For triangle PQR, find equation of altitude PS if co-ordinates of P, Q and R are (1, 2), (2, -1) and (0,5) respectively?

Answer: Option C

Given,
P(1, 2), Q(2, -1) and R(0, 5)
Slope of line QR (m₁) $$=\frac{y_2-y_1}{x_2-x_1}=\frac{5-\left(-1\right)}{0-2}=-3$$
If the lines are perpendicular then product of slopes is equal to -1
m₁ × m₂ = -1
-3 × m₂ = -1
m₂ = $$\frac{1}{3}$$
∴ Equation of line passes through the point (1, 2) whose slope (m₂) = $$\frac{1}{3}$$
y - y₁ = m₂(x - x₁)
y - 2 = $$\frac{1}{3}$$(x - 1)
y - $$\frac{1}{3}$$x = 2 - $$\frac{1}{3}$$
3y - x = $$\frac{5}{3}$$ × 3
x - 3y = -5