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 $-\frac{2}{1}$        

Perpendicular line's slope: $\frac{1}{2}$        (flip the sign from - to +, and flip the fraction)

Slope = $\frac{1}{3}$

Perpendicular line's slope = $-\frac{3}{1}$  or  -3     (flip the sign from + to -, flip fraction)

y = 8x - 2    The slope is 8, so the perpendicular line's slope is $-\frac{1}{8}$.

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

y = mx + b

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

$-6=-\frac{1}{8} (-5)+b$     (Two negative signs cancel each other out and become positive)

$-6=\frac{5}{8} +b$    Subtract 5/8 on both sides to get "b" by itself

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

$(\frac{8}{8}) (-6)-\frac{5}{8} =b$

$-\frac{48}{8} -\frac{5}{8}$ = b      Now combine the fractions

$-\frac{53}{8} =b$

$y=-\frac{1}{8} x-\frac{53}{8}$