We have, 3x – 4y – 16 = 0 ... (i) Slope of the line (i) = $\frac{3}{4}$ Then equation of any line ⊥ from (– 1, 3) to the given line (i) is y - 3 = $-\frac{4}{3}$ [x - (- 1)] ⇒ y − 3 = $-\frac{4}{3}$ (x + 1) ⇒ 3(y – 3) = –4(x + 1) ⇒ 3y – 9 = – 4x – 4 ⇒ 4x + 3y – 5 = 0 ... (ii) On solving (i) and (ii), we get x = $\frac{98}{25}$ , y = $\frac{-49}{25}$ ∴ The required foot of perpendicular is $\left(\frac{68}{25},\frac{-49}{25}\right)$