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.

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Straight Lines Class 11 Questions And Answers
Straight Lines Class 11 Questions And Answers PDF Free Download
Straight Lines Class 11 Solutions - Important Formulae
Straight Lines Class 11 NCERT Solutions (Intext Questions and Exercise)
Class 11 maths chapter 10 NCERT solutions - Topics
NCERT solutions for class 11 mathematics - Chapter Wise
Key Feature of class 11 maths chapter 10 NCERT solutions
NCERT solutions for class 11 - Subject wise

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

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.

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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)

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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 by $A = \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$ Therefore,Area of triangle ABC $= \frac{1}{2}|-4(7+5)+0(-5-5)+5(5-7)|= \frac{1}{2}|-48-10|= \frac{58}{2}=29$ Similarly,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

chapter-1	Sets
chapter-2	Relations and Functions
chapter-3	Trigonometric Functions
chapter-4	Principle of Mathematical Induction
chapter-5	Complex Numbers and Quadratic equations
chapter-6	Linear Inequalities
chapter-7	Permutation and Combinations
chapter-8	Binomial Theorem
chapter-9	Sequences and Series
chapter-10	Straight Lines
chapter-11	Conic Section
chapter-12	Introduction to Three Dimensional Geometry
chapter-13	Limits and Derivatives
chapter-14	Mathematical Reasoning
chapter-15	Statistics
chapter-16	Probability

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

NCERT solutions for class 11 biology

NCERT solutions for class 11 maths

NCERT solutions for class 11 chemistry

NCERT solutions for Class 11 physics

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.

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Frequently Asked Questions (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.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

Straight Lines Class 11 Questions And Answers

Straight Lines Class 11 Questions And Answers PDF Free Download

Straight Lines Class 11 Solutions - Important Formulae

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

Class 11 maths chapter 10 NCERT solutions - Topics

NCERT solutions for class 11 mathematics - Chapter Wise

Key Feature of class 11 maths chapter 10 NCERT solutions

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