NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

Access premium articles, webinars, resources to make the best decisions for career, course, exams, scholarships, study abroad and much more with

Plan, Prepare & Make the Best Career Choices

# 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 â€“ 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.

## 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 = âˆš((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 = âˆš(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 â€“ y3) + x2(y3 â€“ y1) + x3(y1 â€“ 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 (Î¸) between two lines with slopes m1 and m2 is given by:

• tan Î¸ = |(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)/âˆš(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|/âˆš(1 + m2)

Different Forms of Equation of a Line:

Various forms of the equation of a line include:

• Normal form: x cos Î± + y sin Î± = 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

Area of ABCD = Area of ABC + Area of ACDNow, we know that the area of a triangle with vertices is given byTherefore,Area of triangle ABC Similarly,Area of triangle ACD Now,Area of ABCD = Area of ABC + Area of ACD

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
$(0,a),(0,-a) \and \ (\sqrt3a,0) \ \ or \ \ (0,a),(0,-a) \and \ (-\sqrt3a,0)$

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 $|(y_2-y_1)|$

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 $|x_2-x_1|$

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 $( \frac{15}{2},0)$

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 $\frac{-1}{2}$

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

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 $-\sqrt3$

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

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 =

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

= $\frac{2+1}{-3+2} = \frac{3}{-1} = -3$
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

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 $135\degree$

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, $m_1,m_2 = \frac{1}{2},1 \ or \ \frac{-1}{2},-1 \ or \ 1,2 \ or \ -1,-2$

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

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

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

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

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 $x-2y+10 = 0$

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 $y = mx$

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 $(\sqrt3+1)x-(\sqrt3-1)y = 4(\sqrt3-1)$

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 $2x+y+6=0$

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 $x-\sqrt3y+2\sqrt3=0$

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
$(y-1)= \frac{-5}{3}(x-(-1))\\ \\3(y-1)=-5(x+1)\\ 3y-3=-5x-5\\ 5x+3y+2=0$

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 $\sqrt3x+y =10$

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 $3x-4y+8=0$

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 $5x-y+20=0$

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 $x(1+n)+3y(1+n)=n+11$

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 $x+ y = 5$

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

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 $\sqrt3x+y+2=0$

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 $2x-9y+85 = 0$

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

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 $Rs\hspace{1mm}17/litre$

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

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 $2kx+yh=3kh$

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

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

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

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

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

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

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

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

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

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

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

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)

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 $\frac{65}{17} \ units$

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 $\frac{1}{\sqrt2}\left | \frac{p+r}{l} \right |$

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 $3x-4y+18= 0$

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
$(y-0)=-7(x-3)$
$y = -7x+21$
$7x+y-21=0$

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 $30\degree \ and \ 150\degree$

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 $\frac{22}{9}$

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

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 $x(2+\sqrt3)+y(2\sqrt3-1)=1+8\sqrt3$

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 $2x+y=5$

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, $(x_1,y_1) = \left ( \frac{68}{25},-\frac{49}{25} \right )$

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

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

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

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

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

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 $\pm2$

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 $6 \ or \ 1$

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 $\small \theta = \frac{7\pi}{6} \ and \ p =1$

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 $2x-3y=6 \ and \ -3x+2y=6$

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 $\left ( 0,-\frac{8}{3} \right )$

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
$d = \left | \frac{(\sin\phi -\sin\theta).0-(\cos\phi-\cos\theta).0-\sin(\theta-\phi)}{\sqrt{(\sin\phi-\sin\theta)^2+(\cos\phi-\cos\theta)^2}} \right |$
$d = \left | \frac{-\sin(\theta-\phi)}{\sqrt{(\sin^2\phi+\cos^2\phi)+(\sin^2\theta+\cos^2\theta)-2(\cos\theta\cos\phi+\sin\theta\sin\phi)}} \right |$
$d = \left | \frac{-\sin(\theta-\phi)}{1+1-2\cos(\theta-\phi)} \right |$ $(\because \cos a\cos b +\sin a\sin b = \cos(a-b) \ \ and \ \ \sin^2a+\cos^2a=1)$
$d = \left |\frac{ - \sin(\theta-\phi)}{2(1-\cos(\theta-\phi))} \right |$
$d = \left | \frac{-2\sin\frac{\theta-\phi}{2}\cos \frac{\theta-\phi}{2}}{\sqrt{2.(2\\sin^2\frac{\theta-\phi}{2})}}\right |$
$d = \left | \frac{-2\sin\frac{\theta-\phi}{2}\cos \frac{\theta-\phi}{2}}{2\sin\frac{\theta-\phi}{2}}\right |$
$d = \left | \cos\frac{\theta-\phi}{2} \right |$

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 $x = -\frac{5}{22}$

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 $2x-3y+18=0$

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 $k^2 \ square \ units$

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 $5$

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

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 $x+3y=9$

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 $13x+13y=6$

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

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 $1:2$

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 $\frac{23\sqrt5}{18} \ units$

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

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 $m = \infty$
$1-x=0 \ \ \ \ and \ \ \ \ \ 1-y =0$
$x=1 \ \ \ \ and \ \ \ \ \ y =1$

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 $(-1,-4)$

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 $\frac{1\pm5\sqrt2}{7}$

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

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 $18a+12b+11=0$

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 $\left ( \frac{13}{5},0 \right )$

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

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

## 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 $\dpi{100} (x_1,\:y_1)$ and $\dpi{100} (x_2,\:y_2)$ then the slope(m) of the line is given by-

$\dpi{100} m=\frac{y_2-y_1}{x_2-x_1}=\frac{y_1-y_2}{x_1-x_2},\:\:x_1\neq x_2$

• The slope of the line which makes an angle with the positive x-axis is given by $\dpi{100} m = tan\:\alpha \: , \alpha\neq 90^o$ .
• The slope of the horizontal line is zero and the slope of the vertical line is undefined.
• An acute angle ( $\dpi{100} \theta$ ) between lines $\dpi{100} L_1$ and $\dpi{100} L_2$ with slopes $\dpi{100} m_1$ and $\dpi{100} m_2$ is given by-

$\dpi{100} tan \:\theta =|\frac{m_2-m_1}{1+m_1m_2}|\:,\:1+m_1m_2\neq 0$

• Two lines (with slopes $\dpi{100} m_1 \:$ and $\dpi{100} m_2 \:$ ) are parallel if and only if their slopes ( $\dpi{100} m_1=m_2 \:$ ) are equal.
• Two lines (with slopes $\dpi{100} m_1 \:$ and $\dpi{100} m_2 \:$ ) are perpendicular if and only if the product of their slopes is â€“1 or $\dpi{100} m_1.m_2 =-1$ .
• The equation of a line passing through the points $\dpi{100} (x_1,\:y_1)$ and $\dpi{100} (x_2,\:y_2)$ is given by-

$\dpi{100} y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

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

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.

## Upcoming School Exams

#### All India Sainik Schools Entrance Examination

Application Date:06 November,2023 - 15 December,2023

#### National Institute of Open Schooling 12th Examination

Application Date:20 November,2023 - 19 December,2023

#### National Institute of Open Schooling 10th examination

Application Date:20 November,2023 - 19 December,2023

#### National Means Cum-Merit Scholarship

Admit Card Date:22 November,2023 - 02 December,2023

Exam Date:29 November,2023

Get answers from students and experts

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) Option 2) Option 3) Option 4)

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) Option 2) Option 3) Option 4)

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

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, 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
##### Bio Medical Engineer

The field of biomedical engineering opens up a universe of expert chances. An Individual in the biomedical engineering career path work in the field of engineering as well as medicine, in order to find out solutions to common problems of the two fields. The biomedical engineering job opportunities are to collaborate with doctors and researchers to develop medical systems, equipment, or devices that can solve clinical problems. Here we will be discussing jobs after biomedical engineering, how to get a job in biomedical engineering, biomedical engineering scope, and salary.

4 Jobs Available

Database professionals use software to store and organise data such as financial information, and customer shipping records. Individuals who opt for a career as data administrators ensure that data is available for users and secured from unauthorised sales. DB administrators may work in various types of industries. It may involve computer systems design, service firms, insurance companies, banks and hospitals.

4 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
##### GIS Expert

GIS officer work on various GIS software to conduct a study and gather spatial and non-spatial information. GIS experts update the GIS data and maintain it. The databases include aerial or satellite imagery, latitudinal and longitudinal coordinates, and manually digitized images of maps. In a career as GIS expert, one is responsible for creating online and mobile maps.

3 Jobs Available
##### Ethical Hacker

A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.

3 Jobs Available
##### 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

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.

3 Jobs Available

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 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 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.

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

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