Class 11 Maths Ncert Supplementary Exercise 10.4 Solutions Pdf

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines: In earlier classes, you have studied 2D coordinate geometry. This chapter is a continuation of the coordinate geometry to study the simplest geometric figure – a straight line. The word 'straight' means without a bend or not curved. A straight line is a line which is not bent or curved. In this article, you get NCERT solutions for class 11 maths chapter 10 straight lines. A straight line is the simplest figure in the geometry but it is the most important concept of geometry. Important topics like 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 are covered in this chapter. In this chapter, there are 3 exercises with 52 questions. All these questions are explained in NCERT solutions for class 11 maths chapter 10 straight lines. 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. You can take help of NCERT solutions for class 11 maths chapter 10 straight lines which are prepared in a detailed manner. Check all NCERT solutions from class 6 to 12 at a single place to learn science and maths . Here you will get NCERT solutions for class 11 also.

Topics of NCERT Grade 11 Maths Chapter-10 Straight Lines

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

The complete Solutions of NCERT Class 11 Mathematics Chapter 10 is provided below:

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

Answer:

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

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

Answer:

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

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

Answer:

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

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

The slope of BC =

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

The slope of CD =

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

The slope of AD

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

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

Answer:

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

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

Answer:

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

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

Answer:

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

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

Answer:

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

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

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

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

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

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

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

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

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

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

Answer:

exercise 10.3
Let suppose foot of perpendicular is $(x_1,y_1)$
We can say that line passing through point $(x_1,y_1) \ and \ A(2,3)$ is perpendicular to line passing through point $B(4,-1) \ and \ C(1,2)$
Now,
Slope of line passing through point $B(4,-1) \ and \ C(1,2)$ is , $m' = \frac{2+1}{1-4}= \frac{3}{-3}=-1$
And
Slope of line passing through point $(x_1,y_1) \ and \ (2,3)$ is , $m$
lines are perpendicular
Therefore,
$m = -\frac{1}{m'}= 1$
Now, equation of line passing through point (2 ,3) and slope with 1
$(y-3)=1(x-2)$
$x-y+1=0$ -(i)
Now, equation line passing through point $B(4,-1) \ and \ C(1,2)$ is
$(y-2)=-1(x-1)$
$x+y-3=0$
Now, perpendicular distance of (2,3) from the $x+y-3=0$ is
$d= \left | \frac{1\times2+1\times3-3}{\sqrt{1^2+1^2}} \right |= \left | \frac{2+3-3}{\sqrt{1+1}} \right |= \frac{2}{\sqrt{2}}=\sqrt2$ -(ii)

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

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

Answer:

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

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

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

NCERT solutions for class 11 mathematics

NCERT solutions for class 11- Subject wise

Important points to remember from NCERT solutions for class 11 maths chapter 10 straight lines-

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

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

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

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.

Happy Reading !!!

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

Question: What are the important topics of the chapter Straight Lines ?

Answer:

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.

Question: What is the weightage of the chapter sequence and series in Jee Main ?

Answer:

This chapter has 4 % weightage in the Jee main exam.

Question: Does CBSE provides the solutions of NCERT class 11 maths ?

Answer:

No, CBSE doesn't provide NCERT solutions for any class or subject.

Question: Where can I find the complete solutions of NCERT for class 11 maths ?

Question: How many chapters are there in CBSE class 11 maths ?

Answer:

There are 16 chapters starting from set to probability in CBSE class 11 maths.

Question: Which is the official website of NCERT ?

Answer:

http://ncert.nic.in/ is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

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NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

Class 11 Maths Ncert Supplementary Exercise 10.4 Solutions Pdf

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Class 11 Maths Ncert Supplementary Exercise 10.4 Solutions Pdf

Frequently Asked Question (FAQs) - NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

Question: What are the important topics of the chapter Straight Lines ?

Question: What is the weightage of the chapter sequence and series in Jee Main ?

Question: Does CBSE provides the solutions of NCERT class 11 maths ?

Question: Where can I find the complete solutions of NCERT for class 11 maths ?

Question: How many chapters are there in CBSE class 11 maths ?

Question: Which is the official website of NCERT ?

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