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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Unit Exercise 5
Question 1.
$\mathrm{PQRS}$ is a rectangle formed by joining the points $\mathrm{P}(-1,-1), \mathrm{Q}(-1,4), \mathrm{R}(5,4)$ and $\mathrm{S}(5,-1) . \mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ are the mid-points of $\mathrm{PQ}, \mathrm{QR}, \mathrm{RS}$ and $\mathrm{SP}$ respectively. Is the quadrilateral $\mathrm{ABCD}$ a square, a rectangle or a rhombus? Justify your answer.
Solution:
$\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ are mid points of $\mathrm{PQ}, \mathrm{QR}, \mathrm{RS} \& \mathrm{SP}$ respectively.

$
\begin{aligned}
\therefore A & =\left(\frac{-1+-1}{2}, \frac{-1+4}{2}\right)=\left(-1, \frac{3}{2}\right) \\
B & =\left(\frac{-1+5}{2}, \frac{4+4}{2}\right)=(2,4) \\
C & =\left(\frac{5+5}{2}, \frac{4+(-1)}{2}\right)=\left(5, \frac{3}{2}\right)
\end{aligned}
$

$
\begin{aligned}
& \qquad \mathrm{D}=\left(\frac{5+(-1)}{2}, \frac{-1+-1}{2}\right)=(2,-1) \\
& \text { Slope of } \mathrm{AC}=\frac{\frac{3}{2}-\frac{3}{2}}{-1-5}=0 \\
& \text { Slope of } \mathrm{BD}=\frac{4-(-1)}{2-2}=\infty \\
& \therefore \text { AC is perpendicular to } \mathrm{BD} . \\
& \therefore \text { ABCD can be a square or rhombus. } \\
& \text { Slope of } \mathrm{AB}=\frac{4-\frac{3}{2}}{2-(-1)}=\frac{\frac{5}{2}}{3}=\frac{5}{6} \\
& \text { Slope of } \mathrm{BC}=\frac{\frac{3}{2}-4}{5-2}=\frac{\frac{-5}{2}}{3}=\frac{-5}{6}
\end{aligned}
$
$\therefore \mathrm{AB}$ and $\mathrm{BC}$ are not perpendicular
$\Rightarrow \mathrm{ABCD}$ is rhombus as diagonals are perpendicular and sides are not perpendicular.
Question 2 .
The area of a triangle is 5 sq.units. Two of its vertices are $(2,1)$ and $(3,-2)$. The third Vertex is $(\mathrm{x}$, $y)$ where $y=x+3$. Find the coordinates of the third vertex.
Solution:
Area of triangle formed by points $\left(x_1, y_1\right)$,

$
\begin{aligned}
& \left(x_2, y_2\right) \text {, and }\left(x_3, y_3\right)=\frac{1}{2}\left\{\left(x_1 y_2+x_2 y_3+x_3 y_1\right)-\right. \\
& \left.\left.\left(x_2 y_1\right)+x_3 y_2+x_1 y_3\right)\right\} \\
& \begin{array}{|cclc|}
\hline 2 & 3 & x & 2 \\
1 & -2 & x+3 & 1 \\
\hline
\end{array} \\
& \text { A B C } \\
& \Rightarrow(2,1) \quad(3,-2) \quad(x, x+3) \\
& \therefore \text { Area }=\frac{1}{2}\{(-4+3 x+9+x)-(3-2 x+2 x+6)\} \\
& \Rightarrow \quad \text { Area }=5 \text { (given) } \\
& \Rightarrow \quad 5=\frac{1}{2}\{(4 x+5\}-(9)\} \\
& \Rightarrow \quad 4 x-4=10 \\
& \Rightarrow \quad 4 x=14 \\
& \Rightarrow \quad x=\frac{14}{4}=\frac{7}{2} \\
& y=x+3 \\
& \Rightarrow \quad y=\frac{7}{2}+3 \Rightarrow y=\frac{13}{2} \\
& (x, y)=\left(\frac{7}{2}, \frac{13}{2}\right) \\
&
\end{aligned}
$
Question 3.
Find the area of a triangle formed by the lines $3 x+y-2=0,5 x+2 y-3=0$ and $2 x-y-3=0$

Solution:

$
\begin{aligned}
& l_1: 3 x+y-2=0 \\
& l_2: 5 x+2 y-3=0 \\
& l_3: 2 x-y-3=0 \\
& \rightarrow \text { Solving } l_1 \text { and } l_2, l_2-2 l_1 \\
& \Rightarrow 5 x+2 y-3-6 x-2 y+4=0 \\
& \left.\begin{array}{l}
\Rightarrow \quad-x+1=0 \Rightarrow \quad x=1 \\
\therefore 3(1)+y-2=0 \Rightarrow \quad y=-1
\end{array}\right\} \mathbf{B}=(1,-1) \\
& \rightarrow \text { Solving } l_2 \text { and } l_3 \text {, } \\
& l_2+2 l_3 \\
& \Rightarrow 5 x+2 y-3+4 x-2 y-6=0 \\
& \rightarrow \text { Solving } l_1 \text { and } l_3 \text {, } \\
& l_1+l_3 \\
& \Rightarrow 3 x+y-2+2 x-y-3=0 \\
&
\end{aligned}
$
$\therefore l_1, l_2$ and $l_3$ do not form a triangle as they intersect at the same point $(1,-1)$.
$\therefore$ Area is 0 sq. units.

Question 4.
If vertices of a quadrilateral are at $\mathrm{A}(-5,7), \mathrm{B}(-4, \mathrm{k}), \mathrm{C}(-1,-6)$ and $\mathrm{D}(4,5)$ and its area is
72 sq.units. Find the value of $\mathrm{k}$.
Area (quadrilateral $\mathrm{ABCD}$ )
$$
\begin{aligned}
& =\frac{1}{2}\left|\begin{array}{rrrrr}
-5 & -4 & -1 & 4 & -5 \\
7 & k & -6 & 5 & 7
\end{array}\right| \\
& \Rightarrow \frac{1}{2}[(-5 k+24-5+28)-(-28-k-24-25)]=72 \\
& \Rightarrow(47-5 k)-(-77-k)=144 \\
& \Rightarrow \quad 47-5 k+77+k=144 \\
& \Rightarrow \quad 124-4 k=144 \\
& \Rightarrow \quad-4 k=20 \\
& k=-5 \\
&
\end{aligned}
$$
Question 5 .
Without using distance formula, show that the points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ are vertices of a parallelogram.
Solution:

Slope of $\mathrm{AB}=\frac{0-(-1)}{4-(-2)}=\frac{+1}{6}$
Slope of $\mathrm{BC}=\frac{3-0}{3-4}=-3$
Slope of $\mathrm{CD}=\frac{2-3}{-3-3}=\frac{+1}{6}$
Slope of DA $=\frac{-1-2}{-2-(-3)}=-3$
Slope of $\mathrm{AB}=$ Slope of $\mathrm{CD}$
Slope of BC $=$ Slope of DA
Hence $\mathrm{ABCD}$ forms a parallelogram.
Question 6.
Find the equations of the lines, whose sum and product of intercepts are 1 and $-6$ respectively. Let the intercepts be $\mathrm{x}_1, \mathrm{y}_1$ respectively
$
\begin{aligned}
x_1+y_1=1 \Rightarrow y_1=1-x_1 & \\
x_1 \cdot y_1 & =-6 \\
y=m x+c \text { where } m & =\frac{-y_1}{x_1}
\end{aligned}
$
Solving (1) and (2)
$
\begin{aligned}
& \Rightarrow \quad x_1\left(1-x_1\right)=-6 \\
& \Rightarrow \quad x_1-x_1^2+6=0 \\
& \Rightarrow \quad x_1^2-x_1-6=0 \\
& \Rightarrow \quad x_1^2-3 x_1+2 x_1-6=0 \\
& \Rightarrow x_1\left(x_1-3\right)+2\left(x_1-3\right)=0 \\
&
\end{aligned}
$

$\begin{aligned}
\Rightarrow \quad\left(x_1-3\right)\left(x_1+2\right) & =0 \\
\therefore x_1=3 \text { (or) } x_1 & =-2 \\
\left.\therefore y_1=-2 \text { (or }\right) y_1 & =3
\end{aligned}$

Question 7.
The owner of a milk store finds that, he can sell 980 litres of milk each week at $\square 14$ /litre and 1220 litres of milk each week at $\square 16$ litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at $\square 17 /$ litre?
Solution:

Question 8 .
Find the image of the point $(3,8)$ with respect to the line $x+3 y=7$ assuming the line to be a plane mirror.
Solution:

$
\begin{aligned}
\frac{-1}{3} \times \frac{8-b}{3-a} & =-1 \\
8-b & =(3-a) \times 3 \\
8-b & =9-3 a \\
3 a-b & =1 \Rightarrow b=3 a-1 \ldots \text {...(1) }
\end{aligned}
$
Mid point of line joining $(3,8)$ and $(a, b)$ lies on $x+3 y=7$
$
\begin{aligned}
\text { mid point } & =\left(\frac{a+3}{2}, \frac{b+8}{2}\right) \\
\therefore \frac{a+3}{2}+3\left(\frac{b+8}{2}\right) & =7
\end{aligned}
$
$\therefore$ Solving (1) and (2)
$
\begin{aligned}
\frac{a+3}{2}+\frac{3}{2}(3 a-1+8) & =7 \\
\text { as } b & =3 a-1 \text { from (1). } \\
a+3+9 a+21 & =14 \\
10 a & =-10 \\
a & =-1 \\
\therefore b & =3(-1)-1=-4 \\
(a, b) & =(-1,-4)
\end{aligned}
$
Question 9.
Find the equation of a line passing through the point of intersection of the lines $4 x+7 y-3=0$ and $2 x-3 y+1=0$ that has equal intercepts on the axes.
Solution:
$
\begin{aligned}
& 4 x+7 y-3=0 \\
& 2 x-3 y+1=0
\end{aligned}
$

$
\begin{aligned}
& 4 \mathrm{x}+7 \mathrm{y}-3-2(2 \mathrm{x}-3 \mathrm{y}+1)=0 \\
& 4 x+7 y-3=0 \\
& 2 x-3 y+1=0 \\
& 4 x+7 y-3-2(2 x-3 y+1)=0 \\
& \Rightarrow 4 x+7 y-3-4 x+6 y-2=0 \\
& 13 y=5 \Rightarrow y=\frac{5}{13} \\
& \Rightarrow \quad \\
& 2 x-\frac{15}{13}+1=0 \\
& 2 x=\frac{2}{13} \Rightarrow x=\frac{1}{13} \\
&(x, y)=\left(\frac{1}{13}, \frac{5}{13}\right),
\end{aligned}
$
point of intersection. Equal intercepts
$
\begin{array}{rlrl}
\Rightarrow & & \text { Slope } & =-1 \\
& & \therefore y-\frac{5}{13} & =-1\left(x-\frac{1}{13}\right) \\
\Rightarrow & & x+y & =\frac{5}{13}+\frac{1}{13} \\
\Rightarrow & & 13 x+13 y & =6 \\
\Rightarrow & 13 x+13 y-6 & =0
\end{array}
$
Question 10.
A person standing at a junction (crossing) of two straight paths represented by the equations $2 \mathrm{x}-$ $3 y+4=0$ and $3 x+4 y-5=0$ seek to reach the path whose equation is $6 x-7 y+8=0$ in the least time. Find the equation of the path that he should follow.
Solution:

$l_1$ and $l_2$ intersect at $\mathrm{A}(a, b) 3 l_1-2 l_2=0$
$
\begin{aligned}
& 6 x-9 y+12-6 x-8 y+10=0 \\
&-17 y=-22 \\
& y=\frac{22}{7}=b \\
& \Rightarrow \\
& 2 a-3 \times \frac{22}{7}+4=0 \\
& 2 a=\frac{66}{7}-\frac{28}{7} \\
& \Rightarrow a=\frac{19}{7} \\
& \text { Slope of } l_3=\frac{-6}{-7}=\frac{6}{7} \\
& \text { Slope of } l=\frac{-1}{\left(\frac{6}{7}\right)}=\frac{-7}{6}
\end{aligned}
$
Equation of $l: y-b=m(x-a)$

$\begin{aligned}
& \Rightarrow \quad y-\frac{22}{7}=\frac{-7}{6}\left(x-\frac{19}{7}\right) \\
& \Rightarrow \quad \frac{7 y-22}{7}=\frac{-7}{6}\left(\frac{7 x-19}{7}\right) \\
& \Rightarrow \quad 6(7 y-22)=-7(7 x-19) \\
& \Rightarrow 42 y-132=-49 x+133 \\
& \Rightarrow \quad 49 x+42 y=265 \\
& 49 x+42 y-265=0 \\
&
\end{aligned}$