What is the equation of the line that passes through #(-2,1) # and is perpendicular to the line that passes through the following points: #(5,2),(-12,5)#?

1 Answer
Shwetank Mauria
Feb 29, 2016

#17x-3y+37=0#

Explanation:

The slope of the line joining points #(x_1,y_1)# and #(x_1,y_1)# is given by #(y_2-y_1)/(x_2-x_1)^#. Hence slope of line joining #(5,2)# and #(−12,5)# is #(5-2)/(-12-5)=-3/17#

Hence slope of the line perpendicular to line joining #(5,2)# and #(−12,5)# will be #-1/(-3/17)# or #17/3#, as product of slopes of lines perpendicular to each other is #-1#.

Hence equation of line passing through #(-2,1)# and having slope #17/3# will be (using point-slope form)

#(y-1)=17/3(x-(-2))# or #3(y-1)=17(x+2))# or

#17x-3y+37=0#

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