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 $\frac{2}{1}$

Perpendicular line's slope = $-\frac{1}{2}$

m = $-\frac{1}{4}$

Perpendicular line's slope = $\frac{4}{1}$ or 4

Since you know the slope of the line is:

y = -3/4x + 5

m = $-\frac{3}{4}$      

The perpendicular line's slope is $\frac{4}{3}$, so plug it into the equation

y = mx + b

$y=\frac{4}{3} x+b$        To find b, plug in the point (-3, -3) into the equation

$-3=\frac{4}{3} (-3)+b$

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

1 = b

The y-intercept is 1