Check here the Latest detailed Answers and Solutions as per the latest guidelines of Exercise 5.1 - Chapter 5 Coordinate Geometry 10th Maths Guide Samacheer Kalvi Solutions - [2024-2025]

Exercise 5.1 - Chapter 5 Coordinate Geometry 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $5.1$ : Chapter 5 - Coordinate Geometry - 10th Maths Guide Samacheer Kalvi Solutions
Question 1.
Find the area of the triangle formed by the points
(i) $(1,-1),(-4,6)$ and $(-3,-5)$
(ii) $(-10,-4),(-8,-1)$ and $(-3,-5)$
Solution:
(i) $(1,-1),(-4,6)$, and $(-3,-5)$,

$\begin{aligned}
& \mathrm{A}(-4,6), \quad \mathrm{B}(-3,-5), \quad \mathrm{C}(1,-1) \\
& \left(x_1, y_1\right) \quad\left(x_2, y_2\right) \quad\left(x_3, y_3\right) \\
& \therefore \text { Area of the } \triangle \mathrm{ABC}= \\
& \frac{1}{2}\left[\left(x_1 y_2+x_2 y_3+x_3 y_1\right)-\left(x_2 y_1+x_3 y_2+x_1 y_3\right)\right]_{\text {sq }} \text {. units } \\
& =\frac{1}{2}\left[\begin{array}{l}
((-4 \times-5)+(-3 \times-1)+(1 \times 6)) \\
-((-3 \times 6)+(1 \times-5)+(-4 \times-1))
\end{array}\right] \\
& =\frac{1}{2}[(20+3+6)-(-18-5+4)] \\
& =\frac{1}{2}[29-(-19)]=\frac{1}{2}[29+19]=\frac{1}{2} \times 48 \\
& =24 \text { sq. units } \\
&
\end{aligned}$

$\text { (ii) }(-10,-4),(-8,-1) \text { and }(-3,-5)$

Question 2.
Detemine whether the sets of points are collinear?
(i) $\left(-\frac{1}{2}, 3\right),(-5,6)$ and $(-8,8)$
(ii) $(a, b+c),(b, c+a)$ and $(c, a+b)$

Solution:
(i)

$
=\frac{1}{2}[-67+67]=0 \text { sq. units }
$
$\therefore$ The given points are collinear.

Question 3.
Vertices of given triangles are taken in order and their areas are provided aside. In each case, find the value of ' $p$ '.

Solution:
Area 20 sq. units.

Question 4.
In each of the following, find the value of ' $a$ ' for which the given points are collinear.
(i) $(2,3),(4, a)$ and $(6,-3)$
(ii) $(\mathrm{a}, 2-2 \mathrm{a}),(-\mathrm{a}+1,2 \mathrm{a})$ and $(-4-\mathrm{a}, 6-2 \mathrm{a})$
Solution:

$
\begin{aligned}
& {\left[\left(2 a^2+(-a+1) \times(6-2 a)+(-4-a) \times(2-2 a)\right)\right.} \\
&-((2-2 a)(-a+1)+2 a(-4-a)+a(6-2 a))]=0 \\
& \Rightarrow \quad 2 a^2-6 a+6+2 a^2-2 a+(-8)-2 a+8 a+2 a^2 \\
&-\left[-2 a+2 a^2+2-2 a-8 a-2 a^2+6 a-2 a^2\right]=0 \\
& \Rightarrow 2 a^2-6 a+6+2 a^2-2 a-8-2 a+8 a+2 a^2+2 a \\
&-2 a^2-2+2 a+8 a+2 a^2-6 a+2 a^2=0 \\
& 8 a^2+4 a-4=0 \\
& 2 a^2+a-1=0 \\
& 2 a^2+2 a-a-1=0 \\
& 2 a(a+1)-1(a+1)=0 \\
&(2 a-1)(a+1)=0 \\
& 2 a-1=0 \\
& \Rightarrow \quad a=\frac{1}{2} \\
& \Rightarrow \quad a+1=0 \\
& a \quad a=-1
\end{aligned}
$
Question 5.
Find the area of the quadrilateral whose vertices are at
(i) $(-9,-2),(-8,-4),(2,2)$ and $(1,-3)$
(ii) $(-9,0),(-8,6),(-1,-2)$ and $(-6,-3)$
Solution:
(i) $(-9,-2),(-8,-4),(2,2)$, and $(1,-3)$

$\text { (ii) }(-9,0),(-8,6),(-1,-2) \text { and }(-6,-3)$

Question 6.
Find the value of $\mathrm{k}$, if the area ofa quadrilateral is 28 sq.units, whose vertices are $(-4,-2),(-3, \mathrm{k})$, $(3,-2)$ and $(2,3)$
Solution:

Question 7.
If the points $\mathrm{A}(-3,9), \mathrm{B}(\mathrm{a}, \mathrm{b})$ and $\mathrm{C}(4,-5)$ are collinear and if $\mathrm{a}+\mathrm{b}=1$, then find $\mathrm{a}$ and $\mathrm{b}$.

Solution:

Question 8.
Let $\mathrm{P}(11,7), \mathrm{Q}(13.5,4)$ and $\mathrm{R}(9.5,4)$ be the mid-points of the sides $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{AC}$ respectively of $\triangle \mathrm{ABC}$. Find the coordinates of the vertices $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$. Hence find the area of $\triangle \mathrm{ABC}$ and compare this with area of $\triangle \mathrm{PQR}$.
Solution:
$\mathrm{p}(11,7), \mathrm{Q}(13.5,4)$, and $\mathrm{R}(9.5,4)$ are the mid points of the sides of a $\triangle \mathrm{ABC}$.

$\begin{aligned}
\text { Mid point }(x, y) & =\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \\
\therefore \text { Mid point of } \mathrm{AB} & =\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \\
\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) & =(11,7) \\
\Rightarrow \quad \frac{x_1+x_2}{2} & =11 \\
\Rightarrow \quad x_1+x_2 & =22
\end{aligned}$

$\begin{aligned}
& \frac{y_1+y_2}{2}=7 \\
& \Rightarrow \quad y_1+y_2=14 \\
& \text { Mid point of } \mathrm{BC}=\left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right) \\
& =(13.5,4) \\
& \Rightarrow \quad \frac{x_2+x_3}{2}=13.5 \\
& \Rightarrow \quad x_2+x_3=27.0 \\
& \Rightarrow \quad \frac{y_2+y_3}{2}=4 \\
& \Rightarrow \quad y_2+y_3=8 \\
& \text { Mid point } \mathrm{AC}=\left(\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2}\right) \\
& =(9.5,4) \\
& \Rightarrow \quad \frac{x_1+x_3}{2}=9.5 \\
& x_1+x_3=19.0 \\
& \frac{y_1+y_3}{2}=4 \\
& y_1+y_3=8 \\
& \text { (1) }+(3)+(5) \rightarrow 2\left(x_1+x_2+x_3\right)=68 \\
& x_1+x_2+x_3=34 \\
& \text { (7) }-(1) \Rightarrow \quad x_3=34-22=12 \\
& \text { (7) }-(3) \Rightarrow \quad x_1=34-27=7 \\
& \text { (7) }-(5) \Rightarrow \quad x_2=34-19=15 \\
& (2)+(4)+(6) \rightarrow 2\left(y_1+y_2+y_3\right)=30 \\
&
\end{aligned}$

$\Rightarrow \quad y_1+y_2+y_3=15$
(8) $-$ (2) $\rightarrow y_3=15-14=1$
(8) $-$ (4) $\rightarrow y_1=15-8=7$
(8) $-$ (6) $\rightarrow y_2=15-8=7$
$\therefore$ The vertices of the $\triangle \mathrm{ABC}$ are $\mathrm{A}(7,7) \mathrm{B}(15,7) \mathrm{C}(12,1) \Rightarrow \mathrm{A}(7,7) \mathrm{B}(12,1)$ C $(15,7)$

$
\begin{aligned}
& =\frac{1}{2}[(44+38+94.5)-(66.5+54+44)] \\
& =\frac{1}{2}(176.5-164.5) \\
& =\frac{1}{2} \times(-12)-6=6 \text { sq. units }
\end{aligned}
$
(Area cannot be (-ve). Area is always (+ve))
$
\begin{aligned}
\therefore \text { Area of } \triangle \mathrm{PRQ} & =6 \text { sq. units } \\
\text { Area of } \triangle \mathrm{ABC} & =24 \text { sq. units } \\
\therefore \text { Area of } \triangle \mathrm{ABC} & =4 \times \text { Area of } \triangle \mathrm{PRQ}
\end{aligned}
$
Question 9.
In the figure, the quadrilateral swimming pool shown is surrounded by concrete patio. Find the area of the patio.

Solution:
Area of the patio $=$ Area of the quadrilateral $\mathrm{ABCD}$ - Area of the swimming pool EFGFI.

Question 10.
A triangular shaped glass with vertices at $\mathrm{A}(-5,-4), \mathrm{B}(1,6)$ and $\mathrm{C}(7,-4)$ has to be painted. If one bucket of paint covers 6 square feet, how many buckets of paint will be required to paint the whole glass, if only one coat of paint is applied.
Solution: