Exercise 5.4 - Chapter 5 Coordinate Geometry 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\mathbf{E} \times 5.4$
Question 1.
Find the slope of the following straight lines
(i) $5 y-3=0$
(ii) $7 x-\frac{3}{17}=0$
Solution:
(ii)
$
\begin{aligned}
7 x-\frac{3}{17} & =0 \Rightarrow 7 x+0 y-\frac{3}{17}=0 \\
\text { Slope } & =\frac{-\text { Co }-\text { efficient of } x}{\text { Co-efficient of } y} \\
& =\frac{-7}{0}=\infty \text { (undefined) }
\end{aligned}
$
Question 2.
Find the slope of the line which is
(i) parallel to $y=0.7 x-11$
(ii) perpendicular to the line $\mathrm{x}=-11$
Solution:
(i) $y=0.7 x-11$
line parallel to $y=0.7 x-11$ is $y=0.7 x+C$

If the lines are parallel, slopes are equal $\therefore$ The slope of the required line is 0.7.
(ii) $\mathrm{m}=\tan \theta=\tan 90^{\circ}=\infty$ undefined.

Question 3.
Check whether the given lines are parallel or perpendicular
(i) $\frac{x}{3}+\frac{y}{4}+\frac{1}{7}=0$ and $\frac{2 x}{3}+\frac{y}{2}+\frac{1}{10}=0$
(ii) $5 x+23 y+14=0$ and $23 x-5 y+9=0$
Solution:
(i) $\frac{x}{3}+\frac{y}{4}+\frac{1}{7}=0$

and
$
\begin{aligned}
& \frac{2 x}{3}+\frac{y}{2}+\frac{1}{10}=0 \\
& \frac{y}{4}=-\frac{x}{3}-\frac{1}{7} \\
& y=-\frac{4 x}{3}-\frac{4}{7} \\
& m_1=-\frac{4}{3} \\
& \frac{y}{2}=-\frac{2 x}{3}-\frac{1}{10} \\
& y=-\frac{4 x}{3}-\frac{1}{10} \\
& m_2=-\frac{4}{3} \\
& m_1=m_2 \\
&
\end{aligned}
$
$\therefore$ They are Parallel.

(ii)
$
\begin{aligned}
5 x+23 y+14 & =0 \\
23 y & =-5 x-14 \\
y & =-\frac{5 x}{23} x-\frac{14}{23} \\
m_1 & =-\frac{5}{23} \\
23 x-5 y+9 & =0 \\
-5 y & =-23 x-9 \\
y & =\frac{-23}{-5} x-\frac{-9}{-5} \\
y & =\frac{23}{5} x+\frac{9}{5} \\
m_2 & =\frac{23}{5} \\
\therefore m_1 \times m_2 & =\frac{-5}{23} \times \frac{23}{5}=-1 \\
m_1 \times m_2 & =-1
\end{aligned}
$
$\therefore$ They are $\perp^{\mathrm{r}}$.
Question 4.
If the straight lines $12 y=-(p+3) x+12,12 x-7 y=16$ are perpendicular then find ' $p$ '. Solution:
$
\begin{aligned}
12 y & =-(p+3) x+12 \\
y & =-\frac{p+3}{12} x+\frac{12}{12} \\
y & =-\frac{p+3}{12} x+1
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \quad m_1=-\frac{p+3}{12} \\
& 12 x-7 y=16 \\
& -7 y=-12 x+16 \\
& y=\frac{-12}{-7} x+\frac{16}{-7} \\
& y=\frac{12}{7} x-\frac{16}{7} \\
& \Rightarrow \quad m_2=\frac{12}{7} \\
& m_1 \times m_2=-1 \\
& -\frac{p+3}{12} \times \frac{12}{7}=-1 \\
& -\left(\frac{p+3}{7}\right)=-1 \\
& \frac{p+3}{7}=1 \\
& p+3=7 \\
& p=4 \\
&
\end{aligned}
$
Question 5.
Find the equation of a straight line passing through the point $\mathrm{P}(-5,2)$ and parallel to the line joining the points $\mathrm{Q}(3,-2)$ and $\mathrm{R}(-5,4)$.

Solution:
$
\begin{aligned}
& \text { Slope of } \mathrm{QR}=\frac{y_2-y_1}{x_2-x_1} \\
& =\frac{4-(-2)}{-5-3}=\frac{4+2}{-8}=\frac{6}{-8} \\
& m=\frac{-3}{4}
\end{aligned}
$
$\mathrm{P}$ is $(-5,2)$
Require equation is
$
\begin{aligned}
y-y_1 & =m\left(x-x_1\right) \\
y-2 & =\frac{-3}{4}(x+5) \\
4 y-8 & =-3 x-15 \\
3 x+4 y+7 & =0
\end{aligned}
$
Question 6.
Find the equation of a line passing through $;(6,-2)$ and perpendicular to the line joining the points $(6,7)$ and $(2,-3)$.
Solution:
Slope of line joining $(6,7)$ and $(2,-3)$ is
$
=\frac{-3-7}{2-6}=\frac{-10}{-4}=\frac{5}{2}
$
Slope of the $\perp^r$ line $=\frac{-2}{5}$
Required equation is
$
\begin{aligned}
y+2 & =\frac{-2}{5}(x-6) \\
5 y+10 & =-2 x+12 \\
2 x+5 y-2 & =0
\end{aligned}
$

Question 7.
$\mathrm{A}(-3,0) \mathrm{B}(10,-2)$ and $\mathrm{C}(12,3)$ are the vertices of $\triangle \mathrm{ABC}$. Find the equation of the altitude through A and B.
Solution:
$
\mathrm{A}(-3,0), \mathrm{B}(10,-2), \mathrm{C}(12,3)
$
Since $\mathrm{AD} \perp \mathrm{BC}$

$
\text { Slope } \mathrm{AD}=\frac{-1}{\text { Slope of } \mathrm{BC}}
$
$\therefore$ Slope of $\mathrm{BC}=\frac{y_2-y_1}{x_2-x_1}=\frac{3-(-2)}{12-10}=\frac{5}{2}$
$\therefore$ Slope of $\mathrm{AD}=\frac{-1}{5 / 2}=\frac{-2}{5}$
$\therefore$ Equation of $\mathrm{AD} \Rightarrow y-y_1=m\left(x-x_1\right)$
$
\begin{aligned}
y-0 & =\frac{-2}{5}(x-(-3)) \\
5 y & =-2 x-6 \\
2 x+5 y+6 & =0
\end{aligned}
$
Since $\mathrm{BE} \perp \mathrm{AC}$
Slope of BE $=\frac{-1}{\text { Slope of } \mathrm{AC}}$

$
\begin{aligned}
& \mathrm{B}(10,-2), \text { slope of } \mathrm{BE}=\frac{-1}{1 / 5}=-5 \\
& \therefore \text { Equation of } \mathrm{BE} \Rightarrow y-(-2)=-5(x-10) \\
& y+2=-5 x+50 \\
& 5 x+y+2-50=0 \\
& 5 x+y-48=0
\end{aligned}
$
(1), (2) are the required equations of the altitudes through A and B.
Question 8.
Find the equation of the perpendicular bisector of the line joining the points $\mathrm{A}(-4,2)$ and $\mathrm{B}(6,-4)$.
Solution:
Mid Point $\mathrm{AB}$ is

$\mathrm{C}$ is $\left(\frac{-4+6}{2}, \frac{2+(-4)}{2}\right)=\left(\frac{2}{2}, \frac{-2}{2}\right)=(1,-1)$

$\therefore$ Equation of $\mathrm{CD}$ is
$
\begin{aligned}
& y-(-1)=\frac{5}{3}(x-1) \\
& 3(y+1)=5 x-5 \Rightarrow 3 y+3=5 x-5
\end{aligned}
$
$5 x-3 y-8=0$ is the required equation of the line.
Question 9.
Find the equation of a straight line through the intersection of lines $7 \mathrm{x}+3 \mathrm{y}=10,5 \mathrm{x}-4 \mathrm{y}=1$ and parallel to the line $13 x+5 y+12=0$
Solution:
$
\begin{aligned}
& l_1, \| l_2 \\
& \therefore \text { Slope of } l_1=\frac{-13}{5}
\end{aligned}
$
$1_1$ passes through the intersecting point.

Question 10.
Find the equation of a straight line through the intersection of lines $5 \mathrm{x}-6 \mathrm{y}=2,3 \mathrm{x}+2 \mathrm{y}=10$ and perpendicular to the line $4 x-7 y+13=0$
Solution:

$
\begin{array}{r}
5 x=\frac{14+66}{7} \\
x=\frac{80}{35}=\frac{16}{7} \\
\text { Slope }=\frac{-7}{4} \\
\therefore \text { The intersecting point is }\left(\frac{16}{7}, \frac{11}{7}\right)
\end{array}
$
$\therefore$ The required equation is
$
\begin{aligned}
y-\frac{11}{7} & =-\frac{7}{4}\left(\frac{x}{1}-\frac{16}{7}\right) \\
4 y-4\left(\frac{11}{7}\right) & =-7\left(\frac{7 x-16}{7}\right) \\
4 y-\frac{44}{7} & =-7 x+16 \\
\frac{28 y-44}{7} & =-7 x+16 \\
28 y-44 & =-49 x+112
\end{aligned}
$
$49 x+28 y-156=0$ is the required equation of the line.
Question 11.
Find the equation of a straight line joining the point of intersection of $3 x+y+2=0$ and $x-2 y-4$ $=0$ to the point of intersection of $7 \mathrm{x}-3 \mathrm{y}=-12$ and $2 \mathrm{y}=\mathrm{x}+3$
Solution:

$
\begin{aligned}
& 3 x+y+2=0 \\
& x-2 y-4=0
\end{aligned}
$
Solving (1) and (2) we get the point of intersection of the lines (1) and (2).
$
\begin{aligned}
& 3 x+y=-2 \\
& \text { (2) } \times 3 \Rightarrow \quad 3 x-6 y=12 \\
& \frac{(-) \quad(+) \quad(-)}{7 y=-14} \Rightarrow y=-2 \\
&
\end{aligned}
$
Substitute $y=-2$ in (1), we get
$
\begin{aligned}
3 x+(-2) & =-2 \\
3 x & =-2+2=0 \\
x & =0
\end{aligned}
$
The intersecting point is $(0,-2)$

$
\begin{aligned}
& 7 x-3 y=-12 \\
& x-2 y=-3 \\
& \text { (4) } \times 7 \Rightarrow 7 y-14 y=-21 \\
& \text { (3) } \Rightarrow 7 x-3 y=-12 \\
& \begin{array}{rr}
(-) \quad(+) \quad(+) \\
\hline-11 y=-9
\end{array} \\
& y=\frac{9}{11} \\
&
\end{aligned}
$
Substitute $y=\frac{9}{11}$ in (4), we get
$
\begin{aligned}
x-2 \times \frac{9}{11} & =-3 \\
x-\frac{18}{11} & =-3 \\
\Rightarrow x & =-3+\frac{18}{11}=\frac{-33+18}{11} \\
x & =\frac{-15}{11}
\end{aligned}
$
$\therefore$ The intersecting point of (3) and (4) is $\left(\frac{-15}{11}, \frac{9}{11}\right)$
$\therefore$ The required line passes through $(0,-2)$ and
$
\begin{array}{r}
\left(\frac{-15}{11}, \frac{9}{11}\right) \\
\therefore \text { Equation }=\frac{y-(-2)}{\frac{9}{11}+2}=\frac{x-0}{\frac{-15}{11}-0}
\end{array}
$

$
\begin{array}{ll}
\Rightarrow & \frac{y+2}{\frac{31}{11}}=\frac{x}{\frac{-15}{11}} \\
\Rightarrow & \frac{31}{11} x=\frac{-15}{11}(y+2)
\end{array}
$
$31 x+15 y+30=0$ is the required equation of the line.
Question 12.
Find the equation of a straight line through the point of intersection of the lines $8 x+3+=18,4 x+$ $5+=9$ and bisecting the line segment joining the points $(5,-4)$ and $(-7,6)$.
Solution:
The intersecting point of the lines
$
\begin{aligned}
& 8 x+3 y=18 \\
& 4 x+5 y=9 \\
& \text { solving (1) and (2) } \\
& \text { (1) } \Rightarrow \quad 8 t+3 y=18 \\
& \text { (2) } \times 2 \Rightarrow \quad 8 x+10 y=18 \\
& \Rightarrow \quad \frac{(-)(-)(+)}{-7 y=0} \\
& y=0 \\
&
\end{aligned}
$
Substitute $y=0$ in (1), we get
$
\begin{aligned}
8 x+3(0) & =18 \\
x & =\frac{18}{4}=\frac{9}{4}
\end{aligned}
$
$\therefore$ The intersecting point is $\left(\frac{9}{4}, 0\right)$

The mid point of the line joining the two points $(5,-4)$ and $(-7,6)$ is
$
\left(\frac{5+(-7)}{2}, \frac{-4+6}{2}\right)=\left(\frac{-2}{2}, \frac{2}{2}\right)=(-1,1)
$
The required line is passing through the points
$
\begin{aligned}
& \left(\begin{array}{cc}
\frac{9}{4}, & 0 \\
x_1 & y_1
\end{array}\right) \text { and }\left(\begin{array}{cc}
-1, & 1 \\
x_2 & y_2
\end{array}\right) \\
& \frac{y-0}{1-0}=\frac{x-\frac{9}{4}}{-1-\frac{9}{4}} \\
& \frac{y}{1}=\frac{\frac{4 x-9}{4}}{\frac{-4-9}{4}} \\
& y=\frac{4 x-9}{-13} \\
& -13 y=4 x-9 \\
&
\end{aligned}
$
$\therefore 4 x+13 y-9=0$ is the required equation.