NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 11 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 11 - Straight Lines

Edited By Vishal kumar | Updated on Nov 17, 2023 09:12 AM IST

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Miscellaneous Exercise- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 10 : Straight Lines Miscellaneous Exercise- In this chapter, you have already learned about the slope of the line, parallel lines, perpendicular lines, angles between two lines, colinearity of three points, equations of a line in various forms, the distance of a point from a line, etc. In the Class 11 Maths Chapter 10 miscellaneous exercise solutions, you will get mixed equations from all the concepts given in this Class 11 NCERT syllabus chapter. If you are done with previous exercises, you can start solving problems from the miscellaneous exercise.

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Miscellaneous exercise chapter 10 Class 11 is considered to be tougher than the other exercises of this chapter, so you may not able to solve them by yourself at first. NCERT solutions for Class 11 Maths chapter 10 miscellaneous exercise are here to help you to solve them very easily. Questions from the miscellaneous exercises are not much asked in the CBSE exam but these exercises are very important for engineering entrance exams like JEE Main, SRMJEE, etc. The class 11 maths ch 10 miscellaneous exercise solutions are meticulously crafted by highly qualified subject experts from Careers360. These solutions offer detailed and comprehensive explanations, ensuring a thorough understanding of the concepts. Additionally, the availability of a PDF version allows students the flexibility to access the class 11 chapter 10 maths miscellaneous solutions offline, anytime and anywhere, free of charge. This resource is designed to facilitate convenient and effective learning for students studying Straight Lines in Class 11 Mathematics. If you are looking for NCERT Solutions, you can click on the above link where you will NCERT solutions from Classes 6 to 12 at one place.

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**In the CBSE Syllabus for 2023-24, this miscellaneous exercise class 11 chapter 10 has been renumbered as Chapter 9.

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

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Access Straight Lines Class 11 Chapter 10 Miscellaneous Exercise

Question:1(a) Find the values of k for which the line (k-3)x-(4-k^2)y+k^2-7k+6=0 is

Parallel to the x-axis.

Answer:

Given equation of line is
(k-3)x-(4-k^2)y+k^2-7k+6=0
and equation of x-axis is y=0
it is given that these two lines are parallel to each other
Therefore, their slopes are equal
Slope of y=0 is , m' = 0
and
Slope of line (k-3)x-(4-k^2)y+k^2-7k+6=0 is , m = \frac{k-3}{4-k^2}
Now,
m=m'
\frac{k-3}{4-k^2}=0
k-3=0
k=3
Therefore, value of k is 3

Question:1(b) Find the values of k for which the line (k-3)x-(4-k^2)y+k^2-7k+6=0 is

Parallel to the y-axis.

Answer:

Given equation of line is
(k-3)x-(4-k^2)y+k^2-7k+6=0
and equation of y-axis is x = 0
it is given that these two lines are parallel to each other
Therefore, their slopes are equal
Slope of y=0 is , m' = \infty = \frac{1}{0}
and
Slope of line (k-3)x-(4-k^2)y+k^2-7k+6=0 is , m = \frac{k-3}{4-k^2}
Now,
m=m'
\frac{k-3}{4-k^2}=\frac{1}{0}
4-k^2=0
k=\pm2
Therefore, value of k is \pm2

Question:1(c) Find the values of k for which the line (k-3)x-(4-k^2)y+k^2-7k+6=0 is Passing through the origin.

Answer:

Given equation of line is
(k-3)x-(4-k^2)y+k^2-7k+6=0
It is given that it passes through origin (0,0)
Therefore,
(k-3).0-(4-k^2).0+k^2-7k+6=0
k^2-7k+6=0
k^2-6k-k+6=0
(k-6)(k-1)=0
k = 6 \ or \ 1
Therefore, value of k is 6 \ or \ 1

Question:2 Find the values of \small \theta and \small p, if the equation \small x\cos \theta +y\sin \theta =p is the normal form of the line \small \sqrt{3}x+y+2=0.

Answer:

The normal form of the line is \small x\cos \theta +y\sin \theta =p
Given the equation of lines is
\small \sqrt{3}x+y+2=0
First, we need to convert it into normal form. So, divide both the sides by \small \sqrt{(\sqrt3)^2+1^2}= \sqrt{3+1}= \sqrt4=2
\small -\frac{\sqrt3\cos \theta}{2}-\frac{y}{2}= 1
On comparing both
we will get
\small \cos \theta = -\frac{\sqrt3}{2}, \sin \theta = -\frac{1}{2} \ and \ p = 1
\small \theta = \frac{7\pi}{6} \ and \ p =1
Therefore, the answer is \small \theta = \frac{7\pi}{6} \ and \ p =1

Question:3 Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are \small 1 and \small -6, respectively.

Answer:

Let the intercepts on x and y-axis are a and b respectively
It is given that
a+b = 1 \ \ and \ \ a.b = -6
a= 1-b
\Rightarrow b.(1-b)=-6
\Rightarrow b-b^2=-6
\Rightarrow b^2-b-6=0
\Rightarrow b^2-3b+2b-6=0
\Rightarrow (b+2)(b-3)=0
\Rightarrow b = -2 \ and \ 3
Now, when b=-2\Rightarrow a=3
and when b=3\Rightarrow a=-2
We know that the intercept form of the line is
\frac{x}{a}+\frac{y}{b}=1

Case (i) when a = 3 and b = -2
\frac{x}{3}+\frac{y}{-2}=1
\Rightarrow 2x-3y=6

Case (ii) when a = -2 and b = 3
\frac{x}{-2}+\frac{y}{3}=1
\Rightarrow -3x+2y=6
Therefore, equations of lines are 2x-3y=6 \ and \ -3x+2y=6

Question:4 What are the points on the \small y-axis whose distance from the line \small \frac{x}{3}+\frac{y}{4}=1 is \small 4 units.

Answer:

Given the equation of the line is
\small \frac{x}{3}+\frac{y}{4}=1
we can rewrite it as
4x+3y=12
Let's take point on y-axis is (0,y)
It is given that the distance of the point (0,y) from line 4x+3y=12 is 4 units
Now,
d= \left | \frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}} \right |
In this problem A = 4 , B=3 , C =-12 ,d=4\ \ and \ \ (x_1,y_1) = (0,y)
4 = \left | \frac{4\times 0+3\times y-12}{\sqrt{4^2+3^2}} \right |=\left | \frac{3y-12}{\sqrt{16+9}} \right |=\left | \frac{3y-12}{5} \right |

Case (i)

4 = \frac{3y-12}{5}
20=3y-12
y = \frac{32}{3}
Therefore, the point is \left ( 0,\frac{32}{3} \right ) -(i)

Case (ii)

4=-\left ( \frac{3y-12}{5} \right )
20=-3y+12
y = -\frac{8}{3}
Therefore, the point is \left ( 0,-\frac{8}{3} \right ) -(ii)
Therefore, points on the \small y-axis whose distance from the line \small \frac{x}{3}+\frac{y}{4}=1 is \small 4 units are \left ( 0,\frac{32}{3} \right ) and \left ( 0,-\frac{8}{3} \right )

Question:5 Find perpendicular distance from the origin to the line joining the points (\cos \theta ,\sin \theta ) and (\cos \phi ,\sin \phi )..

Answer:

Equation of line passing through the points (\cos \theta ,\sin \theta ) and (\cos \phi ,\sin \phi ) is
(y-\sin \theta )= \frac{\sin \phi -\sin \theta}{\cos \phi -\cos \theta}(x-\cos\theta)
\Rightarrow (\cos \phi -\cos \theta)(y-\sin \theta )= (\sin \phi -\sin \theta)(x-\cos\theta)
\Rightarrow y(\cos \phi -\cos \theta)-\sin \theta(\cos \phi -\cos \theta)=x (\sin \phi -\sin \theta)-\cos\theta(\sin \phi -\sin \theta)
\Rightarrow x (\sin \phi -\sin \theta)-y(\cos \phi -\cos \theta)=\cos\theta(\sin \phi -\sin \theta)-\sin \theta(\cos \phi -\cos \theta)\Rightarrow x (\sin \phi -\sin \theta)-y(\cos \phi -\cos \theta)=\sin(\theta-\phi)
(\because \cos a\sin b -\sin a\cos b = \sin(a-b) )
Now, distance from origin(0,0) is
d = \left | \frac{(\sin\phi -\sin\theta).0-(\cos\phi-\cos\theta).0-\sin(\theta-\phi)}{\sqrt{(\sin\phi-\sin\theta)^2+(\cos\phi-\cos\theta)^2}} \right |
d = \left | \frac{-\sin(\theta-\phi)}{\sqrt{(\sin^2\phi+\cos^2\phi)+(\sin^2\theta+\cos^2\theta)-2(\cos\theta\cos\phi+\sin\theta\sin\phi)}} \right |
d = \left | \frac{-\sin(\theta-\phi)}{1+1-2\cos(\theta-\phi)} \right | (\because \cos a\cos b +\sin a\sin b = \cos(a-b) \ \ and \ \ \sin^2a+\cos^2a=1)
d = \left |\frac{ - \sin(\theta-\phi)}{2(1-\cos(\theta-\phi))} \right |
d = \left | \frac{-2\sin\frac{\theta-\phi}{2}\cos \frac{\theta-\phi}{2}}{\sqrt{2.(2\\sin^2\frac{\theta-\phi}{2})}}\right |
d = \left | \frac{-2\sin\frac{\theta-\phi}{2}\cos \frac{\theta-\phi}{2}}{2\sin\frac{\theta-\phi}{2}}\right |
d = \left | \cos\frac{\theta-\phi}{2} \right |

Question:6 Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines \small x-7y+5=0 and \small 3x+y=0.

Answer:

Point of intersection of the lines \small x-7y+5=0 and \small 3x+y=0
\left ( -\frac{5}{22},\frac{15}{22} \right )
It is given that this line is parallel to y - axis i.e. x=0 which means their slopes are equal
Slope of x=0 is ,m' = \infty = \frac{1}{0}
Let the Slope of line passing through point \left ( -\frac{5}{22},\frac{15}{22} \right ) is m
Then,
m=m'= \frac{1}{0}
Now, equation of line passing through point \left ( -\frac{5}{22},\frac{15}{22} \right ) and with slope \frac{1}{0} is
(y-\frac{15}{22})= \frac{1}{0}(x+\frac{5}{22})
x = -\frac{5}{22}
Therefore, equation of line is x = -\frac{5}{22}

Question:7 Find the equation of a line drawn perpendicular to the line \small \frac{x}{4}+\frac{y}{6}=1 through the point, where it meets the \small y-axis.

Answer:

given equation of line is
\small \frac{x}{4}+\frac{y}{6}=1
we can rewrite it as
3x+2y=12
Slope of line 3x+2y=12 , m' = -\frac{3}{2}
Let the Slope of perpendicular line is m
m = -\frac{1}{m'}= \frac{2}{3}
Now, the ponit of intersection of 3x+2y=12 and x =0 is (0,6)
Equation of line passing through point (0,6) and with slope \frac{2}{3} is
(y-6)= \frac{2}{3}(x-0)
3(y-6)= 2x
2x-3y+18=0
Therefore, equation of line is 2x-3y+18=0

Question:8 Find the area of the triangle formed by the lines \small y-x=0,x+y=0 and \small x-k=0.

Answer:

Given equations of lines are
y-x=0 \ \ \ \ \ \ \ \ \ \ \ -(i)
x+y=0 \ \ \ \ \ \ \ \ \ \ \ -(ii)
x-k=0 \ \ \ \ \ \ \ \ \ \ \ -(iii)
The point if intersection of (i) and (ii) is (0,0)
The point if intersection of (ii) and (iii) is (k,-k)
The point if intersection of (i) and (iii) is (k,k)
Therefore, the vertices of triangle formed by three lines are (0,0), (k,-k) \ and \ (k,k)
Now, we know that area of triangle whose vertices are (x_1,y_1),(x_2,y_2) \ and \ (x_3,y_3) is
A = \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|
A= \frac{1}{2}|0(-k-k)+k(k-0)+k(0+k)|
A= \frac{1}{2}|k^2+k^2|
A= \frac{1}{2}|2k^2|
A= k^2
Therefore, area of triangle is k^2 \ square \ units

Question:9 Find the value of \small p so that the three lines \small 3x+y-2=0,px+2y-3=0 and \small 2x-y-3=0 may intersect at one point.

Answer:

Point of intersection of lines \small 3x+y-2=0 and \small 2x-y-3=0 is (1,-1)
Now, (1,-1) must satisfy equation px+2y-3=0
Therefore,
p(1)+2(-1)-3=0
p-2-3=0
p=5
Therefore, the value of p is 5

Question:10 If three lines whose equations are y=m_1x+c_1,y=m_2x+c_2 and y=m_3x+c_3 are concurrent, then show that m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0.

Answer:

Concurrent lines means they all intersect at the same point
Now, given equation of lines are
y=m_1x+c_1 \ \ \ \ \ \ \ \ \ \ \ -(i)
y=m_2x+c_2 \ \ \ \ \ \ \ \ \ \ \ -(ii)
y=m_3x+c_3 \ \ \ \ \ \ \ \ \ \ \ -(iii)
Point of intersection of equation (i) and (ii) \left ( \frac{c_2-c_1}{m_1-m_2},\frac{m_1c_2-m_2c_1}{m_1-m_2} \right )

Now, lines are concurrent which means point \left ( \frac{c_2-c_1}{m_1-m_2},\frac{m_1c_2-m_2c_1}{m_1-m_2} \right ) also satisfy equation (iii)
Therefore,

\frac{m_1c_2-m_2c_1}{m_1-m_2}=m_3.\left ( \frac{c_2-c_1}{m_1-m_2} \right )+c_3

m_1c_2-m_2c_1= m_3(c_2-c_1)+c_3(m_1-m_2)

m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0

Hence proved

Question:11 Find the equation of the lines through the point \small (3,2) which make an angle of \small 45^{\circ} with the line \small x-2y=3.

Answer:

Given the equation of the line is
\small x-2y=3
The slope of line \small x-2y=3 , m_2= \frac{1}{2}
Let the slope of the other line is, m_1=m
Now, it is given that both the lines make an angle \small 45^{\circ} with each other
Therefore,
\tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |
\tan 45\degree = \left | \frac{\frac{1}{2}-m}{1+\frac{m}{2}} \right |
1= \left | \frac{1-2m}{2+m} \right |
Now,

Case (i)
1=\frac{1-2m}{2+m}
2+m=1-2m
m = -\frac{1}{3}
Equation of line passing through the point \small (3,2) and with slope -\frac{1}{3}
(y-2)=-\frac{1}{3}(x-3)
3(y-2)=-1(x-3)
x+3y=9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)

Case (ii)
1=-\left ( \frac{1-2m}{2+m} \right )
2+m=-(1-2m)
m= 3
Equation of line passing through the point \small (3,2) and with slope 3 is
(y-2)=3(x-3)
3x-y=7 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)

Therefore, equations of lines are 3x-y=7 and x+3y=9

Question:12 Find the equation of the line passing through the point of intersection of the lines 4x+7y-3=0 and 2x-3y+1=0 that has equal intercepts on the axes.

Answer:

Point of intersection of the lines 4x+7y-3=0 and 2x-3y+1=0 is \left ( \frac{1}{13},\frac{5}{13} \right )
We know that the intercept form of the line is
\frac{x}{a}+\frac{y}{b}= 1
It is given that line make equal intercepts on x and y axis
Therefore,
a = b
Now, the equation reduces to
x+y = a -(i)
It passes through point \left ( \frac{1}{13},\frac{5}{13} \right )
Therefore,
a = \frac{1}{13}+\frac{5}{13}= \frac{6}{13}
Put the value of a in equation (i)
we will get
13x+13y=6
Therefore, equation of line is 13x+13y=6

Question:13 Show that the equation of the line passing through the origin and making an angle \small \theta with the line \small y=mx+c is \small \frac{y}{x}=\frac{m\pm \tan \theta }{1\mp m\tan \theta }.

Answer:

Slope of line \small y=mx+c is m
Let the slope of other line is m'
It is given that both the line makes an angle \small \theta with each other
Therefore,
\tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |
\tan \theta = \left | \frac{m-m'}{1+mm'} \right |
\mp(1+mm')\tan \theta =(m-m')
\mp\tan \theta +m'(\mp m\tan\theta+1)= m
m'= \frac{m\pm \tan \theta}{1\mp m\tan \theta}
Now, equation of line passing through origin (0,0) and with slope \frac{m\pm \tan \theta}{1\mp m\tan \theta} is
(y-0)=\frac{m\pm \tan \theta}{1\mp m\tan \theta}(x-0)
\frac{y}{x}=\frac{m\pm \tan \theta}{1\mp m\tan \theta}
Hence proved

Question:14 In what ratio, the line joining \small (-1,1) and \small (5,7) is divided by the line x+y=4 ?

Answer:

Equation of line joining \small (-1,1) and \small (5,7) is
(y-1)= \frac{7-1}{5+1}(x+1)
\Rightarrow (y-1)= \frac{6}{6}(x+1)
\Rightarrow (y-1)= 1(x+1)
\Rightarrow x-y+2=0
Now, point of intersection of lines x+y=4 and x-y+2=0 is (1,3)
Now, let's suppose point (1,3)divides the line segment joining \small (-1,1) and \small (5,7) in 1:k
Then,
(1,3)= \left ( \frac{k(-1)+1(5)}{k+1},\frac{k(1)+1(7)}{k+1} \right )
1=\frac{-k+5}{k+1} \ \ and \ \ 3 = \frac{k+7}{k+1}
\Rightarrow k =2
Therefore, the line joining \small (-1,1) and \small (5,7) is divided by the line x+y=4 in ratio 1:2

Question:15 Find the distance of the line \small 4x+7y+5=0 from the point \small (1,2) along the line \small 2x-y=0.

Answer:

22229
point \small (1,2) lies on line 2x-y =0
Now, point of intersection of lines 2x-y =0 and \small 4x+7y+5=0 is \left ( -\frac{5}{18},-\frac{5}{9} \right )
Now, we know that the distance between two point is given by
d = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|
d = |\sqrt{(1+\frac{5}{18})^2+(2+\frac{5}{9})^2}|
d = |\sqrt{(\frac{23}{18})^2+(\frac{23}{9})^2}|
d = \left | \sqrt{\frac{529}{324}+\frac{529}{81}} \right |
d = \left | \sqrt{\frac{529+2116}{324}} \right | = \left | \sqrt\frac{2645}{324} \right | =\frac{23\sqrt5}{18}
Therefore, the distance of the line \small 4x+7y+5=0 from the point \small (1,2) along the line \small 2x-y=0 is \frac{23\sqrt5}{18} \ units

Question:16 Find the direction in which a straight line must be drawn through the point \small (-1,2) so that its point of intersection with the line \small x+y=4 may be at a distance of 3 units from this point.

Answer:

Let (x_1,y_1) be the point of intersection
it lies on line \small x+y=4
Therefore,
x_1+y_1=4 \\ x_1=4-y_1\ \ \ \ \ \ \ \ \ \ \ -(i)
Distance of point (x_1,y_1) from \small (-1,2) is 3
Therefore,
3= |\sqrt{(x_1+1)^2+(y_1-2)^2}|
Square both the sides and put value from equation (i)
9= (5-y_1)^2+(y_1-2)^2\\ 9=y_1^2+25-10y_1+y_1^2+4-4y_1\\ 2y_1^2-14y_1+20=0\\ y_1^2-7y_1+10=0\\ y_1^2-5y_1-2y_1+10=0\\ (y_1-2)(y_1-5)=0\\ y_1=2 \ or \ y_1 = 5
When y_1 = 2 \Rightarrow x_1 = 2 point is (2,2)
and
When y_1 = 5 \Rightarrow x_1 = -1 point is (-1,5)
Now, slope of line joining point (2,2) and \small (-1,2) is
m = \frac{2-2}{-1-2}=0
Therefore, line is parallel to x-axis -(i)

or
slope of line joining point (-1,5) and \small (-1,2)
m = \frac{5-2}{-1+2}=\infty
Therefore, line is parallel to y-axis -(ii)

Therefore, line is parallel to x -axis or parallel to y-axis

Question:17 The hypotenuse of a right angled triangle has its ends at the points \small (1,3) and \small (-4,1) Find an equation of the legs (perpendicular sides) of the triangle.

Answer:

22234
Slope of line OA and OB are negative times inverse of each other
Slope of line OA is , m=\frac{3-y}{1-x}\Rightarrow (3-y)=m(1-x)
Slope of line OB is , -\frac{1}{m}= \frac{1-y}{-4-x}\Rightarrow (x+4)=m(1-y)
Now,

Now, for a given value of m we get these equations
If m = \infty
1-x=0 \ \ \ \ and \ \ \ \ \ 1-y =0
x=1 \ \ \ \ and \ \ \ \ \ y =1

Question:18 Find the image of the point \small (3,8) with respect to the line x+3y=7 assuming the line to be a plane mirror.

Answer:

Ch. 10
Let point (a,b) is the image of point \small (3,8) w.r.t. to line x+3y=7
line x+3y=7 is perpendicular bisector of line joining points \small (3,8) and (a,b)
Slope of line x+3y=7 , m' = -\frac{1}{3}
Slope of line joining points \small (3,8) and (a,b) is , m = \frac{8-b}{3-a}
Now,
m = -\frac{1}{m'} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because lines \ are \ perpendicular)
\frac{8-b}{3-a}= 3
8-b=9-3a
3a-b=1 \ \ \ \ \ \ \ \ \ \ \ -(i)
Point of intersection is the midpoint of line joining points \small (3,8) and (a,b)
Therefore,
Point of intersection is \left ( \frac{3+a}{2},\frac{b+8}{2} \right )
Point \left ( \frac{3+a}{2},\frac{b+8}{2} \right ) also satisfy the line x+3y=7
Therefore,
\frac{3+a}{2}+3.\frac{b+8}{2}=7
a+3b=-13 \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
On solving equation (i) and (ii) we will get
(a,b) = (-1,-4)
Therefore, the image of the point \small (3,8) with respect to the line x+3y=7 is (-1,-4)

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

Answer:

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

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

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

Question:20 If the sum of the perpendicular distances of a variable point \small P(x,y) from the lines \small x+y-5=0 and \small 3x-2y+7=0 is always \small 10. Show that \small P must move on a line.

Answer:

Given the equation of line are
x+y-5=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)
3x-2y+7=0 \ \ \ \ \ \ \ \ \ \ \ -(ii)
Now, perpendicular distances of a variable point \small P(x,y) from the lines are

d_1=\left | \frac{1.x+1.y-5}{\sqrt{1^2+1^2}} \right | d_2=\left | \frac{3.x-2.y+7}{\sqrt{3^2+2^2}} \right |
d_1=\left | \frac{x+y-5}{\sqrt2} \right | d_2=\left | \frac{3x-2y+7}{\sqrt{13}} \right |
Now, it is given that
d_1+d_2= 10
Therefore,
\frac{x+y-5}{\sqrt2}+\frac{3x-2y+7}{\sqrt{13}}=10
(assuming \ x+y-5 > 0 \ and \ 3x-2y+7 >0)
(x+y-5)\sqrt{13}+(3x-2y+7)\sqrt2=10\sqrt{26}

x(\sqrt{13}+3\sqrt{2})+y(\sqrt{13}-2\sqrt{2})=10\sqrt{26}+5\sqrt{13}-7\sqrt2

Which is the equation of the line
Hence proved

Question:21 Find equation of the line which is equidistant from parallel lines 9x+6y-7=0 and 3x+2y+6=0.

Answer:

Let's take the point p(a,b) which is equidistance from parallel lines 9x+6y-7=0 and 3x+2y+6=0
Therefore,
d_1= \left | \frac{9.a+6.b-7}{\sqrt{9^2+6^2}} \right | d_2= \left | \frac{3.a+2.b+6}{\sqrt{3^2+2^2}} \right |
d_1= \left | \frac{9a+6b-7}{\sqrt{117}} \right | d_2= \left | \frac{3a+2b+6}{\sqrt{13}} \right |
It is that d_1=d_2
Therefore,
\left | \frac{9a+6b-7}{3\sqrt{13}} \right |= \left | \frac{3a+2b+6}{\sqrt{13}} \right |
(9a+6b-7)=\pm 3(3a+2b+6)
Now, case (i)
(9a+6b-7)= 3(3a+2b+6)
25=0
Therefore, this case is not possible

Case (ii)
(9a+6b-7)= -3(3a+2b+6)
18a+12b+11=0

Therefore, the required equation of the line is 18a+12b+11=0

Question:22 A ray of light passing through the point (1,2) reflects on the x-axis at point A and the reflected ray passes through the point (5,3). Find the coordinates of A.

Answer:

22246
From the figure above we can say that
The slope of line AC (m)= \tan \theta
Therefore,
\tan \theta = \frac{3-0}{5-a} = \frac{3}{5-a} \ \ \ \ \ \ \ \ \ \ (i)
Similarly,
The slope of line AB (m') = \tan(180\degree-\theta)
Therefore,
\tan(180\degree-\theta) = \frac{2-0}{1-a}
-\tan\theta= \frac{2}{1-a}
\tan\theta= \frac{2}{a-1} \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
Now, from equation (i) and (ii) we will get
\frac{3}{5-a} = \frac{2}{a-1}
\Rightarrow 3(a-1)= 2(5-a)
\Rightarrow 3a-3= 10-2a
\Rightarrow 5a=13
\Rightarrow a=\frac{13}{5}
Therefore, the coordinates of A. is \left ( \frac{13}{5},0 \right )

Question:23 Prove that the product of the lengths of the perpendiculars drawn from the points \small (\sqrt{a^2-b^2},0) and \small (-\sqrt{a^2-b^2},0) to the line \small \frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1 is \small b^2.

Answer:

Given equation id line is
\small \frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1
We can rewrite it as
xb\cos \theta +ya\sin \theta =ab
Now, the distance of the line xb\cos \theta +ya\sin \theta =ab from the point \small (\sqrt{a^2-b^2},0) is given by
d_1=\left | \frac{b\cos\theta.\sqrt{a^2-b^2}+a\sin \theta.0-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right | = \left | \frac{b\cos\theta.\sqrt{a^2-b^2}-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |
Similarly,
The distance of the line xb\cos \theta +ya\sin \theta =ab from the point \small (-\sqrt{a^2-b^2},0) is given by
d_2=\left | \frac{b\cos\theta.(-\sqrt{a^2-b^2})+a\sin \theta.0-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right | = \left | \frac{-(b\cos\theta.\sqrt{a^2-b^2}+ab)}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |
d_1.d_2 = \left | \frac{b\cos\theta.\sqrt{a^2-b^2}-ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |.\times\left | \frac{-(b\cos\theta.\sqrt{a^2-b^2}+ab)}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}} \right |
=\left | \frac{-((b\cos\theta.\sqrt{a^2-b^2})^2-(ab)^2)}{(b\cos\theta)^2+(a\sin\theta)^2} \right |
=\left | \frac{-b^2\cos^2\theta.(a^2-b^2)+a^2b^2)}{(b\cos\theta)^2+(a\sin\theta)^2} \right |
=\left | \frac{-a^2b^2\cos^2\theta+b^4\cos^2\theta+a^2b^2)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |
=\left | \frac{-b^2(a^2\cos^2\theta-b^2\cos^2\theta-a^2)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |
=\left | \frac{-b^2(a^2\cos^2\theta-b^2\cos^2\theta-a^2(\sin^2\theta+\cos^2\theta))}{b^2\cos^2\theta+a^2\sin^2\theta} \right | \ \ \ \ (\because \sin^2a+\cos^2a=1)
=\left | \frac{-b^2(a^2\cos^2\theta-b^2\cos^2\theta-a^2\sin^2\theta-a^2\cos^2\theta)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |
=\left | \frac{+b^2(b^2\cos^2\theta+a^2\sin^2\theta)}{b^2\cos^2\theta+a^2\sin^2\theta} \right |
=b^2
Hence proved

Question:24 A person standing at the junction (crossing) of two straight paths represented by the equations \small 2x-3y+4=0 and \small 3x+4y-5=0 wants to reach the path whose equation is \small 6x-7y+8=0 in the least time. Find equation of the path that he should follow.

Answer:

point of intersection of lines \small 2x-3y+4=0 and \small 3x+4y-5=0 (junction) is \left ( -\frac{1}{17},\frac{22}{17} \right )
Now, person reaches to path \small 6x-7y+8=0 in least time when it follow the path perpendicular to it
Now,
Slope of line \small 6x-7y+8=0 is , m'=\frac{6}{7}
let the slope of line perpendicular to it is , m
Then,
m= -\frac{1}{m}= -\frac{7}{6}
Now, equation of line passing through point \left ( -\frac{1}{17},\frac{22}{17} \right ) and with slope -\frac{7}{6} is
\left ( y-\frac{22}{17} \right )= -\frac{7}{6}\left ( x-(-\frac{1}{17}) \right )
\Rightarrow 6(17y-22)=-7(17x+1)
\Rightarrow 102y-132=-119x-7
\Rightarrow 119x+102y=125

Therefore, the required equation of line is 119x+102y=125

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

As the name suggests miscellaneous exercise chapter 10 class 11 contains the mixed kind of questions from all the concepts you have learned in the previous exercises of this chapter. There are six solved examples given in the NCERT textbook before this exercise. You must go through these solved examples before solving this exercise. You must solve problems from this exercise to get conceptual clarity. Also, it will check your understanding of the concepts.

Topic Covered in class 11 Maths ch 10 Miscellaneous Exercise Solutions

The Miscellaneous Exercise in NCERT Solutions for Class 11 Maths Chapter 10 - Straight Lines covers the following topics:

  1. Slope of a Line
  2. Various Forms of the Equation of a Line
  3. General Equation of a Line
  4. Distance of a Point From a Line
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Mathematics, being a subject that benefits from consistent practice, requires a thorough understanding of the best problem-solving methods. By practising with the help of these NCERT Solutions for Class 11 Maths, students can enhance their problem-solving skills, ultimately leading to the potential to score high marks in examinations.

Also Read| Straight Lines Class 11 Notes

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

  • Class 11 Maths chapter 10 miscellaneous exercise solutions are designed by subject matter experts based on the guideline given by CBSE.
  • Class 11 Maths chapter 10 miscellaneous solutions are solved in a step-by-step manner that could be understood by an average student also.

Key Features of Class 11 Chapter 10 Maths Miscellaneous Solutions

  1. Comprehensive Coverage: Solutions cover all themiscellaneous exercise class 11 chapter 10, addressing various aspects of Straight Lines.

  2. Step-by-Step Solutions: Each class 11 maths miscellaneous exercise chapter 10 problem is solved with a step-by-step approach, facilitating a clear and logical progression of concepts for Class 11 students.

  3. Clarity in Language: The class 11 chapter 10 miscellaneous exercise solutions are presented in clear and concise language, making complex geometric concepts associated with Straight Lines more accessible for students.

  4. PDF Version Access: The class 11 maths ch 10 miscellaneous exercise solutions are available in a downloadable PDF format, providing students with offline access for study anytime and anywhere.

Also see-

NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Questions (FAQs)

1. Write the equation of line passing through (5,4) and parallel to y-axis ?

Equation of line parallel to y-axis  =>  x = c

Line passing through the point (5,4)  =>  5 = c

Equation of line   =>  x = 5

2. Write the equation of line passing through (5,4) and parallel to x-axis ?

Equation of line parallel to x-axis  =>  y = c

Line passing through the point (5,4)  =>  4 = c

Equation of line   =>  y = 4

3. Write the equation of line passing through (1,0) and make angle of 45 degree with positive x-axis ?

Line make angle of 45 degree with positive x-axis  => slope of line = 1

Equation of line with slope 1 => y = x + c

Line pass through the (1,0)  =>  0 = 1 + c  =>  c=-1

Equation of line =>  y = x -1

4. Find the slope of line 4y = 3x - 6 ?

Given line 4y = 3x - 6

y = 3x/4 - 6/4

Compare with y = mx + c

Slope (m)  = 3/4

5. Do I need to buy NCERT solutions book for Class 11 Maths chapter 10 straight line ?

No, you don't need to buy the NCERT solutions book for Class 11 Maths chapter 10 straight line. Here you will get NCERT Solutions for Class 11 Maths Chapter 10.

6. Do I need to buy NCERT exemplar solutions book for Class 11 Maths chapter 10 straight line ?

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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)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

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)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

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

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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

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