Question

Find the equation of a line passing through (3, – 2) and perpendicular to the line.
x - 3y + 5 = 0.

Answer 1

x - 3y + 5 = 0
⇒ 3y = x + 5
∴ y = `x/(3) + (5)/(3)`
∴ m₁ = `(1)/(3)`
Since lines are perpendicular to each other
∴ m₁ x m₂ = -1
`(1)/(3) xx m_2` = -1
m₂ = -1 x 3
m₂ = -3
Passing point is (3, -2)
∴ Equation of line
y - y₁ = m(x - x₁)
⇒  y + 2 = -3(x - 3)
⇒  y + 2 = -3x + 9
⇒  3x + y + 2 - 9 = 0
⇒  3x + y =7.