# 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

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

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
$=\frac{121}{2} \ units$

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
$(0,a),(0,-a) \and \ (\sqrt3a,0) \ \ or \ \ (0,a),(0,-a) \and \ (-\sqrt3a,0)$

PQ is parallel to the $y$-axis.

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 $|(y_2-y_1)|$

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 $|x_2-x_1|$

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 $( \frac{15}{2},0)$

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  $\frac{-1}{2}$

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

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 $-\sqrt3$

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

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

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 $135\degree$

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, $m_1,m_2 = \frac{1}{2},1 \ or \ \frac{-1}{2},-1 \ or \ 1,2 \ or \ -1,-2$

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

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

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

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

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

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

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

We know that , equation of line passing through point $(x_1,y_1)$ and with slope m is given by
$(y-y_1)=m(x-x_1)$
Now,  equation of line passing through point (-4,3) and with slope $\frac{1}{2}$ is
$(y-3)=\frac{1}{2}(x-(-4))\\ 2y-6=x+4\\ x-2y+10 = 0$
Therefore, equation of the line  is   $x-2y+10 = 0$

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

We know that the equation of the line passing through the point $(x_1,y_1)$ and with slope m is given by
$(y-y_1)=m(x-x_1)$
Now, the equation of the line passing through the point (0,0) and with slope m is
$(y-0)=m(x-0)\\ y = mx$
Therefore, the equation of the line  is   $y = mx$

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

We know that the equation of the line passing through the point $(x_1,y_1)$ and with slope m is given by
$(y-y_1)=m(x-x_1)$
we know that
$m = \tan \theta$
where $\theta$ is angle made by line with positive x-axis measure in the anti-clockwise direction
$m = \tan75\degree \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \theta=75\degree \ given)$
$m = \frac{\sqrt3+1}{\sqrt3-1}$
Now, the equation of the line passing through the point $(2,2\sqrt3)$ and with slope $m = \frac{\sqrt3+1}{\sqrt3-1}$  is
$(y-2\sqrt3)=\frac{\sqrt3+1}{\sqrt3-1}(x-2)\\ \\ (\sqrt3-1)(y-2\sqrt3)=(\sqrt3+1)(x-2)\\ (\sqrt3-1)y-6+2\sqrt3= (\sqrt3+1)x-2\sqrt3-2\\ (\sqrt3+1)x-(\sqrt3-1)y = 4(\sqrt3-1)$
Therefore, the equation of the line  is   $(\sqrt3+1)x-(\sqrt3-1)y = 4(\sqrt3-1)$

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

We know that the equation of the line passing through the point $(x_1,y_1)$ and with slope m is given by
$(y-y_1)=m(x-x_1)$
Line Intersecting the $x$-axis at a distance of $3$ units to the left of origin which means the point is (-3,0)
Now, the equation of the line passing through the point (-3,0) and with slope -2  is
$(y-0)= -2(x-(-3))\\ y = -2x-6\\ 2x+y+6=0$
Therefore, the equation of the line  is   $2x+y+6=0$

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

We know that , equation of line passing through point $(x_1,y_1)$ and with slope m is given by
$(y-y_1)=m(x-x_1)$
Line Intersecting the y-axis at a distance of 2 units above the origin which means point is (0,2)
we know that
$m = \tan \theta\\ m = \tan 30\degree \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \theta = 30 \degree \ given)\\ m = \frac{1}{\sqrt3}$
Now, the equation of the line passing through the point (0,2) and with slope $\frac{1}{\sqrt3}$  is
$(y-2)= \frac{1}{\sqrt3}(x-0)\\ \sqrt3(y-2)= x\\ x-\sqrt3y+2\sqrt3=0$
Therefore, the equation of the line  is   $x-\sqrt3y+2\sqrt3=0$

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

We know that , equation of line passing through point $(x_1,y_1)$ and with slope m is given by
$(y-y_1)=m(x-x_1)$
Now, it is given that line passes throught point (-1 ,1) and (2 , -4)
$m = \frac{y_2-y_1}{x_2-x_1}\\ \\ m = \frac{-4-1}{2+1}= \frac{-5}{3}$
Now,  equation of line passing through point (-1,1) and with slope $\frac{-5}{3}$  is
$(y-1)= \frac{-5}{3}(x-(-1))\\ \\3(y-1)=-5(x+1)\\ 3y-3=-5x-5\\ 5x+3y+2=0$

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

It is given that length of perpendicular is 5 units  and  angle made by the  perpendicular with the positive $x$-axis is  $30^{\circ}$
Therefore, equation of line is
$x\cos \theta + y \sin \theta = p$
In this case p = 5 and  $\theta = 30\degree$
$x\cos 30\degree + y \sin 30\degree = 5\\ x.\frac{\sqrt3}{2}+\frac{y}{2}= 5\\ \sqrt3x+y =10$
Therefore, equation of the line  is   $\sqrt3x+y =10$

The vertices of  $\Delta \hspace{1mm}PQR$ are  $P(2,1),Q(-2,3)$  and   $R(4,5)$
Let m be RM b the median through vertex R
Coordinates of M (x, y ) = $\left ( \frac{2-2}{2},\frac{1+3}{2} \right )= (0,2)$
Now, slope of line RM
$m = \frac{y_2-y_1}{x_2-x_1} = \frac{5-2}{4-0}= \frac{3}{4}$
Now, equation of line passing through point $(x_1,y_1)$ and with slope m is
$(y-y_1)= m(x-x_1)$
equation of line passing through point (0 , 2) and with slope $\frac{3}{4}$ is
$(y-2)= \frac{3}{4}(x-0)\\ \\ 4(y-2)=3x\\ 4y-8=3x\\ 3x-4y+8=0$
Therefore, equation of median is $3x-4y+8=0$

It is given that the line passing through  $(-3,5)$  and perpendicular to the line through the points  $(2,5)$  and $(-3,6)$
Let the slope of the line passing through the point (-3,5) is m and
Slope of line  passing through points (2,5) and (-3,6)
$m' = \frac{6-5}{-3-2}= \frac{1}{-5}$
Now this line is perpendicular to line passing through point (-3,5)
Therefore,
$m= -\frac{1}{m'} = -\frac{1}{\frac{1}{-5}}= 5$

Now, equation of line passing through point $(x_1,y_1)$ and with slope m is
$(y-y_1)= m(x-x_1)$
equation of line passing through point (-3 , 5) and with slope  5 is
$(y-5)= 5(x-(-3))\\ \\ (y-5)=5(x+3)\\ y-5=5x+15\\ 5x-y+20=0$
Therefore, equation of line  is $5x-y+20=0$

Co-ordinates of point which divide line segment joining the points  $(1,0)$  and  $(2,3)$  in the ratio $1:n$ is
$\left ( \frac{n(1)+1(2)}{1+n},\frac{n.(0)+1.(3)}{1+n} \right )= \left ( \frac{n+2}{1+n},\frac{3}{1+n} \right )$
Let the slope of the perpendicular line is m
And Slope of  line segment joining the points  $(1,0)$  and  $(2,3)$ is
$m'= \frac{3-0}{2-1}= 3$
Now, slope of perpendicular line is
$m = -\frac{1}{m'}= -\frac{1}{3}$
Now, equation of line passing through point $(x_1,y_1)$ and with slope m is
$(y-y_1)= m(x-x_1)$
equation of line passing through point $\left ( \frac{n+2}{1+n},\frac{3}{1+n} \right )$ and with slope  $-\frac{1}{3}$ is
$(y- \frac{3}{1+n})= -\frac{1}{3}(x- (\frac{n+2}{1+n}))\\ 3y(1+n)-9=-x(1+n)+n+2\\ x(1+n)+3y(1+n)=n+11$
Therefore, equation of line  is  $x(1+n)+3y(1+n)=n+11$

Let (a, b) are the intercept on x and y-axis respectively
Then, the equation of the line is given by
$\frac{x}{a}+\frac{y}{b}= 1$
Intercepts are equal which means a = b
$\frac{x}{a}+\frac{y}{a}= 1\\ \\ x+y = a$
Now, it is given that line passes through the point (2,3)
Therefore,
$a = 2+ 3 = 5$
therefore, equation of the line is   $x+ y = 5$

Let (a, b) are the intercept on x and y axis respectively
Then, the equation of line is given by
$\frac{x}{a}+\frac{y}{b}= 1$
It is given that
a + b = 9
b = 9 - a
Now,
$\frac{x}{a}+\frac{y}{9-a } = 1\\ \\ x(9-a)+ay= a(9-a)\\ 9x-ax+ay=9a-a^2$
It is given that line passes through point (2 ,2)
So,
$9(2)-2a+2a=9a-a^2\\ a^2-9a+18=0\\ a^2-6a-3a+18=0\\ (a-6)(a-3)= 0\\ a=6 \ \ \ \ \ \ or \ \ \ \ \ \ a = 3$

case (i)  a = 6  b = 3
$\frac{x}{6}+\frac{y}{3}= 1\\ \\ x+2y = 6$

case (ii)   a = 3 , b = 6
$\frac{x}{3}+\frac{y}{6}= 1\\ \\ 2x+y = 6$
Therefore, equation of line is 2x + y = 6 , x + 2y = 6

We know that
$m = \tan \theta \\ m = \tan \frac{2\pi}{3} = -\sqrt3$
Now, equation of line passing through point (0 , 2) and with slope $-\sqrt3$ is
$(y-2)= -\sqrt3(x-0)\\ \sqrt3x+y-2=0$
Therefore, equation of line is  $\sqrt3x+y-2=0$                 -(i)

Now, It is given that line crossing the $y$-axis at a distance of $2$ units below the origin which means coordinates are  (0 ,-2)
This line is parallel to above line which means slope of both the lines are equal
Now, equation of line passing through point (0 , -2) and with slope $-\sqrt3$ is
$(y-(-2))= -\sqrt3(x-0)\\ \sqrt3x+y+2=0$
Therefore, equation of line is  $\sqrt3x+y+2=0$

Let the slope of the line is m
and slope of a perpendicular line is which passes through the origin (0, 0) and (-2, 9) is
$m' = \frac{9-0}{-2-0}= \frac{9}{-2}$
Now, the slope of the line is
$m = -\frac{1}{m'}= \frac{2}{9}$
Now, the equation of line passes through the point (-2, 9) and with slope $\frac{2}{9}$  is
$(y-9)=\frac{2}{9}(x-(-2))\\ \\ 9(y-9)=2(x+2)\\ 2x-9y+85 = 0$
Therefore, the equation of the line is   $2x-9y+85 = 0$

It is given that
If $C=20$ then $L=124.942$
and  If  $C=110$  then  $L=125.134$
Now, if assume C along x-axis and L along y-axis
Then, we will get coordinates of two points (20 , 124.942)  and (110 , 125.134)
Now, the relation between C and L is given by equation
$(L-124.942)= \frac{125.134-124.942}{110-20}(C-20)$
$(L-124.942)= \frac{0.192}{90}(C-20)$
$L= \frac{0.192}{90}(C-20)+124.942$
Which is the required relation

It is given that the owner of a milk store sell
980 litres milk each week at $Rs\hspace{1mm}14/litre$
and  $1220$  litres of milk each week at  $Rs\hspace{1mm}16/litre$
Now, if we assume the rate of milk as x-axis and Litres of milk as y-axis
Then, we will get coordinates of two points i.e.  (14, 980)  and   (16, 1220)
Now, the relation between  litres of milk and Rs/litres is given by equation
$(L-980)= \frac{1220-980}{16-14}(R-14)$
$(L-980)= \frac{240}{2}(R-14)$
$L-980= 120R-1680$
$L= 120R-700$
Now, at $Rs\hspace{1mm}17/litre$ he could sell
$L= 120\times 17-700= 2040-700= 1340$
He could sell 1340 litres of milk each week at $Rs\hspace{1mm}17/litre$

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

Let the coordinates of Point A is (x,0) and of point B is (0,y)
It is given that point R(h , k) divides the line segment between the axes in the ratio $1:2$
Therefore,
R(h , k) $=\left ( \frac{1\times 0+2\times x}{1+2},\frac{1\times y+2\times 0}{1+2} \right )=\left ( \frac{2x}{3},\frac{y}{3} \right )$
$h = \frac{2x}{3} \ \ and \ \ k = \frac{y}{3}$
$x = \frac{3h}{2} \ \ and \ \ y = 3k$
Therefore, coordinates of point A is $\left ( \frac{3h}{2},0 \right )$  and of point B is $(0,3k)$
Now, slope of line passing through points $\left ( \frac{3h}{2},0 \right )$ and $(0,3k)$  is
$m = \frac{3k-0}{0-\frac{3h}{2}}= \frac{2k}{-h}$
Now, equation of line passing through point $(0,3k)$  and with slope $-\frac{2k}{h}$ is
$(y-3k)=-\frac{2k}{h}(x-0)$
$h(y-3k)=-2k(x)$
$yh-3kh=-2kx$
$2kx+yh=3kh$
Therefore, the equation of line is $2kx+yh=3kh$

Points are collinear means they lies on same line
Now,  given points are   $A(3,0),B(-2,-2)$ and  $C(8,2)$
Equation of line passing through point A and B is
$(y-0)=\frac{0+2}{3+2}(x-3)$
$y=\frac{2}{5}(x-3)\Rightarrow 5y= 2(x-3)$
$2x-5y=6$
Therefore, the equation of line passing through A and B is $2x-5y=6$

Now, Equation of line passing through point B and C is
$(y-2)=\frac{2+2}{8+2}(x-8)$
$(y-2)=\frac{4}{10}(x-8)$
$(y-2)=\frac{2}{5}(x-8) \Rightarrow 5(y-2)=2(x-8)$
$5y-10=2x-16$
$2x-5y=6$
Therefore, Equation of line passing through point B and C is $2x-5y=6$
When can clearly see that  Equation of line passing through point A nd B  and through B and C is the same
By this we can say that points  $A(3,0),B(-2,-2)$ and  $C(8,2)$ are collinear points

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

$x+7y=0$

Given equation is
$x+7y=0$
we can rewrite it as
$y= -\frac{1}{7}x$                            -(i)
Now, we know that the Slope-intercept form of the line is
$y = mx+C$                     -(ii)
Where m is the slope and C is some constant
On comparing equation (i) with equation (ii)
we  will get
$m =- \frac{1}{7}$  and $C = 0$
Therefore, slope and y-intercept are $-\frac{1}{7} \ and \ 0$  respectively

$6x+3y-5=0$

Given equation is
$6x+3y-5=0$
we can rewrite it as
$y= -\frac{6}{3}x+\frac{5}{3}\Rightarrow y = -2x+\frac{5}{3}$                            -(i)
Now, we know that the Slope-intercept form of line is
$y = mx+C$                     -(ii)
Where m is the slope and C is some constant
On comparing equation (i) with equation (ii)
we  will get
$m =- 2$  and $C = \frac{5}{3}$
Therefore, slope and y-intercept are $-2 \ and \ \frac{5}{3}$  respectively

$y=0.$

Given equation is
$y=0$                                  -(i)
Now, we know that the Slope-intercept form of the line is
$y = mx+C$                     -(ii)
Where m is the slope and C is some constant
On comparing equation (i) with equation (ii)
we  will get
$m =0$  and $C = 0$
Therefore, slope and y-intercept are $0 \ and \ 0$  respectively

$3x+2y-12=0$

Given equation is
$3x+2y-12=0$
we can rewrite it as
$\frac{3x}{12}+\frac{2y}{12} = 1$
$\frac{x}{4}+\frac{y}{6} = 1$                         -(i)
Now, we know that the intercept form of line is
$\frac{x}{a}+\frac{y}{b} = 1$                         -(ii)
Where a and b are intercepts on x and y axis respectively
On comparing equation (i) and (ii)
we will get
a = 4 and b = 6
Therefore,  intercepts on x and y axis  are 4 and 6 respectively

$4x-3y=6$

Given equation is
$4x-3y=6$
we can rewrite it as
$\frac{4x}{6}-\frac{3y}{6} = 1$
$\frac{x}{\frac{3}{2}}-\frac{y}{2} = 1$                         -(i)
Now, we know that the intercept form of line is
$\frac{x}{a}+\frac{y}{b} = 1$                         -(ii)
Where a and b are intercepts on x and y axis respectively
On comparing equation (i) and (ii)
we will get
$a = \frac{3}{2}$ and $b = -2$
Therefore,  intercepts on x and y axis  are $\frac{3}{2}$ and -2 respectively

$3y+2=0$

Given equation is
$3y+2=0$
we can rewrite it as
$y = \frac{-2}{3}$
Therefore,  intercepts on y-axis are $\frac{-2}{3}$
and there is no intercept on x-axis

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

Given equation is
$x-\sqrt{3}y+8=0$
we can rewrite it as
$-x+\sqrt3y=8$
Coefficient of x is -1 and y is $\sqrt3$
Therefore, $\sqrt{(-1)^2+(\sqrt3)^2}= \sqrt{1+3}=\sqrt4=2$
Now, Divide both the sides by 2
we will get
$-\frac{x}{2}+\frac{\sqrt3y}{2}= 4$
we can rewrite it as
$x\cos 120\degree + y\sin 120\degree= 4 \ \ \ \ \ \ \ \ \ \ \ -(i)$
Now, we know that the normal form of the line is
$x\cos \theta + y\sin \theta= p \ \ \ \ \ \ \ \ \ \ \ -(ii)$
Where $\theta$ is the angle between perpendicular and the positive x-axis and p is the perpendicular distance  from the origin
On comparing equation (i) and (ii)
we wiil get
$\theta = 120\degree \ \ and \ \ p = 4$
Therefore,  the angle between perpendicular and the positive x-axis and  perpendicular distance  from the origin is $120\degree \ and \ 4$  respectively

$y-2=0$

Given equation is
$y-2=0$
we can rewrite it as
$0.x+y = 2$
Coefficient of x is 0 and y is 1
Therefore, $\sqrt{(0)^2+(1)^2}= \sqrt{0+1}=\sqrt1=1$
Now, Divide both the sides by 1
we will get
$y=2$
we can rewrite it as
$x\cos 90\degree + y\sin 90\degree= 2 \ \ \ \ \ \ \ \ \ \ \ -(i)$
Now, we know that normal form of line is
$x\cos \theta + y\sin \theta= p \ \ \ \ \ \ \ \ \ \ \ -(ii)$
Where $\theta$ is the angle between perpendicular and the positive x-axis and p is the perpendicular distance  from the origin
On comparing equation (i) and (ii)
we wiil get
$\theta = 90\degree \ \ and \ \ p = 2$
Therefore,  the angle between perpendicular and the positive x-axis and  perpendicular distance  from the origin is $90\degree \ and \ 2$  respectively

$x-y=4$

Given equation is
$x-y=4$

Coefficient of x is 1 and y is -1
Therefore, $\sqrt{(1)^2+(-1)^2}= \sqrt{1+1}=\sqrt2$
Now, Divide both the sides by $\sqrt2$
we wiil get
$\frac{x}{\sqrt2}-\frac{y}{\sqrt2}= \frac{4}{\sqrt2}$
we can rewrite it as
$x\cos 315\degree + y\sin 315\degree= 2\sqrt2 \ \ \ \ \ \ \ \ \ \ \ -(i)$
Now, we know that normal form of line is
$x\cos \theta + y\sin \theta= p \ \ \ \ \ \ \ \ \ \ \ -(ii)$
Where $\theta$ is the angle between perpendicular and the positive x-axis and p is the perpendicular distance  from the origin
On compairing equation (i) and (ii)
we wiil get
$\theta = 315\degree \ \ and \ \ p = 2\sqrt2$
Therefore,  the angle between perpendicular and the positive x-axis and  perpendicular distance  from the origin is $315\degree \ and \ 2\sqrt2$  respectively

Given the equation of the line is
$12(x+6)=5(y-2)$
we can rewrite it as
$12x+72=5y-10$
$12x-5y+82=0$
Now, we know that
$d= \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$       where A and B are the coefficients of x and y and  C is some constant  and  $(x_1,y_1)$ is point from which we need to find the distance
In this problem A = 12 , B = -5 , c = 82 and $(x_1,y_1)$ = (-1 , 1)
Therefore,
$d = \frac{|12.(-1)+(-5).1+82|}{\sqrt{12^2+(-5)^2}} = \frac{|-12-5+82|}{\sqrt{144+25}}=\frac{|65|}{\sqrt{169}}=\frac{65}{13}= 5$
Therefore, the distance of the point  $(-1,1)$  from the line   $12(x+6)=5(y-2)$ is 5 units

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

Given equations of lines are
$15x+8y-34=0$  and  $15x+8y+31=0$
it is given that these lines are parallel
Therefore,
$d = \frac{ |C_2-C_1|}{\sqrt{A^2+B^2}}$
$A = 15 , B = 8 , C_1= -34 \ and \ C_2 = 31$
Now,
$d = \frac{|31-(-34)|}{\sqrt{15^2+8^2}}= \frac{|31+34|}{\sqrt{225+64}}= \frac{|65|}{\sqrt{289}} = \frac{65}{17}$
Therefore, the distance between two lines is $\frac{65}{17} \ units$

Given equations of lines are
$l(x+y)+p=0$ and $l(x+y)-r = 0$
it is given that these lines are parallel
Therefore,
$d = \frac{ |C_2-C_1|}{\sqrt{A^2+B^2}}$
$A = l , B = l , C_1= -r \ and \ C_2 = p$
Now,
$d = \frac{|p-(-r)|}{\sqrt{l^2+l^2}}= \frac{|p+r|}{\sqrt{2l^2}}= \frac{|p+r|}{\sqrt{2}|l|}$
Therefore, the distance between two lines is $\frac{1}{\sqrt2}\left | \frac{p+r}{l} \right |$

It is given that line is parallel to line  $3x-4y+2=0$ which implies that the slopes of both the lines are equal
we can rewrite it as
$y = \frac{3x}{4}+\frac{1}{2}$
The slope of line $3x-4y+2=0$  =  $\frac{3}{4}$
Now, the equation of the line passing through the point $(-2,3)$ and with slope $\frac{3}{4}$ is
$(y-3)=\frac{3}{4}(x-(-2))$
$4(y-3)=3(x+2)$
$4y-12=3x+6$
$3x-4y+18= 0$
Therefore, the equation of the line is  $3x-4y+18= 0$

It is given that line is  perpendicular to the line $x-7y+5=0$
we can rewrite it as
$y = \frac{x}{7}+\frac{5}{7}$
Slope of line $x-7y+5=0$   ( m' ) = $\frac{1}{7}$
Now,
The slope of the line is      $m = \frac{-1}{m'} = -7 \ \ \ \ \ \ \ \ \ \ \ (\because lines \ are \ perpendicular)$
Now, the equation of the line with  $x$intercept $3$  i.e. (3, 0) and  with slope -7 is
$(y-0)=-7(x-3)$
$y = -7x+21$
$7x+y-21=0$

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

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

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

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

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

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

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

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

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

Line passing through points ( h ,3) and (4 ,1)

Therefore,Slope of the line is

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

$m =\frac{3-1}{h-4}$

This line intersects the line $7x-9y-19=0$  at right angle
Therefore, the Slope of both the lines are negative times inverse of each other
Slope of line $7x-9y-19=0$ , $m'=\frac{7}{9}$
Now,
$m=-\frac{1}{m'}$
$\frac{2}{h-4}= -\frac{9}{7}$
$14=-9(h-4)$
$14=-9h+36$
$-9h= -22$
$h=\frac{22}{9}$
Therefore, the value of h is  $\frac{22}{9}$

It is given that line is parallel to the line  $Ax+By+C=0$
Therefore, their slopes are equal
The slope of line $Ax+By+C=0$  , $m'= \frac{-A}{B}$
Let the slope of other line be m
Then,
$m =m'= \frac{-A}{B}$
Now, the equation of the line passing through the point $(x_1,y_1)$  and with slope $-\frac{A}{B}$  is
$(y-y_1)= -\frac{A}{B}(x-x_1)$
$B(y-y_1)= -A(x-x_1)$
$A(x-x_1)+B(y-y_1)= 0$
Hence proved

Let the slope of two lines are $m_1 \ and \ m_2$   respectively
It is given the lines intersects each other at an angle of  $60^{\circ}$  and slope of the line is 2
Now,
$m_1 = m\ and \ m_2= 2 \ and \ \theta = 60\degree$
$\tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |$
$\tan 60\degree = \left | \frac{2-m}{1+2m} \right |$
$\sqrt3 = \left | \frac{2-m}{1+2m} \right |$
$\sqrt3 = \frac{2-m}{1+2m} \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \sqrt 3 = -\left ( \frac{2-m}{1+2m} \right )$
$m = \frac{2-\sqrt3}{2\sqrt3+1} \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ m = \frac{-(2+\sqrt3)}{2\sqrt3-1}$
Now, the equation of line passing through point (2 ,3) and with slope  $\frac{2-\sqrt3}{2\sqrt3+1}$  is
$(y-3)= \frac{2-\sqrt3}{2\sqrt3+1}(x-2)$
$(2\sqrt3+1)(y-3)=(2-\sqrt3)(x-2)$
$x(\sqrt3-2)+y(2\sqrt3+1)=-1+8\sqrt3$                        -(i)

Similarly,
Now , equation of line passing through point (2 ,3) and with slope  $\frac{-(2+\sqrt3)}{2\sqrt3-1}$  is
$(y-3)=\frac{-(2+\sqrt3)}{2\sqrt3-1}(x-2)$
$(2\sqrt3-1)(y-3)= -(2+\sqrt3)(x-2)$
$x(2+\sqrt3)+y(2\sqrt3-1)=1+8\sqrt3$                             -(ii)

Therefore, equation of line is    $x(\sqrt3-2)+y(2\sqrt3+1)=-1+8\sqrt3$     or      $x(2+\sqrt3)+y(2\sqrt3-1)=1+8\sqrt3$

Right bisector means perpendicular line which divides the line segment into two equal parts
Now, lines are perpendicular which means their slopes are negative times inverse of each other
Slope of line passing through points   $(3,4)$ and   $(-1,2)$  is
$m'= \frac{4-2}{3+1}= \frac{2}{4}=\frac{1}{2}$
Therefore, Slope of bisector line is
$m = - \frac{1}{m'}= -2$
Now, let (h , k) be the point of intersection of two lines
It is given that point (h,k) divides the line segment joining point  $(3,4)$ and   $(-1,2)$ into two equal part which means it is the mid point
Therefore,
$h = \frac{3-1}{2} = 1\ \ \ and \ \ \ k = \frac{4+2}{2} = 3$
$(h,k) = (1,3)$
Now, equation  of line passing through point (1,3) and with slope -2 is
$(y-3)=-2(x-1)\\ y-3=-2x+2\\ 2x+y=5$
Therefore, equation of line is  $2x+y=5$

Let suppose the foot of perpendicular is $(x_1,y_1)$
We can say that line passing through the point $(x_1,y_1) \ and \ (-1,3)$  is perpendicular to the line $3x-4y-16=0$
Now,
The slope of the line $3x-4y-16=0$ is , $m' = \frac{3}{4}$
And
The slope of the line  passing through the point $(x_1,y_1) \ and \ (-1,3)$is, $m = \frac{y-3}{x+1}$
lines are perpendicular
Therefore,
$m = -\frac{1}{m'}\\ \frac{y_1-3}{_1+1} = -\frac{4}{3}\\ 3(y_1-3)=-4(x_1+1)\\ 4x_1+3y_1=5 \ \ \ \ \ \ \ \ \ -(i)$
Now, the point $(x_1,y_1)$ also lies on the line $3x-4y-16=0$
Therefore,
$3x_1-4y_1=16 \ \ \ \ \ \ \ \ \ \ \ -(ii)$
On solving equation (i) and (ii)
we will get
$x_1 = \frac{68}{25} \ and \ y_1 =-\frac{49}{25}$
Therefore, $(x_1,y_1) = \left ( \frac{68}{25},-\frac{49}{25} \right )$

We can say that line passing through point $(0,0) \ and \ (-1,2)$  is perpendicular to line $y=mx+c$
Now,
The slope of the line  passing through the point $(0,0) \ and \ (-1,2)$ is , $m = \frac{2-0}{-1-0}= -2$
lines are perpendicular
Therefore,
$m = -\frac{1}{m'} = \frac{1}{2}$               - (i)
Now, the point $(-1,2)$ also lies on the line $y=mx+c$
Therefore,
$2=\frac{1}{2}.(-1)+C\\ C = \frac{5}{2} \ \ \ \ \ \ \ \ \ \ \ -(ii)$
Therefore, the value of m and C is $\frac{1}{2} \ and \ \frac{5}{2}$   respectively

Given equations of lines are    $x\cos \theta -y\sin \theta =k\cos 2\theta$  and  $x\sec \theta +y\hspace{1mm}cosec\hspace{1mm}\theta =k$

We can rewrite the equation $x\sec \theta +y\hspace{1mm}cosec\hspace{1mm}\theta =k$ as

$x\sin \theta +y\cos \theta = k\sin\theta\cos\theta$
Now, we know that

$d = \left | \frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}} \right |$

In equation $x\cos \theta -y\sin \theta =k\cos 2\theta$

$A= \cos \theta , B = -\sin \theta , C = - k\cos2\theta \ and \ (x_1,y_1)= (0,0)$

$p= \left | \frac{\cos\theta .0-\sin\theta.0-k\cos2\theta }{\sqrt{\cos^2\theta+(-\sin\theta)^2}} \right | = |-k\cos2\theta|$
Similarly,
in the equation $x\sin \theta +y\cos \theta = k\sin\theta\cos\theta$

$A= \sin \theta , B = \cos \theta , C = -k\sin\theta\cos\theta \ and \ (x_1,y_1)= (0,0)$

$q= \left | \frac{\sin\theta .0+\cos\theta.0-k\sin\theta\cos\theta }{\sqrt{\sin^2\theta+\cos^2\theta}} \right | = |-k\sin\theta\cos\theta|= \left | -\frac{k\sin2\theta}{2} \right |$
Now,

$p^2+4q^2=(|-k\cos2\theta|)^2+4.(|-\frac{k\sin2\theta}{2})^2= k^2\cos^22\theta+4.\frac{k^2\sin^22\theta}{4}$
$=k^2(\cos^22\theta+\sin^22\theta)$
$=k^2$
Hence proved

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

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

we know that intercept form of line is
$\frac{x}{a}+\frac{y}{b} = 1$
we know that
$d = \left | \frac{Ax_1+bx_2+C}{\sqrt{A^2+B^2}} \right |$
In this problem
$A = \frac{1}{a},B = \frac{1}{b}, C =-1 \ and \ (x_1,y_1)= (0,0)$
$p= \left | \frac{\frac{1}{a}\times 0+\frac{1}{b}\times 0-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} \right | = \left | \frac{-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} \right |$
On squaring both the sides
we will get
$\frac{1}{p^2}= \frac{1}{a^2}+\frac{1}{b^2}$
Hence proved

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

Parallel to the x-axis.

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

Parallel to the y-axis.

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 $\pm2$

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 $6 \ or \ 1$

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  $\small \theta = \frac{7\pi}{6} \ and \ p =1$

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  $2x-3y=6 \ and \ -3x+2y=6$

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  $\left ( 0,-\frac{8}{3} \right )$

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
$d = \left | \frac{(\sin\phi -\sin\theta).0-(\cos\phi-\cos\theta).0-\sin(\theta-\phi)}{\sqrt{(\sin\phi-\sin\theta)^2+(\cos\phi-\cos\theta)^2}} \right |$
$d = \left | \frac{-\sin(\theta-\phi)}{\sqrt{(\sin^2\phi+\cos^2\phi)+(\sin^2\theta+\cos^2\theta)-2(\cos\theta\cos\phi+\sin\theta\sin\phi)}} \right |$
$d = \left | \frac{-\sin(\theta-\phi)}{1+1-2\cos(\theta-\phi)} \right |$                                $(\because \cos a\cos b +\sin a\sin b = \cos(a-b) \ \ and \ \ \sin^2a+\cos^2a=1)$
$d = \left |\frac{ - \sin(\theta-\phi)}{2(1-\cos(\theta-\phi))} \right |$
$d = \left | \frac{-2\sin\frac{\theta-\phi}{2}\cos \frac{\theta-\phi}{2}}{\sqrt{2.(2\\sin^2\frac{\theta-\phi}{2})}}\right |$
$d = \left | \frac{-2\sin\frac{\theta-\phi}{2}\cos \frac{\theta-\phi}{2}}{2\sin\frac{\theta-\phi}{2}}\right |$
$d = \left | \cos\frac{\theta-\phi}{2} \right |$

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  $x = -\frac{5}{22}$

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   $2x-3y+18=0$

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  $k^2 \ square \ units$

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 $5$

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

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 $x+3y=9$

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   $13x+13y=6$

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

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 $1:2$

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   $\frac{23\sqrt5}{18} \ units$

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

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 $m = \infty$
$1-x=0 \ \ \ \ and \ \ \ \ \ 1-y =0$
$x=1 \ \ \ \ and \ \ \ \ \ y =1$

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 $(-1,-4)$

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 $\frac{1\pm5\sqrt2}{7}$

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

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  $18a+12b+11=0$

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  $\left ( \frac{13}{5},0 \right )$

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

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 $\dpi{100} (x_1,\:y_1)$ and  $\dpi{100} (x_2,\:y_2)$ then the slope(m) of the line is given by-

$\dpi{100} m=\frac{y_2-y_1}{x_2-x_1}=\frac{y_1-y_2}{x_1-x_2},\:\:x_1\neq x_2$

• The slope of the line which makes an angle with the positive x-axis is given by $\dpi{100} m = tan\:\alpha \: , \alpha\neq 90^o$.
• The slope of the horizontal line is zero and the slope of the vertical line is undefined.
• An acute angle ($\dpi{100} \theta$) between lines $\dpi{100} L_1$  and  $\dpi{100} L_2$ with slopes $\dpi{100} m_1$ and $\dpi{100} m_2$ is given by-

$\dpi{100} tan \:\theta =|\frac{m_2-m_1}{1+m_1m_2}|\:,\:1+m_1m_2\neq 0$

• Two lines (with slopes $\dpi{100} m_1 \:$ and $\dpi{100} m_2 \:$) are parallel if and only if their slopes ($\dpi{100} m_1=m_2 \:$) are equal.
• Two lines (with slopes $\dpi{100} m_1 \:$ and $\dpi{100} m_2 \:$) are perpendicular if and only if the product of their slopes is –1 or $\dpi{100} m_1.m_2 =-1$.
• The equation of a line passing through the points $\dpi{100} (x_1,\:y_1)$ and  $\dpi{100} (x_2,\:y_2)$ is given by-

$\dpi{100} y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

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.