Additional Questions - Chapter 5 - Coordinate Geometry - 10th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Questions
Question 1.
Find a relation between $\mathrm{x}$ and $\mathrm{y}$ such that the point $(\mathrm{x}, \mathrm{y})$ is equidistant from the points $(7,1)$ and $(3,5)$
Solution:
Let $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be equidistant from the points $\mathrm{A}(7,1)$ and $\mathrm{B}(3,5)$.
We are given that $\mathrm{AP}=\mathrm{BP}$. So, $\mathrm{AP}^2=\mathrm{BP}^2$
$
\begin{aligned}
& (x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2 \\
& x^2-14 x+49+y^2-2 y+1=x^2-6 x+9+y^2-10 y+25 \\
& x-y=2
\end{aligned}
$
Which is the required relation

Question 2.
Show that the points $(1,7),(4,2),(-1,-1)$ and $(-4,4)$ are the vertices of a square.
Solution:

Let $\mathrm{A}(1,7), \mathrm{B}(4,2), \mathrm{C}(-1,-1)$ and $\mathrm{D}(-4,4)$ be the given points. To prove that $\mathrm{ABCD}$ is a square, we have to prove that all its sides are equal and both its diagonals are equal.
$
\begin{aligned}
& \mathrm{d}=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\
& \mathrm{AB}=\sqrt{(1-4)^2+(7-4)^2}=\sqrt{9+25}=\sqrt{34} \\
& \mathrm{BC}=\sqrt{(4+1)^2+(2+1)^2}=\sqrt{25+9}=\sqrt{34} \\
& \mathrm{CD}=\sqrt{(-1+4)^2+(-1-4)^2}=\sqrt{9+25}=\sqrt{34} \\
& \mathrm{DA}=\sqrt{(1+4)^2+(7-4)^2}=\sqrt{25+9}=\sqrt{34} \\
& \mathrm{AC}=\sqrt{(1+1)^2+(7+1)^2}=\sqrt{4+64}=\sqrt{68} \\
& \mathrm{BD}=\sqrt{(4+4)^2+(2-4)^2}=\sqrt{64+4}=\sqrt{68}
\end{aligned}
$
Since, $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$ and $\mathrm{AC}=\mathrm{BD}$, all the four sides of the quadrilateral $\mathrm{ABCD}$ are equal and its diagonals $\mathrm{AC}$ and $\mathrm{BD}$ are also equal. Therefore, $\mathrm{ABCD}$ is a square.
Question 3.
If $\mathrm{A}(-5,7), \mathrm{B}(-4,-5), \mathrm{C}(-1,-6)$ and $\mathrm{D}(4,5)$ are the vertices of a quadrilateral, find the area of the quadrilateral $\mathrm{ABCD}$.
Solution:
By joining $\mathrm{B}$ to $\mathrm{D}$, you will get two triangles $\mathrm{ABD}$ and $\mathrm{BCD}$.
Now, the area of $\triangle \mathrm{ABD}$

By joining $\mathrm{B}$ to $\mathrm{D}$, you will get two triangles $\mathrm{ABD}$ and $\mathrm{BCD}$.
Now, the area of $\triangle \mathrm{ABD}$

$=\frac{1}{2}[-5(-5-5)+(-4)(5-7)+4(7+5)]$ $=\frac{1}{2}(50+8+48)$ $=\frac{106}{2}=53$ square units. Also, the area of $\Delta B C D$
$
\begin{aligned}
& =\frac{1}{\cdot 2}[-4(-6-5)-1(5+5)+4(-5+6)] \\
& =\frac{1}{2}(44-10+4) \\
& =19 \text { square units. }
\end{aligned}
$
So, the area of quadrilateral $\mathrm{ABCD}=53+19=72$ square units.
Question 4.
Find the coordinates of the points of trisection (i.e. points dividing in three equal parts) of the line segment joining the points $\mathrm{A}(2,-2)$ and $\mathrm{B}(-7,4)$.
Solution:
Let $P$ and $Q$ be the points of trisection at $A B$. i.e., $A P=P Q=Q B$
Therefore, $\mathrm{P}$ divides $\mathrm{AB}$ internally in the ratio $1: 2$. Therefore, the coordinates at $\mathrm{P}$, by applying the section formula, are
$
\left[\frac{1(-7)+2(2)}{1+2}, \frac{1(7)+2(-2)}{1+2}\right] \text { i.e., }(-1,10)
$
Now, $\mathrm{Q}$ also divides $\mathrm{AB}$ internally in the ratio $2: 1$, so, the coordinates at $\mathrm{Q}$ are
$
\left[\frac{2(-7)+1(2)}{2+1}, \frac{2(4)+(-2)}{2+1}\right] \text { i.e., }(-4,2)
$
Therefore, the coordinates of the points at trisection of the line segment joining $\mathrm{A}$ and $\mathrm{B}$ are $(-1,0)$ and $(-4,2)$.

Question 5.
If the points $\mathrm{A}(6,1), \mathrm{B}(8,2), \mathrm{C}(9,4)$ and $\mathrm{D}(\mathrm{P}, 3)$ are the vertices of a parallelogram, taken in order. Find the value of $P$.
Solution:
We know that diagonals of a parallelogram bisect each other.
So, the coordinates at the mid-point of $\mathrm{AC}=$ coordinates of the mid-point of $\mathrm{BD}$.
$
\text { i.e., } \begin{aligned}
{\left[\frac{6+9}{2}, \frac{1+4}{2}\right] } & =\left[\frac{8+P}{2}, \frac{2+3}{2}\right] \\
{\left[\frac{15}{2}, \frac{5}{2}\right] } & =\left[\frac{8+P}{2}, \frac{5}{2}\right] \\
\frac{15}{2} & =\frac{8+P}{2} \\
\mathrm{P} & =7
\end{aligned}
$
Question 6.
Find the area of a triangle whose vertices are $(1,-1),(-4,6)$ and $(-3,-5)$.
Solution:
The area of the triangle formed by the vertices $\mathrm{A}(1,-1), \mathrm{B}(-4,6)$ and $\mathrm{C}(-3,-5)$, by using the formula
$
\begin{aligned}
& \Delta=\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\
& =\frac{1}{2}[1(6+5)+(-4)(-5+1)+(-3)(-1-6)] \\
& =\frac{1}{2}[11+16+21]=24 \text { square units. }
\end{aligned}
$
So, the area of the triangle is 24 square units.
Question 7.
If $\mathrm{A}(-2,-1), \mathrm{B}(\mathrm{a}, 0), \mathrm{C}(4, \mathrm{~b})$ and $\mathrm{D}(1,2)$ are the vertices of a parallelogram, find the values of a and b.
Solution:
We know that the diagonals of a parallelogram bisect each other. Therefore the co-ordinates of the

midpoint of $\mathrm{AC}$ are same as the co-ordinates of the mid-point of $\mathrm{BD}$. i.e.
$
\begin{aligned}
& \left(\frac{-2+4}{2}, \frac{-1+b}{2}\right)=\left(\frac{a+1}{2}, \frac{0+2}{2}\right) \\
& \Rightarrow \quad\left(1, \frac{b-1}{2}\right)=\left(\frac{a+1}{2}, 1\right) \\
& \Rightarrow \quad \frac{a+1}{2}=1 \Rightarrow a+1=2 \Rightarrow a=1 \\
& \Rightarrow \quad \frac{b-1}{2}=1 \Rightarrow b-1=2 \Rightarrow b=3
\end{aligned}
$
Question 8.
Find the area of the quadrilateral whose vertices, taken in order, are $(-3,2),(5,4),(7,-6)$ and $(-5$, -4).
Solution:
We have Area of the quadrilateral

$
\begin{aligned}
& =\frac{1}{2}[(-12-30-28-10)-(+10+28+30+12)] \\
& \frac{1}{2}[-80-(80)] \\
& \frac{1}{2}[-160]=-80=80 \text { square units. } \\
& \quad(\because \text { Area is always }+\mathrm{ve})
\end{aligned}
$
Question 9.
Find the area of the triangle formed by the points $\mathrm{P}(-1.5,3), \mathrm{Q}(6,-2)$ and $\mathrm{R}(-3,4)$.
Solution:
The area of the triangle formed by the given points is equal to
$
\begin{aligned}
& =\frac{1}{2}[-1.5(-2-4)+6(4-3)+(-3)(3+2)] \\
& =\frac{1}{2}[9+6-15]=0
\end{aligned}
$
Can we have a triangle of area 0 square units? What does this mean? If the area of a triangle is 0 square units, then its vertices will be collinear.
Question 10.
Find the value of $\mathrm{k}$ if the pointsA( $(2,3), \mathrm{B}(4, \mathrm{k})$ and $(6,-3)$ are collinear.
Since the given points are collinear, the area a the triangle formed by them must be 0 , i.e.
$
=\frac{1}{2}[2(k+3)+4(-3-3)+6(3-k)]=0
$

$
\begin{aligned}
& =\frac{1}{2}[-4 k]=0 \\
& k=0
\end{aligned}
$
Area of $\triangle \mathrm{ABC}$
$
=\frac{1}{2}[2(0+3)+4(-3-3)+6(3-0)]=0
$