NCERT Solutions for Exercise 10.3 Class 11 Maths Chapter 10 - Straight Lines

# NCERT Solutions for Exercise 10.3 Class 11 Maths Chapter 10 - Straight Lines

Edited By Vishal kumar | Updated on Nov 13, 2023 11:14 AM IST

## NCERT Solutions for Class 11 Maths Chapter 10: Straight Lines Exercise 10.3- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 10: Straight Lines Exercise 10.3- In the previous exercises, you have already learned about the basic two-dimensional geometry slope of the line, conditions for parallelism and perpendicularity of lines, angles between two lines, collinearity of three points, equations of lines in various forms, etc. In the NCERT syllabus for Class 11 Maths chapter 10 exercise 10.3, you will learn about the general equation of the line in various forms like slope-intercept form, intercept form, normal form, etc. Also, the distance of a point from a line and the distance between two parallel lines are covered in exercise 10.3 Class 11 Maths. The Class 11 Maths chapter 10 exercise 10.3 is a very important exercise from the CBSE exam point of view.

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

You must try to solve NCERT book Class 11th Maths chapter 10 exercise 10.3 problems by yourself. Most of the problems in the as well as exercise 10.3 Class 11 Maths are very basic and straightforward which you will be able to solve easily. You can refer to the Class 11 Maths chapter 10 exercise 10.3 solutions, meticulously crafted by subject experts at Careers360. These solutions are presented in detail and in easy-to-understand language. Additionally, PDF versions of the solutions are available, allowing students to access them at any time without incurring any charges. If you are looking for NCERT Solutions, click on the given link to get NCERT solutions at one place.

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

## Access Straight Lines Class 11 Chapter 10 Exercise 10.3

$x+7y=0$

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

$6x+3y-5=0$

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

$y=0.$

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

$3x+2y-12=0$

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

$4x-3y=6$

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

$3y+2=0$

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

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

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

$y-2=0$

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

$x-y=4$

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

## More About NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3:-

Class 11th Maths chapter 10 exercise 10.3 consists of questions related to changing the equations of lines in different forms, finding the slope of the line given the equation of the line, the distance between two parallel lines, the distance between a point and a line, equation of the line passing through a point and parallel to a line, coordinates of the foot of the perpendicular from the point to a line, etc.

Also Read| Straight Lines Class 11 Notes

## Benefits of NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3:-

• Most of the important concepts related to straight-line are covered in exercise 10.3 Class 11 Maths.

• Although this ex 10.3 class 11 is a bit lengthy, you must try to solve all the problems of Class 11 Maths chapter 10 exercise 10.3 by yourself.

• If you are not able to solve NCERT problems at first, Class 11 Maths chapter 10 exercise 10.3 solutions are here to help you.

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## Key Features of Exercise 10.3 Class 11 Maths Solutions

1. Complete Exercise Coverage: 11th class maths exercise 10.3 answers encompass all exercises 10.3 problems in the Class 11 Mathematics textbook.

2. Step-by-Step Approach: Ex 10.3 class 11 solutions are presented in a clear, step-by-step format to aid understanding and application of problem-solving methods.

3. Clarity and Precision: Written with clarity and precision, ensuring easy comprehension of mathematical concepts and techniques.

4. Proper Mathematical Notation: Utilizes appropriate mathematical notations and terminology, fostering fluency in mathematical language.

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6. Free Accessibility: Class 11 ex 10.3 solutions are typically available free of charge, providing an accessible resource for self-study.

7. Supplementary Learning Tool: Class 11 Maths Chapter 10 exercise 10.3 Solutions serve as supplementary resources to complement classroom instruction and support exam preparation.

8. Homework and Practice: Students can use solutions to cross-verify work, practice problem-solving, and enhance overall math skills.

## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. Find the slope of line 3y + x = 1 ?

Given line 3y  + x = 1

3y = -x + 1

y = -x/3 + 1/3

Compare with y = mx + c

Slope of the line (m) = -1/3

2. Write the equation of line passing through (1,0) and slope is 1 ?

Equation of line with slope 1 =>  y = x  + c
Line pass through (1,0)
0 = 1 + c
c = -1
Equation of line =>  y = x - 1

3. Write equation of line passing through (0,0) and parallel to line y = 2x + 3 ?

Equation of line parallel to line (y = 2x + 3 ) => y = 2x + c

Line pass through (0,0)  =>   0 = 0 + c   =>  c = 0
Equation of line =>  y = 2x

4. Find the slope of line 2y = 3x -1 ?

Given line 2y = 3x -1

y  = 3x/2 - 1/2

Compare with y = mx + c

Slope (m) = 3/2

5. What is the weightage of straight line in the JEE Main ?

Chapter straight lines has 6.6% weightage in the JEE Main exam.

6. Do I need to buy NCERT solutions book for Class 11 Chemistry ?

No, You don't need to buy any NCERT solutions book for any class. Here you will get NCERT Solutions for Class 11 Chemistry.

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 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