What is the equation of the line that is perpendicular to the line passing through #(5,12)# and #(-2,-23)# at midpoint of the two points?

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Answer 1 · 2016-04-13T15:16:47

#x+5y=-26#

We need the negative reciprocal of the slope # m# and the midpoint #M(x_m, y_m)#

#m=(y_2-y_1)/(x_2-x_1)=(-23-12)/(-2-5)=(-35)/(-7)=5#

The midpoint:

#x_m=(5+(-2))/2=3/2#
#y_m=(12+(-23))/2=(-11)/2#

The equation

#(y-y_m)=(-1/m)(x-x_m)#

#(y-(-11)/2)=(-1/5)(x-3/2)#
#5(y+11/2)=-x+3/2#

#5(2y+11)=-2x+3#
#10y+55=-2x+3#
#2x+10y=-52#
#x+5y=-26#

God bless....I hope the explanation is useful.