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)?

Question

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)?

Ethan Adkins · Accepted Answer

#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#

Nathan Axel · Answer

undefined To find the equation of a line passing through (-2, 1) and perpendicular to the line passing through (-1, 11) and (13, 7), first find the slope of the second line using the formula: $ m = \frac{y_2 - y_1}{x_2 - x_1} $. The slope of the perpendicular line will be the negative reciprocal of this slope. Then, use the point-slope form $ y - y_1 = m(x - x_1) $ with the point (-2, 1) to find the equation.