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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Questions
Exercise $5.1$
Question 1.
State whether the following statements are true/false.
(i) $(5,7)$ is a point in the IV quadrant.
(ii) $(-2,-7)$ is a point in the III quadrant.
(iii) $(8,-7)$ lies below the $\mathrm{x}$-axis.
(iv) $(-2,3)$ lies in the II quadrant.
(v) For any point on the $\mathrm{x}$-axis its $\mathrm{y}$-coordinate is zero.
Solution:
(i) False
(ii) True
(iii) True
(iv) True
(v) True

Question $2 .$
Locate the points
(i) $(3,5)$ and $(5,3)$
(ii) $(-2,-5)$ and $(-5,-2)$ in the rectangular coordinate system.
Solution:
$\begin{aligned}
&{ }^{\prime} \mathrm{AB}=\sqrt{(4-7)^{2}+(2-5)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18} \\
&\mathrm{BC}=\sqrt{(9-7)^{2}+(7-5)^{2}}=\sqrt{2^{2}+2^{2}}=\sqrt{4+4}=\sqrt{8}
\end{aligned}$
$\mathrm{CA}=\sqrt{(9-4)^{2}+(7-2)^{2}}=\sqrt{5^{2}+5^{2}}=\sqrt{25+25}=\sqrt{50}$
So, $\mathrm{AB}=\sqrt{18}=\sqrt{9 \times 2}=3 \sqrt{2}$
$\begin{aligned}
&\mathrm{BC}=\sqrt{8}=\sqrt{4 \times 2}=2 \sqrt{2} \\
&\mathrm{CA}=\sqrt{50}=\sqrt{25 \times 2}=5 \sqrt{2}
\end{aligned}$
This gives $\mathrm{AB}+\mathrm{BC}=3 \sqrt{2}+2 \sqrt{2}=5 \sqrt{2}=\mathrm{AC}$
Hence the points $A, B$ and $C$ are collinear.

Question 3.
In which quadrant does the following points lie?
(i) $(5,2)$
(ii) $(-5,-8)$
(iii) $(-7,1)$
(iv) $(8,-3)$
Solution:
(i) I quadrant
(ii) III quadrant
(iii) II quadrant
(iv) IV quadrant.

Question $4 .$
Write down the ordinate of the following points.
(i) $(7,5)$
(ii) $(2,9)$
(iii) $(-5,8)$
(iv) $(7,-4)$
Solution:
(i) 5
(ii) 9

(iii) 8
(iv) $-4$ (ordinate is the $y$-coordinate)
 

Exercise $5.2$
Question $1 .$
Find the distance between the following pairs of points.
(i) $(-4,0)$ and $(3,0)$
(ii) $(-7,2)$ and $(5,2)$
Solution:
(i) The points $(-4,0)$ and $(3,0)$ lie on the $x$-axis. Hence,
$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}=\sqrt{(3+4)^{2}+0^{2}}=\sqrt{7^{2}}=\sqrt{49}=7$
(ii) The points $(5,2)$ and $(-7,2)$ lie on a line parallel to the $\mathrm{x}$-axis. Hence the distance $d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}=\sqrt{(5+7)^{2}+(2-2)^{2}}=\sqrt{12^{2}}=\sqrt{144}=12$

Question $2 .$
Show that the three points $(4,2),(7,5)$ and $(9,7)$ lie on a straight line.
Solution:
Let the points be $\mathrm{A}(4,2), \mathrm{B}(7,5)$ and $\mathrm{C}(9,7)$. By the distance formula.

$\begin{aligned} \mathrm{AB} &=\sqrt{(4-7)^{2}+(2-5)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18} \\ \mathrm{BC} &=\sqrt{(9-7)^{2}+(7-5)^{2}}=\sqrt{2^{2}+2^{2}}=\sqrt{4+4}=\sqrt{8} \\ \mathrm{CA} &=\sqrt{(9-4)^{2}+(7-2)^{2}}=\sqrt{5^{2}+5^{2}}=\sqrt{25+25}=\sqrt{50} \\ \text { So, } \mathrm{AB} &=\sqrt{18}=\sqrt{9 \times 2}=3 \sqrt{2} \\ \mathrm{BC} &=\sqrt{8}=\sqrt{4 \times 2}=2 \sqrt{2} \\ \mathrm{CA} &=\sqrt{50}=\sqrt{25 \times 2}=5 \sqrt{2} \end{aligned}$
This gives $\mathrm{AB}+\mathrm{BC}=3 \sqrt{2}+2 \sqrt{2}=5 \sqrt{2}=\mathrm{AC}$
Hence the points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are collinear.
 

Question $3 .$
Determine whether the points are vertices of a right triangle $A(-3,-4), B(2,6)$ and $C(-6,10)$.
Solution:
Using the distance formula $d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
$\begin{aligned}
\mathrm{AB} &=\sqrt{(2+3)^{2}+(6+4)^{2}}=\sqrt{5^{2}+10^{2}} \\
&=\sqrt{25+100}=\sqrt{125} \\
\mathrm{BC} &=\sqrt{(-6-2)^{2}+(10-6)^{2}}=\sqrt{(-8)^{2}+4^{2}} \\
&=\sqrt{64+16}=\sqrt{80} \\
\mathrm{CA} &=\sqrt{(-6+3)^{2}+(10+4)^{2}}=\sqrt{(-3)^{2}+(14)^{2}} \\
&=\sqrt{9+196}=\sqrt{205} \\
\text { i.e } \mathrm{AB}+\mathrm{BC} &=\sqrt{125}+\sqrt{80}=\sqrt{205}=\mathrm{CA}
\end{aligned}$
Hence $\mathrm{ABC}$ is a right angled triangle since the square of one side is equal to sum of the squares of the other two sides.
 

Question $4 .$
Show that the points $(a, a),(-a,-a)$ and $(-a \sqrt{3}, a \sqrt{3})$ form an equilateral triangle.
Solution:
Let the points be represented by $\mathrm{A}(\mathrm{a}, \mathrm{a}), \mathrm{B}(-\mathrm{a},-\mathrm{a})$ and $\mathrm{C}(-a \sqrt{3}, a \sqrt{3})$ using the distance formula.

$\begin{aligned}
d &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}, \text { we have } \\
\mathrm{AB} &=\sqrt{(a+a)^{2}+(a+a)^{2}}=\sqrt{(2 a)^{2}+(2 a)^{2}}=\sqrt{4 a^{2}+4 a^{2}}=\sqrt{8 a^{2}}=2 a \sqrt{2} \\
\mathrm{BC} &=\sqrt{(-a \sqrt{3}+a)^{2}+(a \sqrt{3}+a)^{2}} \\
&=\sqrt{3 a^{2}+a^{2}-2 a^{2} \sqrt{3}+3 a^{2}+a^{2}+2 a^{2} \sqrt{3}}=\sqrt{8 a^{2}}=\sqrt{4 \times 2 a^{2}}=2 a \sqrt{2} \\
\mathrm{CA} &=\sqrt{(a+a \sqrt{3})^{2}+(a-a \sqrt{3})^{2}}=\sqrt{a^{2}+2 a^{2} \sqrt{3}+3 a^{2}+a^{2}-2 a^{2} \sqrt{3}+3 a^{2}} \\
&=\sqrt{8 a^{2}}=2 a \sqrt{2} \\
\therefore \mathrm{AB} &=\mathrm{BC}=\mathrm{CA}=2 a \sqrt{2}
\end{aligned}$
Since all the sides are equal the points form an equilateral triangle.

Question $5 .$
Prove that the points $(-7,-3),(5,10),(15,8)$ and $(3,-5)$ taken in order are the corners of a parallelogram.

Solution:
Let $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D represent the points $(-7,-3),(5,10),(15,8)$ and $(3,-5)$ respectively.
Using the distance formula $d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$, we find
$\begin{aligned}
& A B=\sqrt{(5+7)^{2}+(10+3)^{2}}=\sqrt{12^{2}+13^{2}}=\sqrt{144+169}=\sqrt{313} \\
B C &=\sqrt{(15-5)^{2}+(8-10)^{2}}=\sqrt{10^{2}+(-2)^{2}}=\sqrt{100+4}=\sqrt{104} \\
C D &=\sqrt{(3-15)^{2}+(-5-8)^{2}}=\sqrt{(-12)^{2}+(-13)^{2}}=\sqrt{144+169}=\sqrt{313} \\
\text { DA } &=\sqrt{(3+7)^{2}+(-5+3)^{2}}=\sqrt{10^{2}+(-2)^{2}}=\sqrt{100+4}=\sqrt{104} \\
\text { So } A B &=C D=\sqrt{313} \text { and } B C=D A=\sqrt{104}
\end{aligned}$
i.e. The opposite sides are equal. Hence $\mathrm{ABCD}$ is a parallelogram.
 

Question $6 .$
Show that the following points $\mathrm{A}(3,1) \mathrm{B}(6,4)$ and $\mathrm{C}(8,6)$ lies on a straight line.
Solution:
Using the distance formula, we have
$\odot$
$\begin{aligned}
A B &=\sqrt{(6-3)^{2}+(4-1)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} \\
B C &=\sqrt{(8-6)^{2}+(6-4)^{2}}=\sqrt{(4+4)}=\sqrt{8}=2 \sqrt{2} \\
A C &=\sqrt{(8-3)^{2}+(6-1)^{2}}=\sqrt{25+25}=\sqrt{50}=5 \sqrt{2} \\
A B+B C &=3 \sqrt{2}+2 \sqrt{2}=5 \sqrt{2}=A C
\end{aligned}$
Therefore the points lie on a straight line.

Question $7 .$
If the distance between the points $(5,-2),(1, a)$ is 5 units. Find the value of a.
Solution:
$\begin{gathered}
d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\
\sqrt{(1-5)^{2}+(a+2)^{2}}=5 \Rightarrow \sqrt{16+(a+2)^{2}}=5 \\
16+(a+2)^{2}=25
\end{gathered}$
(By squaring on both the sides)
$(a+2)^{2}=25-16 \Rightarrow(a+2)^{2}=9 \Rightarrow(a+2)=\pm 3$
(By taking the square root on both side)
$\begin{aligned}
&a=-2 \pm 3 \\
&a=-2+3 \text { (or) } \\
&a=-2-3 \\
&a=1 \text { or }-5 .
\end{aligned}$

Exercise $5.3$
Quéstion 1.
$\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are vertices of $\triangle \mathrm{ABC} . \mathrm{D}, \mathrm{E}$ and $\mathrm{F}$ are mid points of sides $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{AC}$ respectively. If the coordinates of $\mathrm{A}, \mathrm{D}$ and $\mathrm{F}$ are $(-3,5),(5,1)$ and $(-5,-1)$ respectively. Find the coordinates of $\mathrm{B}, \mathrm{C}$ and $\mathrm{E}$.

Solution:

Question $2 .$
If $\mathrm{A}(10,11)$ and $\mathrm{B}(2,3)$ are the coordinates of end points of diameter of circle. Then find the centre of the circle.
Solution:

\text { Centre of the circle }=\left(\frac{10+2}{2}, \frac{11+3}{2}\right)=(6,7)
Centre of the circle $=\left(\frac{10+2}{2}, \frac{11+3}{2}\right)=(6,7)$

Question $3 .$
Find the coordinates of the point which divides the line segment joining the points $(3,1)$ and $(5,13)$ internally in the ratio $3: 5$.
Solution:
$\left(\frac{5(3)+3(5)}{5+3}, \frac{5(1)+3(13)}{5+3}\right)=\left(\frac{15+15}{8}, \frac{5+39}{8}\right)=\left(\frac{30}{8}, \frac{44}{8}\right)=\left(\frac{15}{4}, \frac{11}{2}\right)$

Exercise $5.4$
Question 1.
Using section formula, show that the points $\mathrm{A}(7,-5), \mathrm{B}(9,-3)$ and $\mathrm{C}(13,1)$ are collinear.
Solution:

Question $2 .$
A car travels at an uniform speed. At $2 \mathrm{pm}$ it is at a distance of $5 \mathrm{~km}$ at $6 \mathrm{pm}$ it is at a distance of $120 \mathrm{~km}$. Using section formula, find at what distance it will reach 2 mid night.
Solution:
$\begin{aligned}
120 &=\frac{8(50)+4(y)}{12} \\
1440 &=400+4 y \\
4 y &=1040 \\
y &=\frac{1040}{4}=260 \mathrm{~km}
\end{aligned}$

Question $3 .$
Find the coordinates of the point which divides the line segment joining the point $A(3,7)$ and $B(-11,-2)$ in the ratio $5: 1$.
Solution:
$(x, y)=\left(\frac{3(1)+5(-11)}{5+1}, \frac{1(7)+5(-2)}{6}\right)=\left(\frac{3-55}{6}, \frac{7-10}{6}\right)=\left(\frac{-52}{6}, \frac{-3}{6}\right)=\left(\frac{-26}{3}, \frac{-1}{2}\right)$

Exercise $5.5$
Question 1.
Find the centroid of the triangle whose vertices are $(2,-5),(5,11)$ and $(9,9)$
Solution:
$\mathrm{G}\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)=\left(\frac{2+5+9}{3}, \frac{-5+11+9}{3}\right)=\left(\frac{16}{3}, 5\right)$

Question $2 .$
If the centroid of a triangle is at $(10,-1)$ and two of its vertices are $(3,2)$ and $(5,-11)$. Find the third vertex of the triangle.
Solution:
$(10,-1)=\left(\frac{3+5+x}{3}, \frac{2-11+y}{3}\right)$

Exercise $5.6$
Multiple Choice Questions :
Question $1 .$
The point $(-2,7)$ lies is the quadrant
(1) I
(2) II
(3) III
(4) IV
Hint:
$(-,+)$ lies in $\mathrm{II}^{\text {nd }}$ quadrant
Solution:
(2) II
 

Question $2 .$
The point $(x, 0)$ where $x<0$ lies on
(1) $\mathrm{OX}$
(2) OY
(3) $\mathrm{OX}$
(4) $\mathrm{OY}$
Hint:
$(-, 0)$ lies on $O X$,
Solution:
(3) $\mathrm{OX}^{\prime}$

Question $3 .$
For a point $\mathrm{A}(\mathrm{a}, \mathrm{b})$ lying in quadrant III.
(1) $a>0, b<0$
(2) $a<0, b<0$
(3) $a>0, b>0$
(4) $a<0, b>0$
Hint:
$(-,-)$ lies in $\mathrm{III}^{\mathrm{ed}}$ quadrant
Solution:
(2) $a<0, b<0$
 

Question $4 .$
The diagonal of a square formed by the points $(1,0)(0,1)$ and $(-1,0)$ is
(1) 2
(2) 4
(3) $\sqrt{2}$
(4) 8
Hint:
Distance $=\sqrt{(1-(-1))^{2}+(0-0)^{2}}=\sqrt{2^{2}}=2$
Solution:
(1) 2
 

Question $5 .$
The triangle obtained by joining the points $\mathrm{A}(-5,0) \mathrm{B}(5,0)$ and $\mathrm{C}(0,6)$ is
(1) an isosceles triangle
(2) right triangle
(3) scalene triangle
(4) an equilateral triangle
Hint:
Triangles having two sides equal are called isosceles.
Solution:
(a) an isosceles triangle