Find the equation of the perpendicular drawn from the point P(2, 3) on the line y = 3x + 4. Find the co-ordinates of the foot of the perpendicular.

Given line y = 3x + 4…(1)

Comparing above equation with y = mx + c we get m₁ = 3

Since line is perpendicular to y = 3x + 4, the product of their gradients must be -1.

Let the slope of required line be m ₂

⇒ m₁ × m₂ = -1

⇒ 3 × m₂ = -1

⇒ m₂ = $-\dfrac{1}{3}$

By point slope formula,

Equation of a line,

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = $-\dfrac{1}{3}$ (x - 2)

⇒ 3(y - 3) = -1(x - 2)

⇒ 3y - 9 = -x + 2

⇒ x + 3y - 11 = 0…(2)

The equation of the perpendicular drawn from P(2, 3) is x + 3y - 11 = 0.

Substitute y into x + 3y - 11 = 0:

⇒ x + 3(3x + 4) - 11 = 0

⇒ x + 9x + 12 - 11 = 0

⇒ 10x + 1 = 0

⇒ 10x = -1

⇒ x = $-\dfrac{1}{10}$

Substitute value of x in y = 3x + 4:

⇒ y = 3 $\Big(-\dfrac{1}{10}\Big)$ + 4

⇒ y = $\Big(-\dfrac{3}{10} + \dfrac{40}{10}\Big)$

⇒ y = $\dfrac{37}{10}$

Hence, equation of required line is x + 3y - 11 = 0 and coordinates of foot of perpendicular $\Big(-\dfrac{1}{10}, \dfrac{37}{10}\Big)$.