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

1 Answer

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