Question

If the sum of the distances of a point $P$ from two perpendicular lines in a plane is $1$, then the locus of $P$ is a

• A. rhombus
• B. circle
• C. straight line
• D. pair of straight lines

Answer 1

The correct option is A rhombus

$\Rightarrow$ Let $P(x_1,y_1)$ be a point such that the sum of the distances of $P$ from two perpendicular lines $x+y=0$, $x-y=0$ is $1$.

$\Rightarrow$ Then, $\left|\frac{x_1+y_1}{\sqrt{2}}\right|$ + $\left|\frac{x_1-y_1}{\sqrt{2}}\right|=1$

$\Rightarrow$ $\pm (x_1+y_1)\pm (x_1-y_1)=\sqrt{2}$

Taking square from both sides we get,

$\Rightarrow$ $(x_1+y_1{)}^2$ + $(x_1-y_1{)}^2$ $\pm$ $2(x_1+y_1)$$(x_1-y_1)$ = $2$

$\Rightarrow$ $2(x_1^2+y_1^2)\pm 2(x_1^2+y_1^2)\pm (x_1^2-y_1^2)=1$

$\Rightarrow$ $2x_1^2=1$ or $2y_1^2=1$

$\therefore$ The locus of P is $(2x^2-1)(2y^2-1)=0$ which represents a rhombus.