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

1 Answer
Shwetank Mauria
Dec 15, 2016

Equation of a line passing through #(2, -3)# that is perpendicular to the line #4x-y=22# is #x+4y+10=0#

Explanation:

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#.

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