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

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

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Answer 1 · 2018-02-09T19:28:57

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

Explanation:

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

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

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