What is the equation of the line that passes through #(-2,1) # and is perpendicular to the line that passes through the following points: #(-1,11),(13,7)?

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Answer 1 · 2016-04-04T13:05:18

#7x-2y+16=0#

Slope of the line that passes through two points #(x_1,y_1)# and #(x_2,y_2)# is given by #(y_2-y_1)/(x_2-x_1)#.

Hence slope of line joining points #(-1,11)# and #(13,7)# is #(7-11)/(13-(-1))=-4/14=-2/7#.

Hence line perpendicular to this will have a slope of #-1/(-2/7)=7/2#

As this line passes through #(-2,1)#, using point slope equation of line will be

#y-1=7/2(x-(-2))# or

#2(y-1)=7(x+2)# or #2y-2=7x+14# i.e.

#7x-2y+16=0#