Exercise 5.2 - Chapter 5 Coordinate Geometry 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $5.2$
Question 1.
What is the slope of a line whose inclination with positive direction of $\mathrm{x}$-axis is
(i) $90^{\circ}$
(ii) $0^{\circ}$
Solution:
(i) $\theta=90^{\circ}$
$\mathrm{m}=\tan \theta=\tan 90^{\circ}=\propto$ (undefined)
(ii) $\mathrm{m}=\tan \theta=\tan 0^{\circ}=0$
Question 2.

What is the inclination of a line whose slope is (i) 0 Solution:
(i) Slope $=0$
$\tan \theta=0$
$\tan 0=0$
$\therefore \theta=0^{\circ}$
(ii) Slope $=1$
$\tan \theta=1$
$\tan 45^{\circ}=1$
$\therefore \theta=45^{\circ}$
angle of inclination is $45^{\circ}$
Question 3.
Find the slope of a line joining the points
(i) $(5, \sqrt{5}))$ with origin $(0,0)$
(ii) $(\sin \theta,-\cos \theta)$ and $(-\sin \theta, \cos \theta)$

Question 4.
What is the slope of a line perpendicular to the line joining $\mathrm{A}(5,1)$ and $\mathrm{P}$ where $\mathrm{P}$ is the mid-point of the segment joining $(4,2)$ and $(-6,4)$.
Solution:
$\mathrm{P}$ is the mid point of the segment joining $(4,2)$ and $(-6,4)$
$
\mathrm{P}(x, y)=\left(\frac{4+(-6)}{2}, \frac{2+4}{2}\right)=(-1,3)
$

Question 5.
Show that the given points are collinear $(-3,-4),(7,2)$ and $(12,5)$
Solution:
The verticles are $\mathrm{A}(-3,-4), \mathrm{B}(7,2)$ and $\mathrm{C}(12,5)$
$
\begin{aligned}
& \text { Slope of } A B=\frac{2-(-4)}{7-(-3)}=\frac{6}{10}=\frac{3}{5} \\
& \text { Slope of } B C=\frac{5-2}{12-7}=\frac{3}{5}
\end{aligned}
$
Slope of $\mathrm{AB}=$ Slope of $\mathrm{BC}$
$\therefore$ The points A, B and $\mathrm{C}$ lie on the same line.
$\therefore$ They are collinear.
Question 6.
If the three points $(3,-1),(a, 3),(1,-3)$ are collinear, find the value of $a$.
Solution:
Slope of $\mathrm{AB}=$ slope of $\mathrm{BC}$.
$
\begin{aligned}
\frac{3-(-1)}{a-3} & =\frac{-3-3}{1-a} \\
\frac{4}{a-3} & =\frac{-6}{1-a} \Rightarrow 4(1-a)=(a-3)(-6) \\
4-4 a & =-6 a+18 \\
2 a & =18-4=14 \\
a & =7
\end{aligned}
$

Question 7.
The line through the points $\left(-2\right.$, a) and $(9,3)$ has slope $-\frac{1}{2}$. Find the value of a.
Solution:
A line joining the points $(-2, a)$ and $(9,3)$ has slope $\mathrm{m}=-\frac{1}{2}$.
$
\begin{aligned}
m & =\frac{y_2-y_1}{x_2-x_1}=\frac{3-a}{9-(-2)}=\frac{-1}{2} \\
2(3-a) & =-1(11) \Rightarrow-2 a=-11-6=-17 \\
a & =\frac{17}{2} .
\end{aligned}
$
Question 8.
The line through the points $(-2,6)$ and $(4,8)$ is perpendicular to the line through the points $(8,12)$ and $(x, 24)$. Find the value of $x$.
Solution:
The line through the points A $(-2,6)$, and $\mathrm{B}(4,8)$
Slope of $\mathrm{AB}\left(m_1\right)=\frac{8-6}{4-(-2)}=\frac{2}{6}=\frac{1}{3}$
The line through the points $\mathrm{C}(8,12)$ and $\mathrm{D}(x, 24)$
Slope of $\mathrm{CD}\left(m_2\right)=\frac{24-12}{x-8}=\frac{12}{x-8}$
$\mathrm{AB} \perp^r \mathrm{CD} \Rightarrow m_1 \times m_2=-1$

Question 9.
Show that the given points form a right angled triangle and check whether they satisfies pythagoras theorem
(i) $\mathrm{A}(1,-4), \mathrm{B}(2,-3)$ and $\mathrm{C}(4,-7)$
(ii) $\mathrm{L}(0,5), \mathrm{M}(9,12)$ and $\mathrm{N}(3,14)$
Solution:
(i) Slope of $\mathrm{AB}=\frac{-3-(-4)}{2-1}=\frac{1}{1}=1$
Slope of $\mathrm{BC}=\frac{-7-(-3)}{4-2}=\frac{-3}{2}$
Slope of $\mathrm{AC}=\frac{-7-(-4)}{4-1}=\frac{-7+4}{3}=\frac{-3}{3}=-1$
Slope of $\mathrm{AB} \times$ slope of $\mathrm{AC}=1 \times-1=-1$
$\therefore$ Sol : yes. $\mathrm{AB} \perp^r$ to $\mathrm{AC}$

Slope of $\mathrm{AB}=\frac{-3-(-4)}{2-1}=\frac{1}{1}=1$
Slope of $\mathrm{BC}=\frac{-7-(-3)}{4-2}=\frac{-3}{2}$
Slope of $\mathrm{AC}=\frac{-7-(-4)}{4-1}=\frac{-7+4}{3}=\frac{-3}{3}=-1$
Slope of $\mathrm{AB} \times$ slope of $\mathrm{AC}=1 \times-1=-1$
$\therefore$ Sol : yes. $\mathrm{AB} \perp^r$ to $\mathrm{AC}$

By Pythagoras theorem
$
\mathrm{AB}^2+\mathrm{AC}^2=\mathrm{BC}^2 \text {. }
$
$
\begin{aligned}
\mathrm{AB} & =\sqrt{(-3-(-4))^2+(2-1)^2}=\sqrt{1^2+1^2}=\sqrt{2} \\
\mathrm{AC} & =\sqrt{(-7-(-4))^2+(4-(1))^2}=\sqrt{(-3)^2+3^2} \\
& =\sqrt{9+9}=\sqrt{18}
\end{aligned}
$

$
\begin{aligned}
& \mathrm{BC}=\sqrt{(-7-(-3))^2+(4-2)^2}=\sqrt{-4^2+2^2} \\
& =\sqrt{16+4}=\sqrt{20} \\
& \text { By Pythagoras theorem } \\
& \quad \mathrm{BC}^2=\mathrm{AB}^2+\mathrm{AC}^2 \\
& \sqrt{20}^2=\sqrt{2}^2+\sqrt{18}^2
\end{aligned}
$
$20=2+18=20 \therefore$ Hence it is satisfied

(ii) Slope of $\mathrm{LM}=\frac{12-5}{9-0}=\frac{7}{9}$
Slope of $\mathrm{MN}=\frac{14-12}{3-9}=\frac{2}{-6}=\frac{1}{-3}$
Slope of $\mathrm{LN}=\frac{14-5}{3-0}=\frac{\bigotimes^{\circ}}{\not \beta}=3$
Slope of MN $\times$ slope of $\mathrm{LN}=\frac{-1}{3} \times 3=-1$ Yes; MN $\perp^r$ to LN.
$\therefore \mathrm{L}, \mathrm{M}, \mathrm{N}$ form a right angled triangle.
$
\begin{aligned}
& \mathrm{LM}=\sqrt{(12-5)^2+(9-0)^2}=\sqrt{7^2+9^2}=\sqrt{49+81}=\sqrt{130} \\
& \mathrm{MN}=\sqrt{(14-12)^2+(3-9)^2}=\sqrt{2^2+(-6)^2}=\sqrt{4+36}=\sqrt{40} \\
& \mathrm{LN}=\sqrt{(14-5)^2+(3-0)^2}=\sqrt{9^2+3^2}=\sqrt{81+9}=\sqrt{90}
\end{aligned}
$
By Pythagoras theorem
$
\begin{aligned}
& \mathrm{LM}^2=\mathrm{MN}^2+\mathrm{LN}^2 \\
& \sqrt{130}^2=\sqrt{40}^2+\sqrt{90}^2 \Rightarrow 130=40+90 .
\end{aligned}
$
Hence it is satisfied.
Question 10.
Show that the given points form a parallelogram : $\mathrm{A}(2.5,3.5), \mathrm{B}(10,-4), \mathrm{C}(2.5,-2.5)$ and $\mathrm{D}(-5,5)$

Solution:

$
\begin{aligned}
& \mathrm{B}(10,-4) \\
& \mathrm{C}(2.5,-2.5) \\
& \therefore m_1=m_2 \therefore \mathrm{AD} \| \mathrm{BC} \\
& \text { Slope of } \mathrm{AB}\left(m_3\right)=\frac{-4-3.5}{10-2.5}=\frac{-7.5}{7.5}=-1 . \\
& \mathrm{A}(2.5,3.5) \\
& \mathrm{B}(10,-4) \\
& \text { Slope of } \mathrm{CD}\left(m_4\right)=\frac{5-(-2.5)}{-5-2.5}=\frac{7.5}{-7.5}=-1 \\
& \mathrm{C}(2.5,-2.5) \\
& \mathrm{D}(-5,5) \\
& m_3=m_4 .
\end{aligned}
$
From (1) and (2), the opposite sides of the quadrilateral are parallel to each other.
$
\text { Mid point of } \begin{aligned}
\mathrm{AC} & =\left(\frac{2.5+2.5}{2}, \frac{3.5-2.5}{2}\right) \\
& =(2.5, .5)
\end{aligned}
$
$\&$ mid point of $\mathrm{BD}=\left(\frac{10-5}{2}, \frac{-4+5}{2}\right)$
$
\begin{array}{r}
=(2.5, .5)[\because \text { mid point of AC } \\
=\text { mid point of BD }]
\end{array}
$
$\therefore$ The given points form a parallelogram.

Question 11.
If the points $\mathrm{A}(2,2), \mathrm{B}(-2,-3), \mathrm{C}(1,-3)$ and $\mathrm{D}(\mathrm{x}, \mathrm{y})$ form a parallelogram then find the value of $\mathrm{x}$ and $y$.
Solution:
$
\mathrm{A}(2,2), \mathrm{B}(-2,-3), \mathrm{C}(1,-3), \mathrm{D}(\mathrm{x}, \mathrm{y})
$

Since $\mathrm{ABCD}$ forms a parallelogram, slope of opposite sides are equal and diagonals bisect each other.
Mid point of $\mathrm{BD}=$ Mid point of $\mathrm{AC}$
$
\begin{array}{rlrl}
\left(\frac{x+(-2)}{2},\right. & \left.\frac{y+(-3)}{2}\right)=\left(\frac{2+1}{2}\right. & \left., \frac{2+(-3)}{2}\right) \\
\frac{x-2}{2} & =\frac{3}{2} & \frac{y-3}{2} & =\frac{2-3}{2} \\
x-2 & =3 & y-3 & =-1 \\
x & =5 & y & =2
\end{array}
$

Question 12.
Let $\mathrm{A}(3,-4), \mathrm{B}(9,-4), \mathrm{C}(5,-7)$ and $\mathrm{D}(7,-7)$. Show that $\mathrm{ABCD}$ is a trapezium.
Solution:
A $(3,-4), B(9,-4), C(5,-7)$ and $D(7,-7)$

If only one pair of opposite sides of a quadrilateral are parallel, then it is said to be a trapezium.
$
\therefore \text { Slope of } \mathrm{AB}\left(m_1\right)=\frac{-4-(-4)}{9-3}=\frac{0}{6}=0
$
Slope of CD $\left(m_2\right)=\frac{-7-(-7)}{7-5}=\frac{-7+7}{+2}=\frac{0}{+2}=0$

Question 13.
A quadrilateral has vertices at $\mathrm{A}(-4,-2), \mathrm{B}(5,-1), \mathrm{C}(6,5)$ and $\mathrm{D}(-7,6)$. Show that the mid-points of its sides form a parallelogram
Solution:
Mid point of $\mathrm{AB}=\left(\frac{-4+5}{2}, \frac{-2-1}{2}\right)=\mathrm{E}\left(\frac{1}{2}, \frac{-3}{2}\right)$

Mid point of $\mathrm{BC}=\left(\frac{5+6}{2}, \frac{-1+5}{2}\right)=\mathrm{F}\left(\frac{11}{2}, \frac{4}{2}\right)$
Mid point of $\mathrm{CD}=\left(\frac{6+(-7)}{2}, \frac{5+6}{2}\right)=\mathrm{G}\left(\frac{-1}{2}, \frac{11}{2}\right)$
Mid point of $\mathrm{AD}=\left(\frac{-4-7}{2}, \frac{-2+6}{2}\right)=\mathrm{H}\left(\frac{-11}{2}, \frac{4}{2}\right)$
$\therefore$ Slope of $E F=\left(\frac{\frac{4}{2}-\left(\frac{-3}{2}\right)}{\frac{11}{2}-\frac{1}{2}}\right)=\frac{\frac{4+3}{2}}{\frac{10}{2}}=\frac{7}{10}$
Slope of $F G=\left(\frac{\frac{11}{2}-\frac{2}{1}}{\frac{-1}{2}-\frac{11}{2}}\right)=\left(\frac{\frac{11-4}{2}}{\frac{-12}{2}}\right)=\frac{7}{-12}$
Slope of $\begin{aligned} \mathrm{GH} & =\left(\frac{2-\frac{11}{2}}{\frac{-11}{2}-\left(\frac{-1}{2}\right)}\right)=\left(\frac{\frac{4-11}{2}}{\frac{-10}{2}}\right) \\ & =\frac{+7}{10}\end{aligned}$

$
\begin{aligned}
& \text { Slope of HE }=\left(\frac{2-\left(\frac{-3}{2}\right)}{\frac{-11}{2}-\frac{1}{2}}\right)=\left(\frac{\frac{4+3}{2}}{\frac{-12}{2}}\right)=\frac{7}{-12} \\
& \left(\frac{\frac{1}{2}+\left(\frac{-1}{2}\right)}{2}, \frac{\frac{-3}{2}+\frac{11}{2}}{2}\right)=\left(\frac{\frac{-11}{2}+\frac{11}{2}}{2}, \frac{\frac{4}{2}+\frac{4}{2}}{2}\right) \\
& \left(0, \frac{8}{2}\right)=(0,4)
\end{aligned}
$
In a parallelogram diagonals bisect each other. Opposite sides are parallel as their slopes are equal the mid points of the diagonals are the same.
$\therefore$ Mid points of the sides of a quadrilateral form a parallelogram.
Question 14.
$P Q R S$ is a rhombus. Its diagonals $P R$ and $Q S$ intersect at the point $M$ and satisfy $Q S=2 P R$. If the coordinates of $S$ and $M$ are $(1,1)$ and $(2,-1)$ respectively, find the coordinates of $P$.
Solution:

$\mathrm{M}$ is the mid point of QS.
$
\begin{aligned}
\frac{x_2+1}{2} & =2 \Rightarrow x_2=3 \\
\frac{y_2+1}{2} & =-1 \Rightarrow y_2=-3 \\
\Rightarrow \quad \mathrm{Q} & =(3,-3)
\end{aligned}
$

$\begin{aligned}
& \therefore \frac{x_1+x_3}{2}=2 \Rightarrow x_1+x_3=4 \\
& \frac{y_1+y_3}{2}=-1 \Rightarrow y_1+y_3=-2 \\
& \left(\frac{y_3-y_1}{x_3-x_1}\right) \times\left(\frac{-3-1}{3-1}\right)=-1 \\
& \left(\because m_1 \times m_2=-1\right\} \\
& \Rightarrow \frac{y_3-y_1}{x_3-x_1}=\frac{1}{2} \Rightarrow x_3-x_1=2\left(y_3-y_1\right) \\
& \therefore \mathrm{QS}=2 \mathrm{PR} \\
& \mathrm{QS}=\sqrt{(-3-1)^2+(3-1)^2} \\
& =\sqrt{(-4)^2+(2)^2}=\sqrt{20} \\
& \Rightarrow \quad \mathrm{PR}=\frac{\sqrt{20}}{2} \\
&
\end{aligned}$

$
\begin{aligned}
& \Rightarrow \sqrt{\left(x_3-x_1\right)^2+\left(y_3-y_1\right)^2}=\frac{\sqrt{20}}{2} \\
& \text { from (3), } \\
& \Rightarrow \sqrt{\left[2\left(y_3-y_1\right)\right]^2+\left(y_3-y_1\right)^2}=\frac{\sqrt{20}}{2} \\
& \Rightarrow \quad \sqrt{5\left(y_3-y_1\right)^2}=\frac{\sqrt{20}}{2} \\
& \Rightarrow \quad\left(y_3-y_1\right) \times \sqrt{8}=\frac{\sqrt{8} \times \sqrt{4}}{2} \\
& \Rightarrow \quad y_3-y_1=1 \rightarrow(4) \\
& \therefore x_3-x_1=2 \rightarrow(5) \\
&
\end{aligned}
$
Solving (4) and (2),
(4) $+(2) \Rightarrow 2 y_3=-1 \Rightarrow y_3=\frac{-1}{2}$
(4) $-(2) \Rightarrow-2 y_1=3 \Rightarrow y_1=\frac{-3}{2}$
Solving (5) and (1),
$
\begin{aligned}
& (5)+(1) \Rightarrow 2 x_3=6 \Rightarrow x_3=3 \\
& (5)-(1) \Rightarrow-2 x_1=-2 \Rightarrow x_1=+1 \\
& \therefore \mathrm{P}=\left(+1, \frac{-3}{2}\right) \text { or }
\end{aligned}
$
If $y_3-y_1=-1$ from (4), we get $p=\left(3, \frac{-1}{2}\right)$