How do you determine the equation of a line passing through (2, -3) that is perpendicular to the line 4x-y=22?

Question

Genesis Allen · Accepted Answer

Equation of a line passing through #(2, -3)# that is perpendicular to the line #4x-y=22# is #x+4y+10=0# Equation of a line that is perpendicular to a line #ax+by+c=0#, will be of type #bx-ay+k=0# Note that coefficients of #x# and #y# have changed and sign before #y# has been changed, keeping sign before #x# to be same. Hence, equation of a line that is perpendicular to a line #4x-y=22#, will be of type #x+4y+k=0# As it passes through #(2,-3)#, we have #2+4xx(-3)+k=0# or #2-12+k=0# i.e. #k=12-2=10#. Hence, equation of a line passing through #(2, -3)# that is perpendicular to the line #4x-y=22# is #x+4y+10=0#. graph{(4x-y-22)(x+4y+10)=0 [-7.89, 9.89, -5.564, 3.325]} Note: Equation of a line that is parallel to a line #ax+by+c=0#, will be of type #ax+by+k=0#, i.e. no change in coefficients or sign. Only constant term changes. Also observe that slopes of parallel llines are equal but product of slopes of perpendicular lines is #-1#.

Ethan Adkins · Answer

undefined To determine the equation of a line passing through (2, -3) that is perpendicular to the line 4x - y = 22, first, find the slope of the given line by rearranging it into slope-intercept form. Then, find the negative reciprocal of that slope to get the slope of the perpendicular line. Finally, use the point-slope form of a line with the given point (2, -3) and the perpendicular slope to write the equation.