Question

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

Answer 1

Let $(a,b)$ be the coordinates of the foot of the perpendicular from the point $(-1,3)$ to the line $3x-4y-16=0$
Slope of the line joining $(-1,3)$ and $(a,b)$, is $m_1=\frac{b-3}{a+1}$
Slope of the line $3x-4y-16=0$ or $y=\frac{3}{4}x-4,m_2=\frac{3}{4}$
Since these two lines are perpendicular,$m_1{m}_2=-1$
$\therefore \left\{\frac{b-3}{a+1}\right\}\cdot \left\{\frac{3}{4}\right\}=-1$
$\Rightarrow \frac{3b-9}{4a+4}=-1$
$\Rightarrow 3b-9=4a-4$
$\Rightarrow 4a+3b=5$...(1)
Point(a,b) lies on line $3x-4y=16$
$\therefore 3a-4b=16$....(2)
On solving equation (1) and (2) we obtain
$a=\frac{68}{25}$ and $b=-\frac{49}{25}$
Thus, the required coordinates of the foot of the perpendicular are $\{\frac{68}{25},\frac{49}{25}\}$