Find the coordinates of the foot of perpendicular from the point (–1, 3)

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 $(a, b), m_{1}=\frac{b-3}{a+1}$

Slope of the line $3 x-4 y-16=0$ or $y=\frac{3}{4} x-4, m_{2}=\frac{3}{4}$

Since these two lines are perpendicular, $m_{1} m_{2}=-1$

$\therefore\left(\frac{b-3}{a+1}\right) \times\left(\frac{3}{4}\right)=-1$

$\Rightarrow \frac{3 b-9}{4 a+4}=-1$

$\Rightarrow 3 b-9=-4 a-4$

$\Rightarrow 4 a+3 b=5$ $\ldots(1)$

Point $(a, b)$ lies on line $3 x-4 y=16$

$\therefore 3 a-4 b=16 \ldots(2)$

On solving equations (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 $\left(\frac{68}{25},-\frac{49}{25}\right)$.