Question

Let the perpendiculars from any point on the line $7x+56y=0$ upon $3x+4y=0$ and $5x-12y=0$ be $p$ and $p^{\prime }$, then

• A. $2p=p^{\prime }$
• B. $p=2p^{\prime }$
• C. $p=p^{\prime }$
• D. None of these

Answer 1

Answer: (C)

Any point on the line 7x + 56y = 0 is
$\left(x_1,-\frac{7x_1}{56}\right)i.e.,\left(x_1,\frac{x_1}{8}\right)$
$\therefore$ The perpendicular distance p and p' are
$p=\frac{3x_1-\frac{4x_1}{8}}{5}=\frac{x_1}{2}$
and $p^{\prime }=\frac{5x_1+\frac{12x_1}{8}}{13}=\frac{x_1}{2}\Rightarrow p=p^{\prime }$