Exercise 5.3 - Chapter 5 Coordinate Geometry 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\mathbf{E x} 5.3$
Question 1.
Find the equation of a straight line passing through the mid-point of a line segment joining the points $(1,-5)(4,2)$ and parallel to
(i) $\mathrm{X}$ axis
(ii) $\mathrm{Y}$ axis
Solution:

$y-b=m(x-a) \rightarrow$ equation of line passing through $(a, b)$ and with slope ' $m$ '.
$
\Rightarrow y-\left(\frac{-3}{2}\right)=m\left(x-\frac{5}{2}\right) \Rightarrow y+\frac{3}{2}=m\left(x-\frac{5}{2}\right)
$
(i) Line parallel to $x$ axis:
$
\begin{aligned}
\text { is } y & =c \\
y+\frac{3}{2} & =0 \Rightarrow 2 x+3=0
\end{aligned}
$
(ii) Line parallel to $y$ axis is $x=c$
$
\begin{aligned}
x-\frac{5}{2} & =0 \\
\Rightarrow \quad 2 x-5 & =0
\end{aligned}
$
Question 2.
The equation of a straight line is $2(x-y)+5=0$. Find its slope, inclination and intercept on the $\mathrm{Y}$ axis.
Solution:

$
\begin{aligned}
& 2(x-y)+5=0 \\
& \Rightarrow 2 x-2 y+5=
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow 2 \mathrm{y}=2 \mathrm{x}+5 \\
& \Rightarrow \quad y=x+\frac{5}{2} \quad \therefore \text { slope }=1 \\
& \text { Inclination, } \mathrm{m}=1 \\
& \Rightarrow \quad \tan \theta=1 \\
& \theta=45^{\circ} \\
& \Rightarrow y \text { intercept } \Rightarrow y=\frac{5}{2} \\
&
\end{aligned}
$
Question 3.
Find the equation of a line whose inclination is $30^{\circ}$ and making an intercept $-3$ on the $\mathrm{Y}$ axis.
Solution:
$
\theta=30^{\circ}
$
$y$ intercept $(x=0)=-3$ (i.e) when $x=0, y=-3$
let equation of line be $: y=m x+c$
$
\begin{aligned}
m & =\tan \theta \\
m & =\tan 30^{\circ}=\frac{1}{\sqrt{3}} \\
\therefore y & =\frac{1}{\sqrt{3}} \cdot x+c
\end{aligned}
$
When $x=0, y=-3$
$
\begin{aligned}
& \Rightarrow \quad-3=0+c \Rightarrow c=-3 \\
& \therefore \text { Equation of line: } y=\frac{x}{\sqrt{3}}-3 \\
& \Rightarrow \quad x-\sqrt{3} y-3 \sqrt{3}=0
\end{aligned}
$

Question 4.
Find the slope and $y$ intercept of $\sqrt{3} x+(1-\sqrt{3}) y=3$.
Solution:
$
\begin{aligned}
x \sqrt{3}+(1-\sqrt{3}) y & =3 \\
\Rightarrow \quad(1-\sqrt{3}) y & =-x \sqrt{3}+3
\end{aligned}
$

Question 5.
Find the value of ' $a$ ', if the line through $(-2,3)$ and $(8,5)$ is perpendicular to $y=a x=+2$
Solution:

Question 6.
The hill in the form of a right triangle has its foot at $(19,3)$ The inclination of the hill to the ground is $45^{\circ}$. Find the equation of the hill joining the foot and top.
Solution:
$$
\theta=45^{\circ}
$$
Coordinate of foot of hill $=(19,3)$ let equation of line be $y=m x+c$
$\mathrm{m}=\tan \theta=\tan 45^{\circ}=1$
$\Rightarrow \mathrm{y}=\mathrm{x}+\mathrm{c}$
Substituting $y=3 \& x=19,3=19+c \Rightarrow c=-16$
$\therefore$ Equation of line: $y=x-16=x-y-16=0$
Question 7.
Find the equation of a line through the given pair of points

(ii) $(2,3)$ and $(-7,-1)$
Solution:
(i) Equation of the line in two point form is
$
\begin{aligned}
& \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1} \\
& \frac{y-\frac{2}{3}}{-2-\frac{2}{3}}=\frac{x-2}{-\frac{1}{2}-2}\left[\because\left(x_1, y_1\right) \text { is }\left(2, \frac{2}{3}\right)\right. \\
& \frac{\frac{3 y-2}{\not 3}}{\frac{-6-2}{\not 3}}=\frac{x-2}{\frac{-1-4}{2}} \Rightarrow \frac{3 y-2}{-8}=\frac{2(x-2)}{-5} \\
& \Rightarrow-15 y+10=-16 x+32 \\
& \Rightarrow 16 x-15 y+10-32=0 \\
& \Rightarrow 16 x-15 y-22=0 \\
&
\end{aligned}
$

(ii) $\left(x_1, y_1\right)$ is $(2,3)\left(x_2, y_2\right)$ is $(-7,-1)$
$\therefore$ Equation is $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$
$
\begin{aligned}
& \Rightarrow \frac{y-3}{-1-3}=\frac{x-2}{-7-2} \Rightarrow \frac{y-3}{-4}=\frac{x-2}{-9} \\
& \Rightarrow \frac{y-3}{4}=\frac{x-2}{9} \\
& \Rightarrow 9 \mathrm{y}-27=4 \mathrm{x}-8 \\
& \Rightarrow 4 \mathrm{x}-9 \mathrm{y}-8+27=0 \\
& \Rightarrow 4 \mathrm{x}-9 \mathrm{y}+19=0
\end{aligned}
$
Question 8.
A cat is located at the point $(-6,-4)$ in xy plane. A bottle of milk is kept at $(5,11)$. The cat wish to consume the milk traveling through shortest possible distance. Find the equation of the path it needs to take its milk.
Solution:
$
\begin{aligned}
& \mathrm{A}=\left(\mathrm{x}_1, \mathrm{y}_1\right)=(-6,-4) \\
& \mathrm{B}=\left(\mathrm{x}_2, \mathrm{y}_2\right)=(5,11)
\end{aligned}
$
Shortest path between $\mathrm{A}$ and $\mathrm{B}$ is a line joining $\mathrm{A}$ and $\mathrm{B}$.
$
\begin{aligned}
m & =\frac{y_2-y_1}{x_2-x_1}=\frac{11-(-4)}{5-(-6)}=\frac{15}{11} \\
y-y_1 & =m\left(x-x_1\right) \\
\Rightarrow \quad y-(-4) & =\frac{15}{11}(x-(-6)) \\
\Rightarrow \quad(y+4) \times 11 & =15 \times(x-(-6)) \\
\Rightarrow \quad 11 y+44 & =15 x+90 \\
\Rightarrow 15 x-11 y+46 & =0
\end{aligned}
$
Question 9.
Find the equation of the median and altitude of $\triangle \mathrm{ABC}$ through $\mathrm{A}$ where the vertices are $\mathrm{A}(6,2)$ $\mathrm{B}(-5,-1)$ and $\mathrm{C}(1,9)$
Solution:

Let $\mathrm{AP'}$ be the altitude
$
\Rightarrow \mathrm{AP} \perp \mathrm{BC}
$
Slope of BC $=\frac{9-(-1)}{1-(-5)}$
$
=\frac{10}{6}=\frac{5}{3}
$
Slope of AP $=\frac{-1}{\text { Slope of }(B C)}=\frac{-3}{5}$
quation of line AP: $y-2=\frac{-3}{5}(x-6)$
$
\Rightarrow 5 y-10=-3 x+18
$

$
\Rightarrow 3 x+5 y=28
$
Let $\mathrm{AQ}$ be the median $\Rightarrow \mathrm{Q}$ is mid point of $\mathrm{BC}$
$
\Rightarrow Q=\left(\frac{-5+1}{2}, \frac{-1+9}{2}\right)=(-2,4)
$
Slope of $\mathrm{AQ}=\frac{4-2}{-2-6}$
$
=\frac{2}{-8}=\frac{-1}{4}
$
$\therefore$ equation of line $\mathrm{AQ}: y-2=\frac{-1}{4} \times(x-6)$
$
\begin{aligned}
& \Rightarrow \quad y-2=\frac{-1}{4}(x-6) \\
& 4 y-8=-x+6 \\
& \Rightarrow \quad x+4 y=14 \\
& \Rightarrow x+4 y-14=0 \\
&
\end{aligned}
$
Question 10.
Find the equation of a straight line which $-5$ has slope $\frac{-5}{4}$ and passing through the point $(-1,2)$. Solution:
$
\begin{aligned}
m & =\frac{-5}{4} \quad \text { point }=(-1,2) \\
\Rightarrow \quad y-2 & =\frac{-5}{4}(x-(-1)) \\
\Rightarrow \quad y-2 & =\frac{-5}{4}(x+1) \\
\Rightarrow 4(y-2) & =-5(x+1) \\
\Rightarrow 4 y-8 & =-5 x-5 \\
\Rightarrow 5 x+4 y & =3 \Rightarrow 5 x+4 y-3=0
\end{aligned}
$
Question 11.
You are downloading a song. The percent $y$ (in decimal form) of mega bytes remaining to get downloaded in $\mathrm{x}$ seconds is given by $\mathrm{y}=-0.1 \mathrm{x}+1$.
(i) graph the equation.

(ii) find the total MB of the song.
(iii) after how many seconds will $75 \%$ of the song gets downloaded?
(iv) after how many seconds the song will be downloaded completely?
Solution:
(i) $y=-0.1 x+1$
when $x=0 \Rightarrow y=1$
when $y=0 \Rightarrow y=10$

(ii) Total MB of song can be obtained when time $=0$
$
\begin{aligned}
& \therefore \mathrm{x}=0 \\
& \Rightarrow \mathrm{y}=1 \mathrm{MB}
\end{aligned}
$
(iii) time when $75 \%$ of song is downloaded
$
\begin{aligned}
& \Rightarrow \text { remaining } \%=25 \% \Rightarrow y=0.25 \\
& 0.25=-0.1 \mathrm{x}+1 \\
& \Rightarrow 0.1 \mathrm{x}=0.75 \\
& \Rightarrow 7.5 \text { Seconds }
\end{aligned}
$
(iv) song will downloaded completely when, remaining $\%=0 \Rightarrow y=0$
$
\begin{aligned}
& \Rightarrow 0=-0.1 \mathrm{x}+1 \\
& \Rightarrow \mathrm{x}=10 \\
& \therefore 10 \text { seconds }
\end{aligned}
$
Question 12.
Find the equation of a line whose intercepts on the $\mathrm{x}$ and $\mathrm{y}$ axes are given below.
(i) $4,-6$
(ii) $-5 \frac{3}{4}$

Solution:
$
\text { (i) } \begin{aligned}
\frac{x}{a}+\frac{y}{b}= & 1 \\
& \Rightarrow \frac{x}{4}+\frac{y}{-6}=1 \\
& \Rightarrow \frac{3 x-2 y}{12}=1 \\
& \Rightarrow 3 x-2 y=12 \\
& \Rightarrow 3 x-2 y-12=0
\end{aligned}
$
$
\begin{aligned}
& \text { (ii) } \frac{x}{-5}+\frac{y}{\frac{3}{4}}=1 \\
& \Rightarrow \frac{-x}{5}+\frac{4 y}{3}=1 \\
& \Rightarrow \frac{-3 x+20 y}{15}=1 \\
& \Rightarrow-3 x+20 y=15 \\
& 3 x-20 y+15=0
\end{aligned}
$
Question 13.
Find the intercepts made by the following lines on the coordinate axes,
(i) $3 x-2 y-6=0$
(ii) $4 x+3 y+12=0$
Solution:
(i) The given equation is
$
\begin{aligned}
& 3 \mathrm{x}-2 \mathrm{y}-6=0 \\
& 3 \mathrm{x}-2 \mathrm{y}=6
\end{aligned}
$
Divided by 6
$
\begin{aligned}
& \frac{3 x}{6}-\frac{2 y}{6}=\frac{6}{6} \\
& \frac{x}{2}-\frac{y}{3}=1 \Rightarrow \frac{x}{2}+\frac{y}{-3}=1
\end{aligned}
$
(Comparing with $\frac{x}{a}+\frac{y}{b}=1$ )
$\therefore \mathrm{x}$ intercept $=2 ; \mathrm{y}$ intercept $=-3$
(ii) The given equation is
$
4 x+3 y+12=0
$

$
4 \mathrm{x}+3 \mathrm{y}=-12
$
Divided by $-12$
$
\begin{aligned}
& \frac{4 x}{-12}+\frac{3 y}{-12}=\frac{-12}{-12} \\
& \frac{x}{-3}+\frac{y}{-4}=1
\end{aligned}
$
(Comparing with $\frac{x}{a}+\frac{y}{b}=1$ )
$\therefore \mathrm{x}$ intercept $=-3 ; \mathrm{y}$ intercept $=-4$

Question 14.
Find the equation of a straight line
(i) passing through $(1,-4)$ and has intercepts which are in the ratio $2: 5$
(ii) passing through $(-8,4)$ and making equal intercepts on the coordinate axes Solution:
(i) ratio of intercept $=2: 5$
$
\begin{aligned}
& \therefore \text { Slope of line }=\frac{-5}{2} \Rightarrow m=\frac{-5}{2} \\
& y-y_1=m\left(x-x_1\right) \\
& \Rightarrow \quad y-(-4)=\frac{-5}{2}(x-1) \\
& \Rightarrow \quad 2(y+4)=-5(x-1) \\
& \Rightarrow \quad 2 y+8=-5 x+5 \\
& \Rightarrow \quad 5 x+2 y+3=0 \\
&
\end{aligned}
$
(ii) Slope of line $=\frac{y \text { intercept }}{x \text { intercept }} \times-1=-1$
$
\begin{aligned}
& y-y_1=m\left(x-x_1\right) \\
& \Rightarrow \quad y-4=-1(x-(-8)) \\
& y-4=-x-8 \\
& \Rightarrow \quad x+y+4=0 \\
&
\end{aligned}
$