Question

Find the equation of the line passing through the point (–1, 5) and perpendicular to the line y = – 3x + 4.

A) 3y = –x + 16

B) y = –3x + 8

C) y = –3x + 2

D) 3y = x + 16

Answer 1

D

Explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = - 3x + 4 ← is in slope- intercept form

with slope m = - 3

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

`m_(perpendicular)` = -
`(1)/(m)` = -
`(1)/(-3)` =
`(1)/(3)` , then

y =
`(1)/(3)` x + c ← is the partial equation

to find c substitute (- 1, 5 ) into the partial equation

5 = -
`(1)/(3)` + c ( add
`(1)/(3)` to both sides )

5
`(1)/(3)` = c ⇒ c =
`(16)/(3)`

y =
`(1)/(3)` x +
`(16)/(3)` ( multiply through by 3 to clear the fractions )

3y = x + 16