Exercise 6.3 - Chapter 6 - Two Dimensional Analytical Geometry - 11th Maths Guide Samacheer Kalvi Solutions

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Ex 6.3
Question 1.
Show that the lines are $3 x+2 y+9=0$ and $12 x+8 y-15=0$ are parallel lines.
Solution:
Slope of $\mathrm{I}$ line $=m_1=-\left(\frac{3}{2}\right)=\frac{-3}{2}$
Slope of II line $=m_2=-\left(\frac{12}{8}\right)=\frac{-3}{2}$
Here $\mathrm{m}_1=\mathrm{m}_2 \Rightarrow$ the two lines are parallel.
Question 2.
Find the equation of the straight line parallel to $5 x-4 y+3=0$ and having $x-$ intercept 3 .
Solution:
Equation of a line parallel to $a x+b y+c=0$ will be of the form $a x+b y+k=0$
So equation of a line parallel to $5 x-4 y+3=0$ will be of the form $5 x-4 y=k$
$\Rightarrow \frac{5 x}{k}-\frac{4 y}{k}=1$
(i.e) $\frac{x}{k / 5}+\frac{y}{k /-4}=1$
Here we are given that $x$ intercept $=-3 \Rightarrow \frac{k}{5}=-3 \Rightarrow k=-15$
So equation of the line is $\frac{x}{-3}+\frac{y}{-15 /-4}=1$
(i.e) $\frac{-x}{3}+\frac{4 y}{15}=1 \Rightarrow \frac{-5 x+4 y}{15}=1$
$-5 x+4 y=15 \Rightarrow 5 x-4 y-15=0$
Question 3.
Find the distance between the line $4 x+3 y+4=0$ and a point
(i) $(-2,4)$
(ii) $(7,-3)$
Solution:

The distance between the line $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ and the point $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is given by
$
\pm \frac{a x_1+b y_1+c}{\sqrt{a^2+b^2}}
$
(i) Now the distance between the line $4 x+3 y+4=0$ and $(-2,4)$ is $\pm \frac{4(-2)+3(4)+4}{\sqrt{4^2+3^2}}=\frac{-8+12+4}{5}=8 / 5$ units
(ii) The distance between the line $4 x+3 y+4=0$ and $(7,-3)$ is $\pm \frac{4(7)+3(-3)+4}{\sqrt{4^2+3^2}}=\frac{28-9+4}{5}=\frac{23}{5}$ units

Question 4.
Write the equation of the lines through the point $(1,-1)$
(i) Parallel to $x+3 y-4=0$
(ii) Perpendicular to $3 x+4 y=6$
Solution:
(i) Any line parallel to $x+3 y-4=0$ will be of the form $x+3 y+k=0$.
It passes through $(1,-1) \Rightarrow 1-3+\mathrm{k}=0 \Rightarrow \mathrm{k}=2$
So the required line is $x+3 y+2=0$
(ii) Any line perpendicular to $3 x+4 y-6=0$ will be of the form $4 x-3 y+k=0$.
It passes through $(1,-1) \Rightarrow 4+3+\mathrm{k}=0 \Rightarrow \mathrm{k}=-7$.
So the required line is $4 x-3 y-7=0$
Question 5.
If $(-4,7)$ is one vertex of a rhombus and if the equation of one diagonal is $5 x-y+7=0$, then find the equation of another diagonal.
Solution:
In a rhombus, the diagonal cut at right angles.
The given diagonal is $5 x-y+7=0$ and $(-4,7)$ is not a point on the diagonal.
So it will be a point on the other diagonal which is perpendicular to $5 x-y+7=0$.
The equation of a line perpendicular to $5 x-y+7=0$ will be of the form $x+5 y+k=0$. It passes through $(-4,7) \Rightarrow-4+5(7)+\mathrm{k}=0 \Rightarrow \mathrm{k}=-31$
So the equation of the other diagonal is $x+5 y-31=0$
Question 6.
Find the equation of the lines passing through the point of intersection lines $4 x-y+3=0$ and $5 x+2 y$ $+7=0$, and
(i) Through the point $(-1,2)$
(ii) Parallel to $\mathrm{x}-\mathrm{y}+5=0$
(iii) Perpendicular to $x-2 y+1=0$.
Solution:
To find the point of intersection of the lines we have to solve them

Substituting $x=-1$ in equation (2) we get
$
\begin{aligned}
& -5+2 y=-7 \\
& \Rightarrow 2 y=-7+5=-2 \\
& \Rightarrow y=-1
\end{aligned}
$
So the point of intersection is $(-1,-1)$
(i) $\operatorname{Now}\left(x_1, y_1\right)=(-1,-1) ;\left(x_2, y_2\right)=(-1,2)$.
Equation of the line passing through 2 points is $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$
(i.e) $\frac{y+1}{2+1}=\frac{x+1}{-1+1} \Rightarrow \frac{y+1}{3}=\frac{x+1}{0} \Rightarrow x+1=0 \Rightarrow x=-1$
(ii) Equation of a line parallel to $x-y+5=0$ will be of the form $x-y+k=0$.
It passes through $(-1,-1) \Rightarrow-1+1+\mathrm{k}=0 \Rightarrow \mathrm{k}=0$.
So the required line is $x-y=0 \Rightarrow x=y$.
(iii) Equation of a line perpendicular to $x-2 y+1=0$ will be of the form $2 x+y+k=0$. It passes through $(-1,-1) \Rightarrow-2-1+\mathrm{k}=0 \Rightarrow \mathrm{k}=3$.
So the required line is $2 x+y+3=0$

Question 7.
Find the equations of two straight lines which are parallel to the line $12 x+5 y+2=0$ and at a unit distance from the point $(1,-1)$.
Solution:
Equation of a line parallel to $12 \mathrm{x}+5 \mathrm{y}+2=0$ will be of the form $12 \mathrm{x}+5 \mathrm{y}+\mathrm{k}=0$.
We are given that the perpendicular distance form $(1,-1)$ to the line $12 \mathrm{x}+5 \mathrm{y}+\mathrm{k}=0$ is 1 unit.
$
\begin{aligned}
& \Rightarrow \pm \frac{12(1)+5(-1)+k}{\sqrt{12^2+5^2}}=1(\text { i.e }) \frac{12-5+k}{13}= \pm 1 \Rightarrow 7+k= \pm 13 \\
& 7+k=13 \quad \begin{array}{c}
7+k=-13
\end{array} \\
& \Rightarrow k=13-7=6 \mid \Rightarrow k=-13-7=-20
\end{aligned}
$
So the required line will be $12 x+5 y+6=0$ or $12 x+5 y-20=0$

Question 8.
Find the equations of straight lines which are perpendicular to the line $3 x+4 y-6=0$ and are at a distance of 4 units from $(2,1)$.
Solution:
Given equation of line is $3 x+4 y-6=0$.
Any line perpendicular to $3 x+4 y-6=0$ will be of the form $4 x-3 y+k=0$
Given perpendicular distance is 4 units from $(2,1)$ to line (1)

$
\begin{aligned}
& \therefore 4= \pm \frac{(4(2)-3(1)+k)}{\sqrt{4^2+(-3)^2}} \\
& \Rightarrow \quad 4= \pm \frac{(8-3+k)}{\sqrt{16+9}} \\
& \Rightarrow \quad 4= \pm\left(\frac{5+k}{5}\right) \\
& \therefore 20=+(5+\mathrm{k}) \text { or } 20=-(5+\mathrm{k}) \\
& \Rightarrow \mathrm{k}=20-5 \text { or } \mathrm{k}=-(20+5) \\
& \mathrm{k}=15 \text { or } \mathrm{k}:=-25 \\
& \therefore \text { Required equation of the lines are } 4 \mathrm{x}-3 \mathrm{y}+15=0 \text { and } 4 \mathrm{x}-3 \mathrm{y}-25=0
\end{aligned}
$
Question 9.
Find the equation of a straight line parallel to $2 x+3 y=10$ and which is such that the sum of its intercepts on the axes is 15 .
Solution:
The equation of the line parallel to $2 x+3 y=10$ will be of the form $2 x+3 y=k$.
(i.e) $\frac{2 x}{k}+\frac{3 y}{k}=1 \Rightarrow \frac{x}{k / 2}+\frac{y}{k / 3}=1$.
Sum of the intercepts $=15 \Rightarrow \frac{k}{2}+\frac{k}{3}=15 \Rightarrow \frac{3 k+2 k}{6}=15$
$5 k=90 \Rightarrow k=90 / 5=18$.
So the required line is $2 x+3 y=18$ or $2 x+3 y-18=0$
Question 10.
Find the length of the perpendicular and the co-ordinates of the foot of the perpendicular from $(-10,-2)$ to the line $\mathrm{x}+\mathrm{y}-2=0$.
Solution:

Length of the perpendicular from $(-10,-2)$ to $x+y-2=0$ is $\pm\left(\frac{-10-2-2}{\sqrt{1^2+1^2}}\right)= \pm \frac{14}{\sqrt{2}}=\frac{14 \sqrt{2}}{\sqrt{2} \sqrt{2}}=7 \sqrt{2}$ units
Let $\mathrm{A}(a, b)$ be the foot of the perpendicular the given line is $x+y-2=0 \Rightarrow a+b-2=0$ $\Rightarrow a+b=2$
So of line joining A $(a, b)$ and $\mathrm{B}(-10,-2)$ is $\frac{b+2}{a+10}=m_1$.
Slope of $x+y-2=0$ is $\frac{-1}{1}=-1=m_2$
We are given $m_1 m_2=-1$
$
\Rightarrow\left(\frac{b+2}{a+10}\right)(-1)=-1 \Rightarrow \frac{b+2}{a+10}=1 \Rightarrow b+2=a+10
$
Solving (1) and (2) $\Rightarrow a-b=-8$
(1) $\Rightarrow a+b=2$
(2) $\Rightarrow \frac{a-b=-8}{2 a=-6}$
$a=-3$
Substituting $a=-3$ in (1) we get $-3+b=2 \Rightarrow b=2+3=5$
So the foot of the perpendicular is $(-3,5)$

Question 11.
If $\mathrm{p}_1$ and $\mathrm{p}_2$ are the lengths of the perpendiculars from the origin to the straight lines. $\sec \theta+\mathrm{y} \operatorname{cosec} \theta=$ $2 \mathrm{a}$ and $\mathrm{x} \cos \theta-\mathrm{y} \sin \theta=\mathrm{a} \cos 2 \theta$, then prove that $\mathrm{p}_{-}\{1\}^{\wedge}\{2\}+\mathrm{p}_{-}\{2\}^{\wedge}\{2\}=\mathrm{a}^{\wedge}\{2\}$
Solution:
$\mathrm{p}_1=$ length of perpendicular from $(0,0)$ to $\mathrm{x} \sec \theta+\mathrm{y} \operatorname{cosec} \theta=2 \mathrm{a}$

$
\begin{aligned}
& \Rightarrow p_1= \pm \frac{2 a}{\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}} \\
& \therefore p_1^2=\frac{4 a^2}{\sec ^2 \theta+\operatorname{cosec}^2 \theta}=\frac{4 a^2}{\frac{1}{\cos ^2 \theta}+\frac{1}{\sin ^2 \theta}} \\
& =\frac{4 a^2 \cos ^2 \theta \sin ^2 \theta}{\sin ^2 \theta+\cos ^2 \theta}=(2 a \sin \theta \cos \theta)^2 \\
& =a^2[2 \sin \theta \cos \theta]^2=a^2(\sin 2 \theta)^2=a^2 \sin ^2 2 \theta \\
& p_2=\text { length of perpendicular from }(0,0) \text { to } x \cos \theta-y \sin \theta=a \cos 2 \theta \\
& p_2= \pm \frac{a \cos 2 \theta}{\sqrt{\cos ^2 \theta+\sin ^2 \theta}}=a \cos 2 \theta \\
& \therefore p_2^2=(a \cos 2 \theta)^2=a^2 \cos ^2 2 \theta \\
& \text { LHS }=p_1^2+p_2^2=a^2 \sin ^2 2 \theta+a^2 \cos ^2 2 \theta=a^2\left[\sin ^2 2 \theta+\cos ^2 2 \theta\right]=a^2(1)=a^2=\text { RHS } \\
&
\end{aligned}
$
Question 12.
Find the distance between the parallel lines
(i) $12 x+5 y=7$ and $12 x+5 y+7=0$
(ii) $3 x-4 y+5=0$ and $6 x-8 y-15=0$.
Solution:
(i) The distance between the parallel lines $a x+b y+c=0$ and $a x+b y+d=0$ is $\pm \frac{c-d}{\sqrt{a^2+b^2}}$ The distance between $12 x+5 y-7=0$ and $12 x+5 y+7=0$ is $\pm 7-(-7) / \sqrt{12^2+5^2}=\frac{14}{13}$ units.
(ii) $3 x-4 y+5=0$ (multiplying by 2 ) $6 x-8 y+10=0,6 x-8 y-15=0$.
The distance between the parallel lines is $\pm \frac{10-(-15)}{\sqrt{6^2+8^2}}=\frac{10+15}{\sqrt{36+64}}=\frac{25}{10}=\frac{5}{2}$ units

Question 13.
Find the family of straight lines
(i) Perpendicular
(ii) Parallel to $3 x+4 y-12=0$.
Solution:
Equation of lines perpendicular to $3 x+4 y-12=0$ will be of the form $4 x-3 y+k=0, k \in R$ Equation of lines parallel to $3 x+4 y-12=0$ will be of the form $3 x+4 y+k=0, k \in R$
Question 14.
If the line joining two points $\mathrm{A}(2,0)$ and $\mathrm{B}(3,1)$ is rotated about $\mathrm{A}$ in anti-clockwise direction through an angle of $15^{\circ}$, then find the equation of the line in new position.
Solution:
Slope of $\mathrm{A}(2,0)$ and $\mathrm{B}(3,1)$ is $m=\frac{1-0}{3-2}=\frac{1}{1}=1$
(i.e) $\tan \theta=1 \Rightarrow \theta=45^{\circ}$.
This line is rotated about $15^{\circ}$ in anti clockwise direction
$\Rightarrow$ New slope $=\tan \left(45^{\circ}+15^{\circ}\right)=\tan 60^{\circ}=\sqrt{3}$ (i.e) $\mathrm{m}=\sqrt{3}$.
Point $\mathrm{A}=(2,0)$
Equation of the line is $y-0=\sqrt{3}(x-2)$
$
y=\sqrt{3} x-2 \sqrt{3} \Rightarrow \sqrt{3} x-y-2 \sqrt{3}=0
$

Question 15.
A ray of light coming from the point $(1,2)$ is reflected at a point $\mathrm{A}$ on the $\mathrm{x}$-axis and it passes through the point $(5,3)$. Find the co-ordinates of the point A.
Solution:
The image of the point $P(1,2)$ will be $P^{\prime}(1,-2)$.
Since $\angle \mathrm{OAP}=\angle \mathrm{XAQ}$ (angle of inches $=$ angle of reflection) So $\angle \mathrm{OAP}^{\prime}=\angle \mathrm{XAQ}=$ a (Vertically opposite angles)
$\Rightarrow \mathrm{P}, \mathrm{A}, \mathrm{Q}$ lie on a same line.

Now equation of the line $\mathrm{P}^{\prime}, \mathrm{Q}$ is [where $\left.\mathrm{P}^{\prime}=(1,-2), \mathrm{Q}=(5,3)\right]$
$
\begin{aligned}
& \frac{y+2}{3+2}=\frac{x-1}{5-1} \Rightarrow \frac{y+2}{5}=\frac{x-1}{4} \\
& 4 y+8=5 x-5 \\
& 5 x=4 y+13 \Rightarrow x=\frac{4 y+13}{5}
\end{aligned}
$
Since we find point of intersection with $\mathrm{x}$ axis we put $\mathrm{y}=0$.
So $y=0 \Rightarrow x=\frac{13}{5}$.
$\therefore$ The reflection point is $\left(\frac{13}{5}, 0\right)$

Question 16.
A line is drawn perpendicular to $5 x=y+7$. Find the equation of the line if the area of the triangle formed by this line with co-ordinate axes is 10 sq. units.
Solution:
Equation of the given lines $5 x=y+7 \Rightarrow 5 x-y=7$.
So its perpendicular will be of the form $x+5 y=7$
$
\Rightarrow \frac{x}{k}+\frac{5 y}{k}=1(i . e) \frac{x}{k}+\frac{y}{k / 5}=1 \text {. }
$
Now $x$ intercept $=k$ and $y$ intercept $=k / 5$.
Area of the $\Delta=\frac{1}{2}(k)(k / 5)=\frac{k^2}{10}=10$ (given) $\Rightarrow k^2=100$ or $k=10$
So equation of the line is $x+5 y= \pm 10$
Question 17.
Find the image of the point $(-2,3)$ about the line $x+2 y-9=0$.
Solution:
The coordinates of image of the point $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ with respect to the line $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ can be obtained by the line $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-2\left(a x_1+b y_1+c\right)}{a^2+b^2}$
Here $\left(x_1, y_1\right)=(-2,3)$ and the given line is $x+2 y-9=0$.
So the image is $\frac{x+2}{1}=\frac{y-3}{2}=\frac{-2[-2+2(3)-9]}{1+4}$
$
\begin{aligned}
& \text { (i.e) } \frac{x+2}{1}=\frac{y-3}{2}=\frac{-2(-2+6-9)}{5} \\
& \Rightarrow \frac{x+2}{1}=\frac{y-3}{2}=\frac{-2(-5)}{5}=2 \\
& \Rightarrow \frac{x+2}{1}=2 \quad \frac{y-3}{2}=2 \\
& \Rightarrow x=2-2=0 \Rightarrow y=4+3=7 \\
&
\end{aligned}
$
So the image is $(0,7)$

Question 18 .
A photocopy store charges Rs. 1.50 per copy for the first 10 copies and Rs. 1.00 per copy after the 10th copy. Let $x$ be the number of copies, and let $y$ be the total cost of photocopying.
(i) Draw graph of the cost as $\mathrm{x}$ goes from 0 to 50 copies.
(ii) Find the cost of making 40 copies
Solution:

(i)

$y=f(x)= \begin{cases}1.5 x & 0<x \leq 10 \\ 15+1(x-10) & \text { if } x>10 \\ (\text { i.e }) x+5 & \end{cases}$

(ii) The cost of making 40 copies is $40+5=$ Rs. 45 .
Question 19.
Find atleast two equations of the straight lines in the family of the lines $y=5 x+b$, for which $b$ and the $x$-coordinate of the point of intersection of the lines with $3 x-4 y=6$ are integers.
Solution:
$
\begin{aligned}
& y=5 x+b \\
& 3 x-4 y=6
\end{aligned}
$
Solving (1) and (2)
Substituting $y$ value from (1) in (2) we get

$
\begin{aligned}
& y=5 x+b \\
& 3 x-4 y=6
\end{aligned}
$
Solving (1) and (2)
Substituting $y$ value from (1) in (2) we get
$
\begin{aligned}
3 x-4(5 x+b) & =6 \\
3 x-20 x-4 b & =6 \\
-17 x & =6+4 b \\
\Rightarrow \quad x & =\frac{6+4 b}{-17}
\end{aligned}
$
So, $y=5\left(\frac{6+4 b}{-17}\right)+b \Rightarrow y=\frac{30+3 b}{-17}$
So, $(x, y)=\left(\frac{6+4 b}{-17}, \frac{30+3 b}{-17}\right)$
Since, $x$ coordinate and 6 are integers $6+46$ must be multiple of 17
$
\text { (i.e) } \pm 17, \pm 34
$

$\therefore$ Equation of lines $y=5 x+b$ will be $y=5 x+7, y=5 x-10$
Question 20.
Find all the equations of the straight lines in the family of the lines $\mathrm{y}=\mathrm{mx}-3$, for which $\mathrm{m}$ and the $\mathrm{x}$ coordinate of the point of intersection of the lines with $x-y=6$ are integers.
Solution:
Equation of the given lines are
$y=\mathrm{mx}-3 \ldots \ldots(1)$
and $x-y=6$
(2)
Solving (1) and (2)
$
x-(m x-3)=6
$

(i.e)
$
x-m x+3=6
$
$
\begin{array}{rlrl}
\Rightarrow & x(1-m) & =6-3=3 \\
& \text { Now (2) } \Rightarrow & =\frac{3}{1-m} \\
(\text { i.e }) & x-y & =6 \\
\Rightarrow & y & =x-6 \\
y & =\frac{3}{1-m}-6=\frac{3-6+6 m}{1-m}=\frac{6 m-3}{1-m} \\
& \therefore(x, y) & =\left(\frac{3}{1-m}, \frac{6 m-3}{1-m}\right)
\end{array}
$
Since $\mathrm{m}$ and $\mathrm{x}$ coordinates are integers
$1-m$ is the divisor of 3 (i.e) $\pm 1, \pm 3$
$
\begin{array}{c|c|c|c}
1-m=+1 & 1-m=-1 & 1-m=+3 & 1-m=-3 \\
\Rightarrow m=0 & \Rightarrow m=2 & \Rightarrow m=-2 & \Rightarrow m=4
\end{array}
$
So equation of lines are $(y=m x-3), y=m x-3$
(i) When $\mathrm{m}=0, \mathrm{y}=-3$
(ii) When $\mathrm{m}=2, \mathrm{y}=2 \mathrm{x}-3$
(iii) When $\mathrm{m}=-2, \mathrm{y}=-2 \mathrm{x}-3$ or $2 \mathrm{x}+\mathrm{y}+3=0$
(iv) When $\mathrm{m}=4, \mathrm{y}=4 \mathrm{x}-3$