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), $m_1=\frac{b-3}{a+1}$

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, $m_1m_2 = -1$

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

$⇒\frac{\left(3b-9\right)}{\left(4a+4\right)} = -1$

$⇒3b–9=-4a – 4$
$⇒4a+3b=5 …….(1)$
Point $(a, b)$ lies on line $3x \- 4y = 16.$
$∴3a \- 4b = 16 … (2)$
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$\left(\frac{68}{25},\frac{ -49}{25}\right)$