Question

What is the y-intercept of the line perpendicular to the line y = _~x+ 5 that includes the point (-3, -3)?

Answer 1

Slope-intercept form: y = mx + b

[m is the slope, b is the y-intercept or the y value when x = 0 ---> (0, y) or the point where the line crosses through the y-axis]

For lines to be perpendicular, their slopes have to be negative reciprocals of each other. [basically changing the sign (+/-) and flipping the fraction/or switching the numerator and the denominator]

For example:

m = 2 or
`(2)/(1)`

Perpendicular line's slope =
`-(1)/(2)`

m =
`-(1)/(4)`

Perpendicular line's slope =
`(4)/(1)` or 4

Since you know the slope of the line is:

y = -3/4x + 5

m =
`-(3)/(4)`

The perpendicular line's slope is
`(4)/(3)`, so plug it into the equation

y = mx + b

`y=(4)/(3) x+b` To find b, plug in the point (-3, -3) into the equation

`-3=(4)/(3) (-3)+b`

-3 = -4 + b Add 4 on both sides to get "b" by itself

1 = b

The y-intercept is 1