The line $A B$ cuts off equal intercepts $2 a$ from the axes. From any point $P$ on the line $A B$ perpendiculars $P R$ and $P S$ are drawn on the axes. Locus of mid-point of $R S$ is

• A

  $x-y=\frac{a}{2}$
• Correct answer
  B

  $x+y=a$
• C

  $x^{2}+y^{2}=4 a^{2}$
• D

  $x^{2}-y^{2}=2 a^{2}$

Solution

$\Rightarrow$ $$ \begin{array}{l} \frac{x}{2 a}+\frac{y}{2 a}=1 \\ x+y=2 a \end{array} $$ Let the coordinates of the mid-point of $\mathrm{RS}$ be $(\mathrm{h}, \mathrm{k})$ So. $R$ and $S$ are $(2 h, 0)$ and $(0,2 k)$ Hence, the mid-point of $R S$ is $(h, k)$ P lies on line AB Then. from Eq we have $$ 2 h+2 k=2 a $$ $\Rightarrow \quad h+k=a$ So, the locus of $(h, k)$ is $x+y=a$.

Hence, the correct answer is option: 'B'.