How do you find a standard form equation for the line perpendicular to line x-7y=9 and point(5,4)?

1 Answer
Shiva Prakash M V
Feb 9, 2018

The standard form equation for the line perpendicular to line x-7y=9 and point (5,4) is given by
#7x+y-39=0#

Explanation:

Given line:
#x-7y=9#
If #ax+by+c=0#
slope of the line is given by
#=-a/b#
Expressing the given line in the standard form
#x-7y-9=0#
Comparing
#a=1; b=-7; c=-9#
Thus the slope of the line is
#=-1/(-7)=1/7#
The normal has the slope which is the negative reciprocal of the slope of the given line
#=-7#
Further, it is given that the normal passes through the point
#(5,4)#
Equation of the line passing through the point #(5,4)# and having slope #-7# is given by
#(y-4)/(x-5)=-7#
Simplifying
#y-4=-7(x-5)#
#y-4=-7x+35#
#y-4+7x-35=0#
#y+7x-39=0#
Rearranging in the standard form
#7x+y-39=0#

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