Question

Y=8x-2, find the equation of the line that is perpendicular to this line and passes through the point (-5, -6)

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 (flip the sign +/- and the fraction/switch the numerator and the denominator)

For example:

Slope = -2 or
`-(2)/(1)`

Perpendicular line's slope:
`(1)/(2)` (flip the sign from - to +, and flip the fraction)

Slope =
`(1)/(3)`

Perpendicular line's slope =
`-(3)/(1)` or -3 (flip the sign from + to -, flip fraction)

y = 8x - 2 The slope is 8, so the perpendicular line's slope is
`-(1)/(8)`.

Now that you know the slope, substitute/plug it into the equation:

y = mx + b

`y=-(1)/(8) x+b` To find b, plug in the point (-5, -6) into the equation, then isolate/get the variable "b" by itself

`-6=-(1)/(8) (-5)+b` (Two negative signs cancel each other out and become positive)

`-6=(5)/(8) +b` Subtract 5/8 on both sides to get "b" by itself

`-6-(5)/(8) =b` (To combine fractions, they need to have the same denominator, so multiply -6 by 8/8 so that they will have the same denominator)

`((8)/(8)) (-6)-(5)/(8) =b`

`-(48)/(8) -(5)/(8)` = b Now combine the fractions

`-(53)/(8) =b`

`y=-(1)/(8) x-(53)/(8)`