Question

What is an equation of the line that passes through the point (-6, -7) and is

perpendicular to the line 6x + 5y = 30?

Answer 1

Explanation:

Product of slope of perpendicular lines = -1

6x + 5y = 30

Write this equation in y = mx + b form

5y = -6x + 30

y =
`(-6)/(5)x+(30)/(5)`

`y=(-6)/(5)x + 6`

Slope of this line m₁ = -6/5

m₁ * m₂ = -1

m₂ = -1÷m₁ = -1 *
`(5)/(-6)`

`m_(2)=(5)/(6)` & (-6 , -7)

Equation of the required line: y - y₁ = m (x - x₁)

`y - (-7) = (5)/(6)(x - [-6])\\\\y + 7 = (5)/(6)x + 6 *(5)/(6)\\\\y = (5)/(6)x +5-7\\\\y=(5)/(6)x-2`