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 $3x-4y-16=0$

$3y-4k-16=0...\left(i\right)$

Lines $3x-4y-16=0$ and AM are perpendicular

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

$\Rightarrow 4h+3k-5=0...\left(ii\right)$

Solving eq. (i) and eq (ii) by cross multiplication, we get:

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

$a=\frac{68}{25},b=\frac{-49}{25}$

Hence, the coordinates of the foot of perpendicular are $(\frac{68}{25},\frac{-49}{25})$