Question

Write down the equation of the line perpendicular to 3x + 8y = 12 and passing through the point (−1, −2).

Answer 1

⇒ Rewriting the given equation 3x + 8y = 12 into the slope-intercept form, y = mx + c to find the slope (m₁):

8y = −3x + 12

`y = - 3/8 x + 12/8`

∴ The slope (m₁) of the given line is `- 3/8`.

⇒ For perpendicular lines, the product of their slopes is −1. Let’s find slope (m₂):

m₁ × m₂ = −1

`m_2 = - 1/m_1`

`m_2 = - 1/(- 3/8)`

∴ `m_2 = 8/3`

⇒ The line passes through the point (x₁, y₁) = (−1, −2), using the point-slope formula:

y − y₁ = m(x − x₁)

`y - (-2) = 8/3 (x - (-1))`

`y + 2 = 8/3 (x + 1)`

3(y + 2) = 8(x + 1)   ...[Multiplied both sides by 3.]

3y + 6 = 8x + 8

⇒ Rearranging the above equation in the standard form (Ax + By + C = 0),

8x − 3y + 2 = 0

Hence, the equation of the perpendicular line is 8x − 3y + 2 = 0.