Question

If the lines `x/3 + y/4 = 7` and 3x + ky = 11 are perpendicular to each other, find the value of k.

Answer 1

⇒ Let’s multiply the equation `x/3 + y/4 = 7` by the common (LCM) of 3 and 4, which is 12:

`12(x/3) + 12(y/4) = 12(7)`

4x + 3y = 84

⇒ Rearranging the equation into the slope-intercept form, y = mx + c to find the slope (m₁):

3y = −4x + 84

`y = -4/3x + 28`

∴ `m_1 = -4/3`

⇒ For the line 3x + ky = 11, solve for y to find its slope (m₂):

ky = −3x + 11

`y = -3/kx + 11/k`

∴ `m_2 = -3/k`

⇒ Since the lines are perpendicular, the product of their slopes must be −1:

m₁ × m₂ = −1

`(-4/3) xx (-3/k) = -1`

`12/(3k) = -1`

`4/k = -1`

∴ k = −4

Hence, the value of k is −4.