# Trig

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How do you find an equation of a line perpendicular to y=x at the point (5,5) in standard form?

• Trig -

put the original equation in the form
y = m x + b
in this case
y = 1 x + 0
so m, the original slope, = 1
Now m' = -1/m
so our new slope = -1/1 = -1
so our new function looks like
y = -1 x + b
we need b
use the point
5 = -5 + b
b = 10
so
y = -x + 10

• Trig -

The slope of y = x is 1, so the slope of a perpendicular line must be -1. (This is becasue the product of the slopes must be -1).

The equation can therefore be written
(y-5)/(x-5) = -1
y-5 = -x +5
y = -x +10

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