NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

 

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 solutions of NCERT 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 CBSE 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.

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:

NCERT solutions for class 11 maths chapter 10 straight lines-Exercise: 10.1

Question:1 Draw a quadrilateral in the Cartesian plane, whose vertices are (-4,5),(0,7),(5,-5)  and  (-4,-2). Also, find its area. 

Answer:



Area of ABCD = Area of ABC + Area of ACD
Now, we know that the area of a triangle with vertices (x_1,y_1),(x_2,y_2) \ and \ (x_3,y_3)  is given by
A = \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|
Therefore,
Area of triangle ABC = \frac{1}{2}|-4(7+5)+0(-5-5)+5(5-7)|= \frac{1}{2}|-48-10|= \frac{58}{2}=29
Similarly,
Area of triangle ACD = \frac{1}{2}|-4(-5+2)+5(-2-5)+-4(5+5)|= \frac{1}{2}|12-35-40|= \frac{63}{2}
Now,
Area of ABCD = Area of ABC + Area of ACD
=\frac{121}{2} \ units

Question:2 The base of an equilateral triangle with side 2a  lies along the y-axis such that the mid-point of the base is at the origin.  Find vertices of the triangle. 

Answer:


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)

Question:3(i) Find the distance between P(x_1,y_1) and  Q(x_2,y_2)  when :

 PQ is parallel to the y-axis.

Answer:

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)|

Question:3(ii) Find the distance between  P(x_1,y_1)  and  Q(x_2,y_2)  when :

PQ is parallel to the x-axis. 

Answer:

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|

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:5 Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P(0,-4) and  B(8,0).

Answer:

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}

Question:6 Without using the Pythagoras theorem, show that the points  (4,4),(3,5)  and  (-1,-1),  are  the vertices of a right angled triangle.

Answer:

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

Question:7 Find the slope of the line, which makes an angle of  30^{\circ} with the positive direction of  y-axis measured anticlockwise.

Answer:

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

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:10 Find the angle between the x-axis and the line joining the points (3,-1) and  (4,-2)

Answer:

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

Question:11 The slope of a line is double of the slope of another line. If tangent of the angle between them is    \frac{1}{3},  find the slopes of the lines

Answer:

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

Question:12 A line passes through  (x_1,y_1)  and  (h,k) .  If slope of the line is m, show that k-y_1=m(h-x_1).

Answer:

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

Question:13 If three points  (h,0),(a,b)  and  (0,k)  lie on a line, show that   \frac{a}{h}+\frac{b}{k}=1.

Answer:

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 

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

NCERT solutions for class 11 maths chapter 10 straight lines-Exercise: 10.2

Question:1 Find the equation of the line which satisfy the given conditions: 

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

Answer:

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

Question:2 Find the equation of the line which satisfy the given conditions: 

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

Answer:

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

Question:3 Find the equation of the line which satisfy the given conditions: 

Passing through (0,0) with slope m.

Answer:

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

Question:4 Find the equation of the line which satisfy the given conditions: 

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

Answer:

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)

Question:5 Find the equation of the line which satisfy the given conditions: 

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

Answer:

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

Question:6 Find the equation of the line which satisfy the given conditions: 

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. 

Answer:

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

Question:7 Find the equation of the line which satisfy the given conditions: 

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

Answer:

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

Question:8 Find the equation of the line which satisfy the given conditions:

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

Answer:

 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

Question:9 The vertices of  \Delta \hspace{1mm}PQR are  P(2,1),Q(-2,3)  and   R(4,5). Find equation of the median through the vertex R

Answer:


 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

Question:10 Find the equation of the line passing through  (-3,5)  and perpendicular to the line through the points  (2,5)  and (-3,6).

Answer:

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 

Question:11  A line perpendicular to the line segment joining the points  (1,0)  and  (2,3)  divides it in the ratio 1:n. Find the equation of the line. 

Answer:

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

Question:12 Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point  (2,3).

Answer:

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

Question:13 Find equation of the line passing through the point  (2,2) and cutting off intercepts on the axes whose sum is 9.

Answer:

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

Question:14 Find equation of the line through the point  (0,2)  making an angle \frac{2\pi }{3} with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin

Answer:

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

Question:15 The perpendicular from the origin to a line meets it at the point (-2,9) , find the equation of the line. 

Answer:

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

Question:16 The length  L(in centimetre) of a copper rod is a linear function of its Celsius  temperature C.  In an experiment, if  L=124.942  when C=20 and  L=125.134 when C=110, express L in terms of C

Answer:

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

Question:17 The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs\hspace{1mm}14/litre  and  1220  litres of milk each week at  Rs\hspace{1mm}16/litre .  Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at  Rs\hspace{1mm}17/litre 

Answer:

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

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:19 Point  R(h,k)  divides a line segment between the axes in the ratio 1:2 .  Find  equation of the line.

Answer:


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

Question:20 By using the concept of equation of a line, prove that the three points  (3,0),(-2,-2) and  (8,2)  are collinear.

Answer:

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

CBSE NCERT solutions for class 11 maths chapter 10 straight lines-Exercise: 10.3

Question:1(i) Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts.

              x+7y=0

Answer:

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

Question:1(ii) Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts. 

   6x+3y-5=0

Answer:

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

Question:1(iii) Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts. 

        y=0.

Answer:

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

Question:2(i) Reduce the following equations into intercept form and find their intercepts on the axes. 

      3x+2y-12=0

Answer:

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

Question:2(ii)Reduce the following equations into intercept form and find their intercepts on the axes. 

  4x-3y=6

Answer:

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

Question:2(iii) Reduce the following equations into intercept form and find their intercepts on the axes. 

  3y+2=0

Answer:

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

Question:3(i) Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

    x-\sqrt{3}y+8=0

Answer:

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

Question:3(ii) Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

   y-2=0 

Answer:

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

Question:3(iii) Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

  x-y=4

Answer:

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

Question:4 Find the distance of the point  (-1,1)  from the line   12(x+6)=5(y-2).

Answer:

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

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:6(i) Find the distance between parallel lines 15x+8y-34=0  and  15x+8y+31=0.

Answer:

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

Question:6(ii) Find the distance between parallel lines  l(x+y)+p=0 and l(x+y)-r = 0

Answer:

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 |

Question:7 Find equation of the line parallel to the line  3x-4y+2=0  and passing through  the point (-2,3)

Answer:

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

Question:8 Find equation of the line perpendicular to the line x-7y+5=0 and having  xintercept 3.

Answer:

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  xintercept 3  i.e. (3, 0) and  with slope -7 is 
(y-0)=-7(x-3)
y = -7x+21
7x+y-21=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:10 The line through the points  (h,3)  and  (4,1)  intersects the line 7x-9y-19=0  at right angle. Find the value of  h

Answer:

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} 

Question:11 Prove that the line through the point  (x_1,y_1)  and parallel to the line   Ax+By+C=0   is   A(x-x_1)+B(y-y_1)=0.

Answer:

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 

Question:12 Two lines passing through the point  (2,3)  intersects each other at an angle of  60^{\circ}. If slope of one line is 2, find equation of the other line

Answer:

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

Question:13 Find the equation of the right bisector of the line segment joining the points  (3,4) and   (-1,2)

Answer:

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

Question:14 Find the coordinates of the foot of perpendicular from the point  (-1,3)  to the line 3x-4y-16=0.

Answer:

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 )

Question:15 The perpendicular from the origin to the line  y=mx+c  meets it at the point  (-1,2). Find the values of m and c. 

Answer:

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

Question:16 If p and q are the lengths of perpendiculars from the origin to the lines    x\cos \theta -y\sin \theta =k\cos 2\theta  and  x\sec \theta +y\hspace{1mm}cosec\hspace{1mm}\theta =k , respectively,  prove that  p^2+4q^2=k^2

Answer:

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

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:18 If  p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b,  then show that  \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}.

Answer:

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

NCERT solutions for class 11 maths chapter 10 straight lines-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:


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:


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:


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

NCERT solutions for class 11 mathematics

chapter-1

NCERT solutions for class 11 maths chapter 1 Sets

chapter-2

Solutions of NCERT for class 11 chapter 2 Relations and Functions

chapter-3

CBSE NCERT solutions for class 11 chapter 3 Trigonometric Functions

chapter-4

NCERT solutions for class 11 chapter 4 Principle of Mathematical Induction

chapter-5

Solutions of NCERT for class 11 chapter 5 Complex Numbers and Quadratic equations

chapter-6

CBSE NCERT solutions for class 11 maths chapter 6 Linear Inequalities

chapter-7

NCERT solutions for class 11 maths chapter 7 Permutation and Combinations

chapter-8

Solutions of NCERT for class 11 maths chapter 8 Binomial Theorem

chapter-9

CBSE NCERT solutions for class 11 maths chapter 9 Sequences and Series

chapter-10

NCERT solutions for class 11 maths chapter 10 Straight Lines

chapter-11

Solutions of NCERT for class 11 maths chapter 11 Conic Section

chapter-12

CBSE NCERT solutions for class 11 maths chapter 12 Introduction to Three Dimensional Geometry

chapter-13

NCERT solutions for class 11 maths chapter 13 Limits and Derivatives

chapter-14

Solutions of NCERT for class 11 maths chapter 14 Mathematical Reasoning

chapter-15

CBSE NCERT solutions for class 11 maths chapter 15 Statistics

chapter-16

NCERT solutions for class 11 maths chapter 16 Probability

NCERT solutions for class 11- Subject wise

Solutions of NCERT for class 11 biology

CBSE NCERT solutions for class 11 maths

NCERT solutions for class 11 chemistry

Solutions of NCERT for Class 11 physics

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

  • If a non-vertical line passing through the points (x_1,\:y_1) and  (x_2,\:y_2) then the slope(m) of the line is given by-

          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 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 (\theta) between lines L_1  and  L_2 with slopes m_1 and m_2 is given by-

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

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

         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 CBSE NCERT solutions for class 11 maths chapter 10 straight lines.

Happy Reading !!! 

 

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