Question

The line $AB$ cuts off equal intercepts $2a$ from the axes. From any point $P$ on the line $AB$ perpendiculars $PR$ and $PS$ are drawn on the axes. Locus of mid-point of $RS$ is

• A. $x-y=\frac{a}{2}$
• B. $x+y=a$
• C. $x^2+y^2=4a^2$
• D. $x^2-y^2=2a^2$

Answer 1

The correct option is B $x+y=a$

Equation of line $AB$,
$x+y=2a$


Let a on the line be,
$P(t,2a-t)$
So the mid-point of $RS$ is,
$M(h,k)=(\frac{t}{2},\frac{2a-t}{2})$
$h=\frac{t}{2},k=\frac{2a-t}{2}$
The locus of the midpoint will be,
$k=\frac{2a-2h}{2}$
$\Rightarrow h+k=a\Rightarrow x+y=a$