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
$\frac{|2h+k-3|}{\sqrt{5}}+\frac{|h-2k+1|}{\sqrt{5}}=2$
Hence, the locus is
$\frac{|2x+y-3|}{\sqrt{5}}+\frac{|x-2y+1|}{\sqrt{5}}=2$
$\Rightarrow \left|2x+y-3\right|+\left|x-2y+1\right|=2\sqrt{5}$
Locus is a square with centre $(1,1)$ land side length $2\sqrt{2}$ .
So, area is ${\left(2\sqrt{2}\right)}^2=8$