Question

Let $P$ be a moving point such that sum of its perpendicular distances from $2x+y=3$ and $x-2y+1=0$ is always $2$ units then area bounded by locus of point $P$ is

• A. $10$
• B. $8$
• C. $6$
• D. $4$

Answer 1

Answer: (B)

We have,
$2x+y=3$
$x-2y+1=0$
Let coordinates of $P\left(h , k\right)$ .
Now, we know that the perpendicular distance of a point $\left(x_{1} , y_{1}\right)$ from the line $ax+by+c=0$ is given by $\left|\frac{a x_{1} + b y_{1} + c}{\sqrt{a^{2} + b^{2}}}\right|$ .
Therefore, sum of perpendicular distance of $P\left(h , k\right)$ from the given lines is
$\frac{\left|2 h + k - 3\right|}{\sqrt{5}}+\frac{\left|h - 2 k + 1\right|}{\sqrt{5}}=2$
Hence, the locus is
$\frac{\left|2 x + y - 3\right|}{\sqrt{5}}+\frac{\left|x - 2 y + 1\right|}{\sqrt{5}}=2$
$\Rightarrow \left|2 x + y - 3\right|+\left|x - 2 y + 1\right|=2\sqrt{5}$
Locus is a square with centre $\left(\right.1,1\left.\right)$ land side length $2\sqrt{2}$ .
So, area is $\left(2 \sqrt{2}\right)^{2}=8$