Question

What is the equation of a line that is perpendicular to the line y - 1 = -2 (x + 5) and

that passes through the point (-3,2).

Answer 1

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

Explanation:

The equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

y - 1 = - 2(x + 5) ← is in point- slope form

with slope m = - 2

Given a line with slope m then the slope of a line perpendicular to it is

`m_(perpendicular)` = -
`(1)/(m)` = -
`(1)/(-2)` =
`(1)/(2)`

and (a, b) = (- 3, 2), thus

y - 2 =
`(1)/(2)`(x - (- 3) ), that is

y - 2 =
`(1)/(2)`(x + 3) ← equation of perpendicular line

Answer 2

Explanation:

Write the equation in the form of Y=mX+C so that it is easuer to find the gradientor slope

`y - 1 = - 2(x + 5) \\ y - 1 = - 2x - 10 \\ y = - 2x - 9`

the slope is -2 .

since we need to find a perpendicular equation, the gradient or slope should be negative reciprocal of the gradient we have now which is -2

the perpendicular gradient = 1/2

coordinates: (-3,2)

`y - y1 = m(x - x1) \\ y - 2 = (1)/(2) (x - ( - 3)) \\ 2y - 4 = x + 3 \\ 2y = x + 7`