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Q.
Let the perpendiculars from any point on the line $7x + 56y = 0$ upon $3x + 4y = 0$ and $5x - 12y = 0$ be $p$ and $p'$, then
Straight Lines
Solution:
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 $p' = \frac{5x_{1} + \frac{12x_{1}}{8}}{13} = \frac{x_{1}}{2} \Rightarrow p = p'$