# Find a line that passes through (3,5) perpendicular to a line whose slope is -1/9

Find a line that passes through (3,5) perpendicular to a line whose slope is -1/9
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RamPatWeese2w
The standardised equation of a strait line is y=mx+c where m is the gradient.
Any line perpendicular to this will have the gradient of:
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\left(-1\right)×\frac{1}{m}$
So in this case as $m=-\frac{1}{9}$ the gradient of the perpendicular is:
$\left(-1\right)×-\frac{9}{1}=+9$
Thus the equation of the line we are after is y=9x+c
This line passes through the point $\left(x,y\right)\to \left(3,5\right)$
so by substitution we have:
$y=9x+c\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\to \phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}5=9\left(3\right)+c$
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}5=27+c$
subtract 27 from both sides
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}5-27=27-27+c$
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}c=-22$
Thus we have:
$y=9x-22$