How do you write an equation of a line perpendicular to y= 3/4x - 2 and passes through (-12,7)?

Question

Ethan Adkins · Accepted Answer

#y = -4/3x -9#

First we need to get the slope of the perpendicular line

The slope of a perpendicular line is equal to the negative inverse of the slope of the given line

#y = mx + b#

#y = 3/4x - 2#

#=> m = 3/4#

#m' = -1/m#

#=> m' = -1 / (3/4)#

#=> m' = -4/3#

Now that we have the slope, we need to find the y-intercept. To find the y-intercept, we need to plug-in values of #x# and #y# that the line passes through

#y' = m'x' + b#

#=> y' = -4/3x' + b#

#=> 7 = -4/3(-12) + b#

#=> 7 = 16 + b#

#=> b = -9#

Therefore, the equation of the line is

#y = -4/3x -9#

Eva Adamson · Answer

undefined To write the equation of a line perpendicular to $ y = \frac{3}{4}x - 2 $ and passing through the point $(-12, 7)$, you need to find the negative reciprocal of the slope of the given line. The given line has a slope of $ \frac{3}{4} $, so the perpendicular line will have a slope of $ -\frac{4}{3} $. Then, you can use the point-slope form of a linear equation, $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) $ is the given point and $ m $ is the slope. Substituting the given point and the perpendicular slope into the equation will give you the equation of the perpendicular line.

Therefore, the equation of the line perpendicular to $ y = \frac{3}{4}x - 2 $ and passing through $(-12, 7)$ is $ y - 7 = -\frac{4}{3}(x + 12) $.