How do you graph the line that passes through (-4,1) perpendicular to the line whose slope is -3/2?

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Answer 1 · 2017-07-22T18:12:30

# 2x-3y+11=0.#

The product of the slopes of two #bot# lines is #-1.#

Given that, the slope of a line is #-3/2,# that of a line #bot# to this

reqd. one must be, #2/3.#

The reqd. line passes through the pt. #(-4,1).#

Using the Slope-Pt. Form, we find the eqn. of the reqd. line is,

#y-1=2/3(x-(-4)), i.e., 3(y-1)=2(x+4), or, #

# 2x-3y+11=0.#