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

Question

The pedal points of a perpendicular drawn from origin on the line 3x + 4y - 5 = 0, is (A) ((3/5), 2) (B) ((3/5), (4/5)) (C) (-(3/5),-(4/5)) (D) ((30/17), (19/17))

Tardigrade Admin · Accepted Answer

d=(5/√32+42)=1 ; Slope of perpendicular =(4/3) ⇒ x=± 1 . cos θ=± (3/5) and y=± 1 sin θ=± (4/5) Hence ((3/5), (4/5)) lies on straight line.

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

$(\frac{3}{5}, 2)$

$(\frac{3}{5}, \frac{4}{5})$

$(- \frac{3}{5}, - \frac{4}{5})$

$(\frac{30}{17}, \frac{19}{17})$

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

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

(53​,2)

(53​,54​)

(−53​,−54​)

(1730​,1719​)