What is the equation of the line that passes through #(-1,1) # and is perpendicular to the line that passes through the following points: #(13,1),(-2,3) #?

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Answer 1 · 2017-05-29T13:50:25

# 15x-2y+17=0.#

Slope #m'# of the line through points #P(13,1) & Q(-2,3)# is,

# m'=(1-3)/(13-(-2))=-2/15.#

So, if the slope of the reqd. line is #m#, then, as the reqd. line is

#bot# to the line #PQ, mm'=-1 rArr m=15/2.#

Now, we use the Slope-Point Formula for the reqd. line, known to

be passing through the point #(-1,1).#

Thus, the eqn. of the reqd. line, is,

#y-1=15/2(x-(-1)), or, 2y-2=15x+15.#

# rArr 15x-2y+17=0.#