Question

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

Answer 1

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`