Question

Find the coordinates of the foot of the perpendicular from the point (−1, 3) to the line 3x − 4y − 16 = 0.

Answer 1

Let  A (−1, 3) be the given point.
Also, let M (h, k) be the foot of the perpendicular drawn from A (−1, 3) to the line 3x − 4y − 16 = 0

Point M (h, k) lies on the line 3x − 4y − 16 = 0
3h − 4k − 16 = 0                      ... (1)
Lines 3x − 4y − 16 = 0 and AM are perpendicular.

\[\therefore\] \[\frac{k - 3}{h + 1} \times \frac{3}{4} = - 1\]

\[\Rightarrow 4h + 3k - 5 = 0\]                 ... (2)

Solving eq (1) and eq (2) by cross multiplication, we get:

\[\frac{h}{20 + 48} = \frac{k}{- 64 + 15} = \frac{1}{9 + 16}\]

\[ \Rightarrow a = \frac{68}{25}, b = - \frac{49}{25}\]

Hence, the coordinates of the foot of perpendicular are \[\left( \frac{68}{25}, - \frac{49}{25} \right)\].