Question 14 Marks
Drones are used by military for surveillance purposes. These days, drones are also used by individual entrepreneurs, SMEs and large companies to accomplish various other tasks.
Adrone is flying over a rectangular field with vertices at $A(-100,0), B(100,0), C(100,150)$ and $D(-100,150)$. The drone captures an image at a location $(x, y)$.
(i) Find the dimensions of the rectangular field.
(ii) Find the distance between points A and C .
(iii) (a) If a drone captures the image of an object $P(x, y)$ on the rectangular field, find the relation between $x$ and $y$ such that $P A=P C$.
OR
(b) If a drone captures the image of an object at a point $Q$ whose $x$ coordinate is 0 and it is equidistant from points $A$ and $D$, find the coordinates of $Q$.
Adrone is flying over a rectangular field with vertices at $A(-100,0), B(100,0), C(100,150)$ and $D(-100,150)$. The drone captures an image at a location $(x, y)$.
(i) Find the dimensions of the rectangular field.
(ii) Find the distance between points A and C .
(iii) (a) If a drone captures the image of an object $P(x, y)$ on the rectangular field, find the relation between $x$ and $y$ such that $P A=P C$.
OR
(b) If a drone captures the image of an object at a point $Q$ whose $x$ coordinate is 0 and it is equidistant from points $A$ and $D$, find the coordinates of $Q$.
Answer
(i) We find that
$
\begin{array}{l}
A B=\sqrt{(100-(-100))^2+(0-0)^2}=\sqrt{200^2}=200 \\
B C=\sqrt{(100-100)^2+(150-0)^2}=\sqrt{150^2}=150
\end{array}
$
So, dimensions of the rectangular field are : Length $=200$, Breadth $=150$.
(ii) $A C=\sqrt{(100-(-100))^2+(150-0)^2}=\sqrt{40000+22500}=\sqrt{62500}=250$
$\begin{array}{l}\text { (iii) (a) } P A=P C \Rightarrow P A^2=P C^2 \Rightarrow\left(x+100^2\right)+(y-0)^2=(x-100)^2+(y-150)^2 \\ \Rightarrow \quad 400 x+300 y=22500 \Rightarrow 4 x+3 y=225\end{array}$
OR
(b) Let the coordinates of $Q$ be $(0, y)$. Then,
$
\begin{array}{ll}
& Q A=Q D \\
\Rightarrow & Q A^2=Q D^2 \\
\Rightarrow & (0-100)^2+(y-0)^2=(0+100)^2+(y-150)^2 \\
\Rightarrow & 100^2+y^2=100^2+y^2-300 y+22500 \Rightarrow 300 y=22500 \Rightarrow y=75
\end{array}
$
Hence, the ordinate of $D$ is 75 .
View full question & answer→(i) We find that
$
\begin{array}{l}
A B=\sqrt{(100-(-100))^2+(0-0)^2}=\sqrt{200^2}=200 \\
B C=\sqrt{(100-100)^2+(150-0)^2}=\sqrt{150^2}=150
\end{array}
$
So, dimensions of the rectangular field are : Length $=200$, Breadth $=150$.
(ii) $A C=\sqrt{(100-(-100))^2+(150-0)^2}=\sqrt{40000+22500}=\sqrt{62500}=250$
$\begin{array}{l}\text { (iii) (a) } P A=P C \Rightarrow P A^2=P C^2 \Rightarrow\left(x+100^2\right)+(y-0)^2=(x-100)^2+(y-150)^2 \\ \Rightarrow \quad 400 x+300 y=22500 \Rightarrow 4 x+3 y=225\end{array}$
OR
(b) Let the coordinates of $Q$ be $(0, y)$. Then,
$
\begin{array}{ll}
& Q A=Q D \\
\Rightarrow & Q A^2=Q D^2 \\
\Rightarrow & (0-100)^2+(y-0)^2=(0+100)^2+(y-150)^2 \\
\Rightarrow & 100^2+y^2=100^2+y^2-300 y+22500 \Rightarrow 300 y=22500 \Rightarrow y=75
\end{array}
$
Hence, the ordinate of $D$ is 75 .
