Question

What is the equation of a line that is perpendicular to y= (-2/3x) +5 and goes through the point (-8,3)?

What are the steps to completing this? Thank you!

Answer 1

y =
`(3)/(2)`x + 15

Explanation:

We are to find the equation of a line (
`l_(2)`) which is perpendicular to line (
`l_(1)`) whose equation is; y =
`(-2)/(3)`x + 5 and that;

`l_(2)` goes through (-8,3)

The slope of
`l_(2)` is
`-(2)/(3)`

The product of slope of
`l_(1)` and that of
`l_(2)` = -1 (The product of two perpendicular slopes is -1)

Therefore,
`-(2)/(3)` times slope of
`l_(2)` = -1

Slope of
`l_(2)` = -1 ÷
`-(2)/(3)` =
`(3)/(2)`

Taking another point (x,y) on
`l_(2)`;

Slope = Change in y ÷ change in x =
`(y - 3)/(x - -8)` =
`(3)/(2)`

y - 3 =
`(3)/(2)`x + 12

y =
`(3)/(2)`x + 12 + 3

Finally,

y =
`(3)/(2)`x + 15