NCERT Solutions for Exercise 10.2 Class 11 Maths Chapter 10 - Straight Lines

# NCERT Solutions for Exercise 10.2 Class 11 Maths Chapter 10 - Straight Lines

Edited By Vishal kumar | Updated on Nov 09, 2023 11:49 AM IST

## NCERT Solutions for Class 11 Maths Chapter 10: Straight Lines Exercise 10.2- Download Free PDF

In the previous exercise, you have already learned about finding the minimum distance between two points, the slope of the line given two coordinate points on the line, the angles between two lines, etc. In this article, you will get NCERT syllabus for Class 11 Maths chapter 10 exercise 10.2 where you will learn about the various forms of the equation of the line. Equations of the coordinate axis, equation of the line in the point-slope form, equation of the line in the two-point form, equation of the line in the slope-intersect form, equation of the line in intersect form, equation of the line in the normal form are also covered in the NCERT book Class 11 Maths chapter 10 exercise 10.2 solutions.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

Exercise 10.2 Class 11 Maths is a very useful exercise in two-dimensional geometry and very important for the CBSE exam as well. If you have basic knowledge of two-dimensional geometry, you will easily grasp the concepts of Class 11 Maths chapter 10 exercise 10.2. Also, check NCERT Solutions link if you are looking for NCERT solutions explained in a detailed manner.

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**As per the CBSE Syllabus for 2023-24, this chapter has been renumbered as Chapter 9.

## NCERT Solutions for Class 11 Maths Chapter 10– Straight Lines Exercise 10.2

Access Straight Lines Class 11 Chapter 10 Exercise: 10.2

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

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

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

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 $x-2y+10 = 0$

Passing through $(0,0)$ with slope $m$.

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 $y = mx$

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

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

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

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 $2x+y+6=0$

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

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

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

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

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

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

## More About NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.2:-

Class 11th Maths chapter 10 exercise 10.2 consists of questions related to finding equations of lines in various forms like two-point form, point-slope form, etc. Equation of line is the most basic as well as very important concept of two-dimensional geometry. There are few solved examples and some important concepts given before the Class 11 Maths chapter 10 exercise 10.2. You must go through the theory related equations of lines given before solving the exercise 10.2 Class 11 Maths problems.

Also Read| Straight Lines Class 11 Notes

## Benefits of NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.2:-

• Exercise 10.2 Class 11 maths is very important to understand two-dimensional geometry.
• You must try to solve Class 11 Maths chapter 10 exercise 10.2 problems by yourself.
• Class 11 Maths chapter 10 exercise 10.2 solutions are here to assist you if you face any difficulty while solving them.
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## Key Features of Exercise 10.2 Class 11 Maths Solutions

1. Complete Exercise Coverage: These 11th class maths exercise 10.2 answers encompass all exercises and problems found in Chapter 10.2 of the Class 11 Mathematics textbook.

2. Step-by-Step Approach: The ex 10.2 class 11 solutions are presented in a clear, step-by-step format, aiding students in understanding and applying problem-solving methods effectively.

3. Clarity and Precision: The class 11 maths ex 10.2 solutions are written with clarity and precision, ensuring that students can easily comprehend the mathematical concepts and techniques required for successful problem-solving.

4. Correct Mathematical Notation: Appropriate mathematical notations and terminology are used, helping students become proficient in the language of mathematics.

5. Emphasis on Conceptual Understanding: The class 11 ex 10.2 solutions aim to foster a deep understanding of mathematical concepts, encouraging critical thinking and problem-solving skills.

6. Free Accessibility: Typically, these ex 10.2 class 11 solutions are available free of charge, making them an accessible and valuable resource for self-study.

7. Supplementary Learning Tool: These class 11 ex 10.2 solutions can be used as supplementary resources to complement classroom instruction and support students in exam preparation.

8. Homework and Practice: Students can utilize these solutions to cross-verify their work, practice problem-solving, and enhance their overall performance in mathematics.

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## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. Write the equation of line passing through points (0,0) and (1,1) ?

Equation of line passing through points (0,0) and (1,1) =>  y-0 = (1-0)/(1-0) ( x -0)

y = x

2. Write the equation of line passing through points (0,0) and slope is 2 ?

Equation of line passing through points (0,0) and slope is 2 => y = 2x + c

Line is passing through (0,0) =>  0 = 0 +c =>  c=0

Equation of the line =>  y = 2x

3. Write the equation of line parallel to x-axis and passing through (2,4) ?

The equation of line parallel to x-axis=>  y = c

Line is passing through (2,4)  =>  c =4

Equation of the line  =>  y = 4

4. Write the equation of line parallel to y-axis and passing through (2,4) ?

The equation of line parallel to y-axis=>  x = c

Line is passing through (2,4)  =>  c =2

Equation of the line  =>  x = 2

5. Write the equation of x-axis ?

Equation of x-axis =>  y = 0

6. Write the equation of y-axis ?

Equation of y-axis =>  x = 0

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