{3^2+2^2}} \right |
    d_1=\left | \frac{x+y-5}{\sqrt2} \right | d_2=\left | \frac{3x-2y+7}{\sqrt{13}} \right |
    Now, it is given that
    d_1+d_2= 10
    Therefore,
    \frac{x+y-5}{\sqrt2}+\frac{3x-2y+7}{\sqrt{13}}=10
    (assuming \ x+y-5 > 0 \ and \ 3x-2y+7 >0)
    (x+y-5)\sqrt{13}+(3x-2y+7)\sqrt2=10\sqrt{26}

    x(\sqrt{13}+3\sqrt{2})+y(\sqrt{13}-2\sqrt{2})=10\sqrt{26}+5\sqrt{13}-7\sqrt2

    Which is the equation of the line
    Hence proved

    Question:21 Find equation of the line which is equidistant from parallel lines 9x+6y-7=0 and 3x+2y+6=0 .

    Answer:

    Let's take the point p(a,b) which is equidistance from parallel lines 9x+6y-7=0 and 3x+2y+6=0
    Therefore,
    d_1= \left | \frac{9.a+6.b-7}{\sqrt{9^2+6^2}} \right | d_2= \left | \frac{3.a+2.b+6}{\sqrt{3^2+2^2}} \right |
    d_1= \left | \frac{9a+6b-7}{\sqrt{117}} \right | d_2= \left | \frac{3a+2b+6}{\sqrt{13}} \right |
    It is that d_1=d_2
    Therefore,
    \left | \frac{9a+6b-7}{3\sqrt{13}} \right |= \left | \frac{3a+2b+6}{\sqrt{13}} \right |
    (9a+6b-7)=\pm 3(3a+2b+6)
    Now, case (i)
    (9a+6b-7)= 3(3a+2b+6)
    25=0
    Therefore, this case is not possible

    Case (ii)
    (9a+6b-7)= -3(3a+2b+6)
    18a+12b+11=0

    Therefore, the required equation of the line is 18a+12b+11=0

    Question:22 A ray of light passing through the point (1,2) reflects on the x -axis at point A and the reflected ray passes through the point (5,3) . Find the coordinates of A .

    Answer:

    1646805488994 From the figure above we can say that
    The slope of line AC (m)= \tan \theta
    Therefore,
    \tan \theta = \frac{3-0}{5-a} = \frac{3}{5-a} \ \ \ \ \ \ \ \ \ \ (i)
    Similarly,
    The slope of line AB (m') = \tan(180\degree-\theta)
    Therefore,
    \tan(180\degree-\theta) = \frac{2-0}{1-a}
    -\tan\theta= \frac{2}{1-a}
    \tan\theta= \frac{2}{a-1} \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
    Now, from equation (i) and (ii) we will get
    \frac{3}{5-a} = \frac{2}{a-1}
    \Rightarrow 3(a-1)= 2(5-a)
    \Rightarrow 3a-3= 10-2a
    \Rightarrow 5a=13
    \Rightarrow a=\frac{13}{5}
    Therefore, the coordinates of A . is \left ( \frac{13}{5},0 \right )

    Question:23 Prove that the product of the lengths of the perpendiculars drawn from the points \small (\sqrt{a^2-b^2},0) and \small (-\sqrt{a^2-b^2},0) to the line \small \frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1 is \small b^2 .

    Answer:

    Given equation id line is
    \small \frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1
    We can rewrite it as
    xb\cos \theta +ya\sin \theta =ab
    Now, the distance of the line xb\cos \theta +ya\sin \theta =ab from the point \small (\sqrt{a^2-b^2},0) is given by
    d_1=\left | \frac{b\cos\theta.\sqrt{a^2-b^2}+a\sin \theta.0-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right | = \left | \frac{b\cos\theta.\sqrt{a^2-b^2}-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |
    Similarly,
    The distance of the line xb\cos \theta +ya\sin \theta =ab from the point \small (-\sqrt{a^2-b^2},0) is given by
    d_2=\left | \frac{b\cos\theta.(-\sqrt{a^2-b^2})+a\sin \theta.0-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right | = \left | \frac{-(b\cos\theta.\sqrt{a^2-b^2}+ab)}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |
    d_1.d_2 = \left | \frac{b\cos\theta.\sqrt{a^2-b^2}-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |.\times\left | \frac{-(b\cos\theta.\sqrt{a^2-b^2}+ab)}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |
    =\left | \frac{-((b\cos\theta.\sqrt{a^2-b^2})^2-(ab)^2)}{(b\cos\theta)^2+(a\sin\theta)^2} \right |
    =\left | \frac{-b^2\cos^2\theta.(a^2-b^2)+a^2b^2)}{(b\cos\theta)^2+(a\sin\theta)^2} \right |
    =\left | \frac{-a^2b^2\cos^2\theta+b^4\cos^2\theta+a^2b^2)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |
    =\left | \frac{-b^2(a^2\cos^2\theta-b^2\cos^2\theta-a^2)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |
    =\left | \frac{-b^2(a^2\cos^2\theta-b^2\cos^2\theta-a^2(\sin^2\theta+\cos^2\theta))}{b^2\cos^2\theta+a^2\sin^2\theta} \right | \ \ \ \ (\because \sin^2a+\cos^2a=1)
    =\left | \frac{-b^2(a^2\cos^2\theta-b^2\cos^2\theta-a^2\sin^2\theta-a^2\cos^2\theta)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |
    =\left | \frac{+b^2(b^2\cos^2\theta+a^2\sin^2\theta)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |
    =b^2
    Hence proved

    Question:24 A person standing at the junction (crossing) of two straight paths represented by the equations \small 2x-3y+4=0 and \small 3x+4y-5=0 wants to reach the path whose equation is \small 6x-7y+8=0 in the least time. Find equation of the path that he should follow.

    Answer:

    point of intersection of lines \small 2x-3y+4=0 and \small 3x+4y-5=0 (junction) is \left ( -\frac{1}{17},\frac{22}{17} \right )
    Now, person reaches to path \small 6x-7y+8=0 in least time when it follow the path perpendicular to it
    Now,
    Slope of line \small 6x-7y+8=0 is , m'=\frac{6}{7}
    let the slope of line perpendicular to it is , m
    Then,
    m= -\frac{1}{m}= -\frac{7}{6}
    Now, equation of line passing through point \left ( -\frac{1}{17},\frac{22}{17} \right ) and with slope -\frac{7}{6} is
    \left ( y-\frac{22}{17} \right )= -\frac{7}{6}\left ( x-(-\frac{1}{17}) \right )
    \Rightarrow 6(17y-22)=-7(17x+1)
    \Rightarrow 102y-132=-119x-7
    \Rightarrow 119x+102y=125

    Therefore, the required equation of line is 119x+102y=125

    Class 11 maths chapter 10 NCERT solutions - Topics

    10.1 Introduction

    10.2 Slope of a Line

    10.3 Various Forms of the Equation of a Line

    10.4 General Equation of a Line

    10.5 Distance of a Point From a Line

    If you are interested in class 11 maths chapter 10 question answer of exercises then these are listed below.

    NCERT solutions for class 11 mathematics - Chapter Wise

    Key Feature of class 11 maths chapter 10 NCERT solutions

    Clear and Concise Explanations: The NCERT Solution of ch 10 maths class 11 provide clear and concise explanations for all the concepts covered in the chapter. This makes it easier for students to understand and retain the information.

    Step-by-Step Solutions: Each problem in the NCERT Solutions of maths chapter 10 class 11 is solved step-by-step, making it easy for students to follow the logic and understand the solution.

    Revision Notes: The class 11 maths chapter 10 question answer also include revision notes that summarize the key concepts covered in the chapter. This makes it easier for students to revise the chapter before exams.

    NCERT solutions for class 11 - Subject wise

    Important Formulas

    • If a non-vertical line passing through the points (x_1,\:y_1) and (x_2,\:y_2) then the slope(m) of the line is given by-

    m=\frac{y_2-y_1}{x_2-x_1}=\frac{y_1-y_2}{x_1-x_2},\:\:x_1\neq x_2

    • The slope of the line which makes an angle with the positive x-axis is given by m = tan\:\alpha \: , \alpha\neq 90^o .
    • The slope of the horizontal line is zero and the slope of the vertical line is undefined.
    • An acute angle ( \theta ) between lines L_1 and L_2 with slopes m_1 and m_2 is given by-

    tan \:\theta =|\frac{m_2-m_1}{1+m_1m_2}|\:,\:1+m_1m_2\neq 0

    • Two lines (with slopes m_1 \: and m_2 \: ) are parallel if and only if their slopes ( m_1=m_2 \: ) are equal.
    • Two lines (with slopes m_1 \: and m_2 \: ) are perpendicular if and only if the product of their slopes is –1 or m_1.m_2 =-1 .
    • The equation of a line passing through the points (x_1,\:y_1) and (x_2,\:y_2) is given by-

    y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

    Tip- If you facing difficulties in the memorizing the formulas, you should write formula every time when you are solving the problem. You should try to solve every problem on your own and reading the solutions won't be much helpful. You can take help from NCERT solutions for class 11 maths chapter 10 straight lines.

    NCERT Books and NCERT Syllabus

    Happy Reading !!!

    Frequently Asked Question (FAQs)

    1. What are the important topics of the chapter Straight Lines ?

    The slope of a line, various forms of the equation of a line, the general equation of a line, the distance of a point from a line are the important chapters of this chapter. Important topics are enumerated in NCERT syllabus.

    2. Mention the topics and subtopics covered in NCERT solution of class 11th maths chapter 10.

    The contents of straight lines class 11 NCERT solutions are structured as follows:

    10.1 Introduction 

    10.2 Slope of a Line 

    • 10.2.1 Calculation of slope when coordinates of any two points on the line are given 
    • 10.2.2 Criteria for parallelism and perpendicularity of lines based on their slopes 
    • 10.2.3 Calculation of angle between two lines
    • 10.2.4 Determination of collinearity of three points 

    10.3 Various Forms of the Equation of a Line 

    • 10.3.1 Discussion of horizontal and vertical lines 
    • 10.3.2 Derivation of point-slope form 
    • 10.3.3 Derivation of two-point form 
    • 10.3.4 Derivation of slope-intercept form 
    • 10.3.5 Derivation of intercept-form 
    • 10.3.6 Derivation of normal form

    10.4 General Equation of a Line 

    • 10.4.1 Explanation of different forms of Ax + By + C = 0

    10.5 Distance of a Point From a Line 

    • 10.5.1 Computation of distance between two parallel lines.
    3. Explain the equation of a line in general form as per class 11 chapter 10 maths.

    Class 11 maths ch 10 question answer can be expressed in various forms to represent a graphical line. Among these forms, the most prevalent one is the general equation, which is used to represent a line in two variables, namely, x and y, of the first degree. The general equation, Ax + By + C = 0, where A and B are non-zero constants and C is a constant that belongs to the set of real numbers, is commonly used. The variables x and y represent the respective axes' coordinates.

    4. Where can I find the complete solutions of NCERT for class 11 maths ?

    Here you will get the detailed NCERT solutions for class 11 maths  by clicking on the link. Both careers360 official website and this article listed NCERT book solutions. Interested students refer them and also for easy they can study straight lines class 11 pdf both online and offline mode.

    Articles

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    A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

    Option 1)

    0.34\; J

    Option 2)

    0.16\; J

    Option 3)

    1.00\; J

    Option 4)

    0.67\; J

    A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

    Option 1)

    2.45×10−3 kg

    Option 2)

     6.45×10−3 kg

    Option 3)

     9.89×10−3 kg

    Option 4)

    12.89×10−3 kg

     

    An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

    Option 1)

    2,000 \; J - 5,000\; J

    Option 2)

    200 \, \, J - 500 \, \, J

    Option 3)

    2\times 10^{5}J-3\times 10^{5}J

    Option 4)

    20,000 \, \, J - 50,000 \, \, J

    A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

    Option 1)

    K/2\,

    Option 2)

    \; K\;

    Option 3)

    zero\;

    Option 4)

    K/4

    In the reaction,

    2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

    Option 1)

    11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

    Option 2)

    6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

    Option 3)

    33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

    Option 4)

    67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

    How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

    Option 1)

    0.02

    Option 2)

    3.125 × 10-2

    Option 3)

    1.25 × 10-2

    Option 4)

    2.5 × 10-2

    If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

    Option 1)

    decrease twice

    Option 2)

    increase two fold

    Option 3)

    remain unchanged

    Option 4)

    be a function of the molecular mass of the substance.

    With increase of temperature, which of these changes?

    Option 1)

    Molality

    Option 2)

    Weight fraction of solute

    Option 3)

    Fraction of solute present in water

    Option 4)

    Mole fraction.

    Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

    Option 1)

    twice that in 60 g carbon

    Option 2)

    6.023 × 1022

    Option 3)

    half that in 8 g He

    Option 4)

    558.5 × 6.023 × 1023

    A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

    Option 1)

    less than 3

    Option 2)

    more than 3 but less than 6

    Option 3)

    more than 6 but less than 9

    Option 4)

    more than 9

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    Database Architect

    If you are intrigued by the programming world and are interested in developing communications networks then a career as database architect may be a good option for you. Data architect roles and responsibilities include building design models for data communication networks. Wide Area Networks (WANs), local area networks (LANs), and intranets are included in the database networks. It is expected that database architects will have in-depth knowledge of a company's business to develop a network to fulfil the requirements of the organisation. Stay tuned as we look at the larger picture and give you more information on what is db architecture, why you should pursue database architecture, what to expect from such a degree and what your job opportunities will be after graduation. Here, we will be discussing how to become a data architect. Students can visit NIT Trichy, IIT Kharagpur, JMI New Delhi

    3 Jobs Available
    Data Analyst

    The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

    Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

    3 Jobs Available
    Geothermal Engineer

    Individuals who opt for a career as geothermal engineers are the professionals involved in the processing of geothermal energy. The responsibilities of geothermal engineers may vary depending on the workplace location. Those who work in fields design facilities to process and distribute geothermal energy. They oversee the functioning of machinery used in the field.

    3 Jobs Available
    Finance Executive

    A career as a Finance Executive requires one to be responsible for monitoring an organisation's income, investments and expenses to create and evaluate financial reports. His or her role involves performing audits, invoices, and budget preparations. He or she manages accounting activities, bank reconciliations, and payable and receivable accounts.  

    3 Jobs Available
    Investment Banker

    An Investment Banking career involves the invention and generation of capital for other organizations, governments, and other entities. Individuals who opt for a career as Investment Bankers are the head of a team dedicated to raising capital by issuing bonds. Investment bankers are termed as the experts who have their fingers on the pulse of the current financial and investing climate. Students can pursue various Investment Banker courses, such as Banking and Insurance, and Economics to opt for an Investment Banking career path.

    3 Jobs Available
    Bank Branch Manager

    Bank Branch Managers work in a specific section of banking related to the invention and generation of capital for other organisations, governments, and other entities. Bank Branch Managers work for the organisations and underwrite new debts and equity securities for all type of companies, aid in the sale of securities, as well as help to facilitate mergers and acquisitions, reorganisations, and broker trades for both institutions and private investors.

    3 Jobs Available
    Treasurer

    Treasury analyst career path is often regarded as certified treasury specialist in some business situations, is a finance expert who specifically manages a company or organisation's long-term and short-term financial targets. Treasurer synonym could be a financial officer, which is one of the reputed positions in the corporate world. In a large company, the corporate treasury jobs hold power over the financial decision-making of the total investment and development strategy of the organisation.

    3 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    Underwriter

    An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

    3 Jobs Available
    Bank Probationary Officer (PO)

    A career as Bank Probationary Officer (PO) is seen as a promising career opportunity and a white-collar career. Each year aspirants take the Bank PO exam. This career provides plenty of career development and opportunities for a successful banking future. If you have more questions about a career as  Bank Probationary Officer (PO), what is probationary officer or how to become a Bank Probationary Officer (PO) then you can read the article and clear all your doubts. 

    3 Jobs Available
    Operations Manager

    Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

    3 Jobs Available
    Transportation Planner

    A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

    3 Jobs Available
    Conservation Architect

    A Conservation Architect is a professional responsible for conserving and restoring buildings or monuments having a historic value. He or she applies techniques to document and stabilise the object’s state without any further damage. A Conservation Architect restores the monuments and heritage buildings to bring them back to their original state.

    2 Jobs Available
    Safety Manager

    A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

    2 Jobs Available
    Team Lead

    A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

    2 Jobs Available
    Structural Engineer

    A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software. 

    2 Jobs Available
    Architect

    Individuals in the architecture career are the building designers who plan the whole construction keeping the safety and requirements of the people. Individuals in architect career in India provides professional services for new constructions, alterations, renovations and several other activities. Individuals in architectural careers in India visit site locations to visualize their projects and prepare scaled drawings to submit to a client or employer as a design. Individuals in architecture careers also estimate build costs, materials needed, and the projected time frame to complete a build.

    2 Jobs Available
    Landscape Architect

    Having a landscape architecture career, you are involved in site analysis, site inventory, land planning, planting design, grading, stormwater management, suitable design, and construction specification. Frederick Law Olmsted, the designer of Central Park in New York introduced the title “landscape architect”. The Australian Institute of Landscape Architects (AILA) proclaims that "Landscape Architects research, plan, design and advise on the stewardship, conservation and sustainability of development of the environment and spaces, both within and beyond the built environment". Therefore, individuals who opt for a career as a landscape architect are those who are educated and experienced in landscape architecture. Students need to pursue various landscape architecture degrees, such as M.Des, M.Plan to become landscape architects. If you have more questions regarding a career as a landscape architect or how to become a landscape architect then you can read the article to get your doubts cleared. 

    2 Jobs Available
    Plumber

    An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

    2 Jobs Available
    Orthotist and Prosthetist

    Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

    6 Jobs Available
    Veterinary Doctor

    A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

    5 Jobs Available
    Pathologist

    A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

    5 Jobs Available
    Gynaecologist

    Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

    4 Jobs Available
    Surgical Technologist

    When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications. 

    Also Read: Career as Nurse

    3 Jobs Available
    Radiation Therapist

    People might think that a radiation therapist only spends most of his/her time in a radiation operation unit but that’s not the case. In reality, a radiation therapist’s job is not as easy as it seems. The job of radiation therapist requires him/her to be attentive, hardworking, and dedicated to his/her work hours. A radiation therapist is on his/her feet for a long duration and might be required to lift or turn disabled patients. Because a career as a radiation therapist involves working with radiation and radioactive material, a radiation therapist is required to follow the safety procedures in order to make sure that he/she is not exposed to a potentially harmful amount of radiation.

    3 Jobs Available
    Recreational Worker

    A recreational worker is a professional who designs and leads activities to provide assistance to people to adopt a healthy lifestyle. He or she instructs physical exercises and games to have fun and improve fitness. A recreational worker may work in summer camps, fitness and recreational sports centres, nature parks, nursing care facilities, and other settings. He or she may lead crafts, sports, music, games, drama and other activities.

    3 Jobs Available
    Oncologist

    An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

    3 Jobs Available
    Actor

    For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

    4 Jobs Available
    Acrobat

    Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

    3 Jobs Available
    Video Game Designer

    Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

    3 Jobs Available
    Talent Agent

    The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

    If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

    3 Jobs Available
    Radio Jockey

    Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

    A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

    3 Jobs Available
    Talent Director

    Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots. 

    2 Jobs Available
    Videographer

    Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production. 

    2 Jobs Available
    Multimedia Specialist

    A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

    2 Jobs Available
    Copy Writer

    In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

    5 Jobs Available
    Journalist

    Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

    3 Jobs Available
    Publisher

    For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

    3 Jobs Available
    Vlogger

    In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

    3 Jobs Available
    Editor

    Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

    3 Jobs Available
    Advertising Manager

    Advertising managers consult with the financial department to plan a marketing strategy schedule and cost estimates. We often see advertisements that attract us a lot, not every advertisement is just to promote a business but some of them provide a social message as well. There was an advertisement for a washing machine brand that implies a story that even a man can do household activities. And of course, how could we even forget those jingles which we often sing while working?

    2 Jobs Available
    Photographer

    Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

    2 Jobs Available
    Social Media Manager

    A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

    2 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    Quality Controller

    A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

    A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

    3 Jobs Available
    Production Manager

    Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers. 

    Resource Links for Online MBA 

    3 Jobs Available
    QA Manager

    Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes. 

    2 Jobs Available
    QA Lead

    A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

    2 Jobs Available
    Reliability Engineer

    Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment. 

    2 Jobs Available
    Safety Manager

    A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

    2 Jobs Available
    Corporate Executive

    Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

    2 Jobs Available
    Computer Programmer

    Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

    3 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    ITSM Manager

    ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks. 

    3 Jobs Available
    Information Security Manager

    Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

    3 Jobs Available
    Big Data Analytics Engineer

    Big Data Analytics Engineer Job Description: A Big Data Analytics Engineer is responsible for collecting data from various sources. He or she has to sort the organised and chaotic data to find out patterns. The role of Big Data Engineer involves converting messy information into useful data that is clean, accurate and actionable. 

    2 Jobs Available
    Cloud Solution Developer

    A Cloud Solutions Developer is basically a Software Engineer with specialisation in cloud computing. He or she possesses a solid understanding of cloud systems including their operations, deployment with security and efficiency with no little downtime. 

    2 Jobs Available
    CRM Technology Consultant

    A Customer Relationship Management Technology Consultant or CRM Technology Consultant is responsible for monitoring and providing strategy for performance improvement with logged calls, performance metrics and revenue metrics. His or her role involves accessing data for team meetings, goal setting analytics as well as reporting to executives.  

    2 Jobs Available
    IT Manager

    Career as IT Manager  requires managing the various aspects of an organization's information technology systems. He or she is responsible for increasing productivity and solving problems related to software and hardware. While this role is typically one of the lower-level positions within an organisation, it comes with responsibilities related to people and ownership of systems.

    2 Jobs Available
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