Question

The pedal points of a perpendicular drawn from origin on the line $3x + 4y - 5 = 0$, is

• A. $\left(\frac{3}{5}, 2\right)$
• B. $\left(\frac{3}{5}, \frac{4}{5}\right)$
• C. $\left(-\frac{3}{5},-\frac{4}{5}\right)$
• D. $\left(\frac{30}{17}, \frac{19}{17}\right)$

Answer 1

Answer: (B)

$d=\frac{5}{\sqrt{3^{2}+4^{2}}}=1 ;$
Slope of perpendicular $=\frac{4}{3}$
$\Rightarrow x=\pm 1 . \cos \,\theta=\pm \frac{3}{5}$ and
$y=\pm 1\, \sin\, \theta=\pm \frac{4}{5}$
Hence $\left(\frac{3}{5}, \frac{4}{5}\right)$ lies on straight line.