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

Answer: (B)

$x+y=a$

The equation of the line $AB$ using intercept form is
$\frac{x}{2a}+\frac{y}{2a}=1$ or $x+y=2a$ ...$\left(i\right)$
Thus any point $P$ on the line $x+y=2a$ can be written as $P=(x_1,2a-x_1)$.
Thus perpendicular from $P$ cuts the $x$-axis at $R=(x_1,0)$ and the $y$-axis at $S=(0,2a-x_1)$. Now the midpoint $RS$ will be
$Q=(\frac{x_1}{2},\frac{2a-x_1}{2})$.
Therefore $x=\frac{x_1}{2}$ and $y=\frac{2a-x_1}{2}$. Hence
$x+y=\frac{x_1+2a-x_1}{2}$ or $x+y=\frac{2a}{2}$ or $x+y=a$.
Thus the required equation is $x+y=a$.