Question

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

Answer 1

Let Q be the foot of perpendicular drawn from P(- 1, 3) on the line 3x - 4y -16 = 0.

$\therefore$ Equation of a line $\perp$ to 3x- 4y - 16 = 0 is 4x+ 3y + k = 0.

Since this line passes through point (-1, 3)

$\therefore$ 4 $\times$ -1+ 3 $\times$ 3 +k = 0

$\Rightarrow$ -4 + 9+k= 0 $\Rightarrow$ k=-5

Thus Q is a point of intersection of the lines

3x-4y- 16= 0

and 4x + 3y - 5 = 0

Solving these equations by cross multiplication, we have

$\frac{x}{-68}=\frac{y}{49}=-\frac{1}{25}$

$\therefore \frac{x}{-68}=\frac{-1}{25}\Rightarrow x=\frac{68}{25}$

$\frac{y}{49}=\frac{-1}{25}\Rightarrow y=\frac{-49}{25}$

Thus coordinates of foot of perpendicular are

$(\frac{68}{25},\frac{-49}{25})$