Exercise 5.2 - Chapter 5 Coordinate Geometry 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $5.2$
Question 1.

Find the distance between the following pairs of points.
(i) $(1,2)$ and $(4,3)$
(ii) $(3,4)$ and $(-7,2)$
(ii) $(a, b)$ and (c, b)
(iv) $(3,-9)$ and $(-2,3)$
Solution:
We know that distance,
(i) $A(1,2)$ and $B(4,3)$
$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
(ii) $(3,4)$ and $(-7,2)$
$\begin{aligned}
d &=\sqrt{(4-1)^{2}+(3-2)^{2}}=\sqrt{(3)^{2}+(1)^{2}} \\
&=\sqrt{9+1}=\sqrt{10} \text { units }
\end{aligned}$
$\begin{aligned}
d &=\sqrt{(-7-3)^{2}+(2-4)^{2}}=\sqrt{(-10)^{2}+(-2)^{2}} \\
&=\sqrt{100+4}=\sqrt{104}=2 \sqrt{26} \text { units }
\end{aligned}$
(iii) $(a, b)$ and $(c, b)$
$d=\sqrt{(c-a)^{2}+(b-b)^{2}}=\sqrt{(c-a)^{2}}=(c-a) \text { units }$
(iv) $(3,-9)$ and $(-2,3)$
$\begin{aligned}
d &=\sqrt{(-2-3)^{2}+(3+9)^{2}}=\sqrt{(-5)^{2}+(12)^{2}}=\sqrt{25+144} \\
&=\sqrt{169}=13 \text { units }
\end{aligned}$

Question $2 .$
Determine whether the given set of points in each case are collinear or not.
(i) $(7,-2),(5,1),(3,4)$
(ii) $(a,-2),(a, 3),(a, 0)$
Solution:
(i) Let the points be $\mathrm{A}(7,-2), \mathrm{B}(5,1)$ and $\mathrm{C}(3,4)$. By the distance formula.

$ \mathrm{AB}=\sqrt{(5-7)^{2}+(1+2)^{2}}=\sqrt{(-2)^{2}+(3)^{2}}=\sqrt{4+9}=\sqrt{13} $$ $$ \mathrm{BC}=\sqrt{(3-5)^{2}+(4-1)^{2}}=\sqrt{(-2)^{2}+(3)^{2}}=\sqrt{4+9}=\sqrt{13} $$ $$ \mathrm{CA}=\sqrt{(7-3)^{2}+(-2-4)^{2}}=\sqrt{(4)^{2}+(-6)^{2}} $ We see that $\mathrm{AB}+\mathrm{BC}=\sqrt{16+36}=\sqrt{52}=2 \sqrt{13}$
Show that the following points taken in order form an isosceles triangle.
(i) $\mathrm{A}(5,4), \mathrm{B}(2,0), \mathrm{C}(-2,3)$
(ii) $\mathrm{A}(6,-4), \mathrm{B}(-2,-4), \mathrm{C}(2,10)$
Solution:
(i) Let the points be A $(5,4), \mathrm{B}(2,0)$ and $\mathrm{C}(-2,3)$

$\begin{aligned} \mathrm{AB} &=\sqrt{(2-5)^{2}+(0-4)^{2}}=\sqrt{(-3)^{2}+(-4)^{2}} \\ &=\sqrt{9+16}=\sqrt{25}=5 \\ \mathrm{BC} &=\sqrt{(-2-2)^{2}+(3-0)^{2}}=\sqrt{(-4)^{2}+(3)^{2}} \\ &=\sqrt{16+9}=\sqrt{25}=5 \\ \mathrm{CA} &=\sqrt{(5+2)^{2}+(4-3)^{2}}=\sqrt{(7)^{2}+(1)^{2}} \\ &=\sqrt{49+1}=\sqrt{50}=5 \sqrt{2} \end{aligned}$

(ii) Let the points be A $(6,-4), \mathrm{B}(-2,-4)$ and $\mathrm{C}(2,10)$.
$\begin{aligned}
\mathrm{AB} &=\sqrt{(-2-6)^{2}+(-4+4)^{2}}=\sqrt{(-8)^{2}+0}=\sqrt{64}=8 \\
\mathrm{BC} &=\sqrt{(2+2)^{2}+(10+4)^{2}}=\sqrt{(4)^{2}+(14)^{2}}=\sqrt{16+196} \\
&=\sqrt{212}=2 \sqrt{53} \\
\mathrm{CA} &=\sqrt{(6-2)^{2}+(-4-10)^{2}}=\sqrt{(4)^{2}+(-14)^{2}} \\
&=\sqrt{16+196}=\sqrt{212}=2 \sqrt{53}
\end{aligned}$
Here $\mathrm{BC}+\mathrm{BA}>\mathrm{CA}$ and $\mathrm{BC}=\mathrm{CA}$. Two sides are equal, so $\triangle \mathrm{ABC}$ is an isosceles triangle

Question $4 .$
Show that the following points taken in order form an equilateral triangle in each case.
(i) $\mathrm{A}(2,2), \mathrm{B}(-2,-2), \mathrm{C}(-2 \sqrt{3}, 2 \sqrt{3})$
(ii) $\mathrm{A}(\sqrt{3}, 2), \mathrm{B}(0,1), \mathrm{C}(0,3)$
Solution:
(i) Let the points be $\mathrm{A}(2,2) \mathrm{B}(-2,-2)$ and $\mathrm{C}(-2 \sqrt{3}, 2 \sqrt{3})$
$\begin{aligned} \mathrm{AB} &=\sqrt{(-2-2)^{2}+(-2-2)^{2}}=\sqrt{(-4)^{2}+(-4)^{2}} \\ &=\sqrt{16+16}=\sqrt{32}=4 \sqrt{2} \\ \mathrm{BC} &=\sqrt{(-2 \sqrt{3}+2)^{2}+(2 \sqrt{3}+2)^{2}} \\ &=\sqrt{(4 \times 3-8 \sqrt{3}+4)+(4 \times 3+8 \sqrt{3}+4)} \\ &=\sqrt{16+16} \sqrt{32}=4 \sqrt{2} \end{aligned}$
$=\sqrt{16+16} \quad \sqrt{32}=4 \sqrt{2}$
$\begin{aligned}
\mathrm{CA} &=\sqrt{(2+2 \sqrt{3})^{2}+(2-2 \sqrt{3})^{2}}=\sqrt{\left(2^{2}+8 \sqrt{3}+4 \times 3\right)+\left(2^{2}-8 \sqrt{3}+4 \times 3\right)} \\
&=\sqrt{16+16} \sqrt{32}=4 \sqrt{2}
\end{aligned}$
$\mathrm{AB}=\mathrm{BC}=\mathrm{CA}=4 \sqrt{2}$
All the 3 sides of $\triangle \mathrm{ABC}$ are equal, Hence $\triangle \mathrm{ABC}$ is an equilateral triangle.

(ii) Let the points be $\mathrm{A}(\sqrt{3}, 2), \mathrm{B}(0,1)$ and $\mathrm{C}(0,3)$.
$\begin{aligned}
\mathrm{AB} &=\sqrt{(0-\sqrt{3})^{2}+(1-2)^{2}}=\sqrt{(-\sqrt{3})^{2}+(-1)^{2}} \\
&=\sqrt{3+1}=\sqrt{4}=2 \\
\mathrm{BC} &=\sqrt{(0-0)^{2}+(3-1)^{2}}=\sqrt{0+(2)^{2}}=\sqrt{4}=2 \\
\mathrm{CA} &=\sqrt{(\sqrt{3}-0)^{2}+(2-3)^{2}}=\sqrt{(\sqrt{3})^{2}+(-1)^{2}} \quad(0.1) \\
&=\sqrt{3+1}=\sqrt{4}=2
\end{aligned}$
$\therefore \mathrm{AB}=\mathrm{BC}=\mathrm{CA}=2$
All the 3 sides of $\triangle \mathrm{ABC}$ are equal. Hence $\triangle \mathrm{ABC}$ is an equilateral triangle.

Question $5 .$
Show that the following points taken in order form the vertices of a parallelogram.
(i) $\mathrm{A}(-3,1), \mathrm{B}(-6,-7), \mathrm{C}(3,-9)$ and $\mathrm{D}(6,-1)$
(ii) $\mathrm{A}(-7,-3), \mathrm{B}(5,10), \mathrm{C}(15,8)$ and $\mathrm{D}(3,-5)$
Solution:
(i) Let $A, B, C$ and D represent the points $(-3,1),(-6,-7)(3,-9)$ and $(6,-1)$ respectively.
$\begin{aligned}
\mathrm{AB} &=(-3,1)(-6,-7)=\sqrt{(-6+3)^{2}+(-7-1)^{2}}=\sqrt{(-3)^{2}+(-8)^{2}} \\
&=\sqrt{9+64}=\sqrt{73} \\
\mathrm{BC} &=(-6,-7)(3,-9) \\
&=\sqrt{(3+6)^{2}+(-9+7)^{2}}=\sqrt{(9)^{2}+(-2)^{2}} \\
&=\sqrt{81+4}=\sqrt{85} \\
\mathrm{CD} &=(3,-9)(6,-1) \\
&=\sqrt{(6-3)^{2}+(-1+9)^{2}}=\sqrt{(3)^{2}+(8)^{2}}=\sqrt{9+64}=\sqrt{73} \\
\mathrm{DA} &=(6,-1)(-3,1)=\sqrt{(-3-6)^{2}+(1+1)^{2}}=\sqrt{(-9)^{2}+(2)^{2}}=\sqrt{81+4}=\sqrt{85} \\
\text { So, } \mathrm{AB} &=\mathrm{CD}=\sqrt{73} \text { and } \mathrm{BC}=\mathrm{DA}=\sqrt{85}
\end{aligned}$
The opposite sides are equal. Hence $\mathrm{ABCD}$ is a parallelogram.

(ii) Let $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D represent the points $(-7,-3),(5,10)(15,8)$ and $(3,-5)$
$\begin{aligned}
\mathrm{AB} &=(-7,-3)(5,10) \\
&=\sqrt{(5+7)^{2}+(10+3)^{2}}=\sqrt{(12)^{2}+(13)^{2}}=\sqrt{144+169}=\sqrt{313} \\
\mathrm{BC} &=(5,10)(15,8)=\sqrt{(15-5)^{2}+(8-10)^{2}} \\
&=\sqrt{(10)^{2}+(-2)^{2}}=\sqrt{100+4}=\sqrt{104} \\
\mathrm{CD} &=(15,8)(3,-5)=\sqrt{(3-15)^{2}+(-5-8)^{2}} \\
&=\sqrt{(-12)^{2}+(-13)^{2}}=\sqrt{144+169}=\sqrt{313} \\
\mathrm{DA} &=(3,-5)(-7,-3) \\
&=\sqrt{(-7-3)^{2}+(-3+5)^{2}}=\sqrt{(-10)^{2}+(2)^{2}}=\sqrt{100+4}=\sqrt{104} \\
\text { So, } \mathrm{AB} &=\mathrm{CD}=\sqrt{313} \text { and } \mathrm{BC}=\mathrm{DA}=\sqrt{104}
\end{aligned}$
The opposite sides are equal. Hence $A B C D$ is a parallelogram.

Question $6 .$
Verify that the following points taken in order form the vertices of a rhombus.
(i) $\mathrm{A}(3,-2), \mathrm{B}(7,6), \mathrm{C}(-1,2)$ and $\mathrm{D}(-5,-6)$
(ii) $\mathrm{A}(1,1), \mathrm{B}(2,1), \mathrm{C}(2,2)$ and $\mathrm{D}(1,2)$
Solution:
(i) Let the points be $\mathrm{A}(3,-2), \mathrm{B}(7,6), \mathrm{C}(-1,2)$ and $\mathrm{D}(-5,-6)$
$\begin{aligned}
\mathrm{AB} &=(3,-2)(7,6)=\sqrt{(7-3)^{2}+(6+2)^{2}}=\sqrt{(4)^{2}+(8)^{2}} \\
&=\sqrt{16+64}=\sqrt{80}=4 \sqrt{5}
\end{aligned}$

$\mathrm{BC}=(7,6)(-1,2)=\sqrt{(-1-7)^{2}+(2-6)^{2}}$
$=\sqrt{(-8)^{2}+(-4)^{2}}$
$=\sqrt{64+16}=\sqrt{80}=4 \sqrt{5}$
$C D=(-1,2)(-5,-6)$
$=\sqrt{(-5+1)^{2}+(-6-2)^{2}}$
$=\sqrt{(-4)^{2}+(-8)^{2}}$
$=\sqrt{16+64}=\sqrt{80}=4 \sqrt{5}$
$\mathrm{DA}=(-5,-6)(3,-2)$
$=\sqrt{(3+5)^{2}+(-2+6)^{2}}=\sqrt{(8)^{2}+(4)^{2}}=\sqrt{64+16}=\sqrt{80}=4 \sqrt{5}$
All the four sides of quadrilateral ABCD are equal. Hence ABCD is a rhombus.

(ii) Let the points be $\mathrm{A}(1,1), \mathrm{B}(2,1), \mathrm{C}(2,2)$ and $\mathrm{D}(1,2)$
$\begin{aligned}
&\mathrm{AB}=(1,1)(2,1)=\sqrt{(2-1)^{2}+(1-1)^{2}}=\sqrt{1^{2}}=1 \\
&\mathrm{BC}=(2,1)(2,2)=\sqrt{(2-2)^{2}+(2-1)^{2}}=\sqrt{1^{2}}=1 \\
&\mathrm{CD}=(2,2)(1,2)=\sqrt{(1-2)^{2}+(2-2)^{2}}=\sqrt{(-1)^{2}}=1 \\
&\mathrm{DA}=(1,2)(1,1)=\sqrt{(1-1)^{2}+(1-2)^{2}}=\sqrt{(-1)^{2}}=1
\end{aligned}$
$\therefore$ All the four sides of quadrilateral $\mathrm{ABCD}$ are equal. Hence $\mathrm{ABCD}$ is a rhombus.

Question 7.
If $\mathrm{A}(-1,1), \mathrm{B}(1,3)$ and $\mathrm{C}(3, \mathrm{a})$ are points and if $\mathrm{AB}=\mathrm{BC}$, then find " $\mathrm{a}$ '
Solution:
$\begin{aligned}
\mathrm{A}(-1,1), \mathrm{B}(1,3), \mathrm{C}(3, a) \\
\mathrm{AB} &=\mathrm{BC} \\
\mathrm{AB} &=\sqrt{(1+1)^{2}+(3-1)^{2}}=\sqrt{2^{2}+2^{2}}=\sqrt{8} \\
\mathrm{BC} &=\sqrt{(3-1)^{2}+(a-3)^{2}}=\sqrt{2^{2}+a^{2}-6 a+9} \\
&=\sqrt{4+a^{2}-6 a+9} \\
\mathrm{AB} &=\mathrm{BC} \Rightarrow \sqrt{8}=\sqrt{a^{2}-6 a+13} \\
a^{2}-6 a+13 &=8 \\
a^{2}-6 a+5 &=0 \\
(a-1)(a-5) &=0 \\
a &=1,5
\end{aligned}$

Question $8 .$
The abscissa of a point $A$ is equal to its ordinate, and its distance from the point $B(1,3)$ is 10 units, what are the coordinates of A?
Solution:

$\begin{aligned}
\mathrm{BA}=\sqrt{\left(x_{2}-1\right)^{2}+\left(y_{2}-3\right)^{2}} &=10 \\
\sqrt{x_{2}^{2}-2 x_{2}+1+y_{2}^{2}-6 y+9} &=10 \\
\text { Since } x_{2} &=y_{2} \\
\sqrt{x_{2}^{2}-2 x_{2}+1+x_{2}^{2}-6 x_{2}+9} &=10 \\
\sqrt{2 x_{2}^{2}-8 x_{2}+10} &=100 \\
\text { Squaring on both sides } \quad 2 x_{2}{ }^{2}-8 x_{2}+10 &=100 \\
2 x_{2}{ }^{2}-8 x_{2}-90 &=0 \\
x_{2}{ }^{2}-4 x_{2}-45 &=0 \\
\left(x_{2}+5\right)\left(x_{2}-0\right) &=0
\end{aligned}$
Co-ordinates of A are $(-5,-5)$ or $(9,9)$
 

Question 9.
The point $(x, y)$ is equidistant from the points $(3,4)$ and $(-5,6)$. Find a relation between $x$ and $y$.
Solution:
$\mathrm{P}(\mathrm{x}, \mathrm{y})$ is equidistant from the points $\mathrm{A}(3,4)$ and $\mathrm{B}(-5,6)$
$\begin{aligned}
\mathrm{PA} &=\mathrm{PB} \\
\sqrt{(x-3)^{2}+(y-4)^{2}} &=\sqrt{(x-(-5))^{2}+(y-6)^{2}} \\
x^{\prime}-6 x+9+y^{z}-8 y+16 &=x^{x}+10 x+25+y^{2}-12 y+36 \\
10 x+25-12 y+36+6 x+8 y-9-16 &=0 \\
16 x-4 y+36 &=0 \\
4 x-y+9 &=0 \\
y &=4 x+9
\end{aligned}$

Question $10 .$
Let $\mathrm{A}(2,3)$ and $\mathrm{B}(2,-4)$ be two points. If $\mathrm{P}$ lies on the $\mathrm{x}$-axis, such that $\mathrm{AP}=\frac{3}{7} \mathrm{AB}$, find the coordinates of P.

Solution:
$\mathrm{A}(2,3), \mathrm{B}(2,-4), \mathrm{P}(x, y)$ lies on the $x$ axis. $\mathrm{P}(x, 0)$
$\begin{aligned}
\mathrm{AP} &=\frac{3}{7} \mathrm{AB} \\
\sqrt{(x-2)^{2}+(0-3)^{2}} &=\sqrt{(x-2)^{2}+(0-3)^{2}} \\
\sqrt{x^{2}-4 x+4+9} &=\frac{3}{7} \sqrt{(-7)^{2}} \\
7 \sqrt{x^{2}-4 x+13} &=3 \times 7 \Rightarrow \sqrt{x^{2}-4 x+13}=3 \\
x^{2}-4 x+13 &=9 \\
x^{2}-4 x+13-9 &=0 \Rightarrow x^{2}-4 x+4=0 \\
(x-2)^{2} &=0 \\
x &=2,2 \\
\text { The co-ordinates of } \mathrm{P}(x, 0) &=\mathrm{P}(2,0)
\end{aligned}$

Question $11 .$
Show that the point $(11,2)$ is the centre of the circle passing through the points $(1,2),(3,-4)$ and $(5,-6)$

Solution:
We have to show that $\mathrm{SA}=\mathrm{SB}=\mathrm{SC}$ $$ \begin{aligned} d &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\ &=\sqrt{(1-11)^{2}+(2-2)^{2}}=\sqrt{(-10)^{2}+0^{2}} \\ \mathrm{SA} &=\sqrt{(3-11)^{2}+(-4-2)^{2}} \\ &=\sqrt{(-8)^{2}+(-6)^{2}}=\sqrt{64+36} \end{aligned} $$ $\begin{aligned} \mathrm{SB} &=\sqrt{100}=10 \text { units } \\ \therefore \mathrm{SC}=\sqrt{(5-11)^{2}+(-6-2)^{2}}=\sqrt{(-6)^{2}+(-8)^{2}}=\sqrt{36+64}=\sqrt{100}=10 \text { units } \end{aligned}$ $\mathrm{SA}=\mathrm{SB}=\mathrm{SC}=10$ units.
Therefore $S$ is the centre of the circle, passing through $A, B$ and $C$.

Question $12 .$
The radius of a circle with centre at origin is 30 units. Write the coordinates of the points where the circle intersects the axes. Find the distance between any two such points.

Solution:

Equation of the circle $=\sqrt{x^{2}+y^{2}}=30$, Radius $=\mathrm{SA}=30$ units.
$\sqrt{(x-0)^{2}+(0-0)^{2}}=30$
$\sqrt{x^{2}+0^{2}}=30$
$x^{2}+0^{2}=900$
$x^{2}=900$ (where it cuts $x$ axis)
$x=\pm 30$
$\sqrt{(0-0)^{2}+(y-0)^{2}}=30$
$\sqrt{y^{2}}=30$
$y^{2}=900$
$y=\pm 30$
$\therefore$ The distance between the points $(30,0)$ and $(0,30)$ is
$\sqrt{(0-30)^{2}+(30-0)^{2}}=\sqrt{(-30)^{2}+30^{2}}=\sqrt{900+900}=\sqrt{1800}=\sqrt{2 \times 900}=30 \sqrt{2}$