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3. What is the slope of a line perpendicular to the line with equation y = 4x + 5? (Points : 4) 4 –4 –5
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\[ y=4x+5 \implies -4x+1y = 5\implies \mathbf n = \langle -4,1\rangle \]
A perpendicular line's normal vector would be \(\langle 1,4\rangle\).
So we have: \[ x + 4y=b\implies 4y=-x+b \implies y=-\frac x4+\frac b4 \]So the slope we are getting is \(-1/4\).
In general, we have: \[ y=mx+b \implies \mathbf n = \langle -m,1 \rangle \]And the perpendicular normal vector will be \( \langle 1,m\rangle \). This means: \[ x+my =c \implies y = -\frac{x}{n}+\frac cm \]So the slope of the perpendicular line is always \(-1/m\), i.e. the negative inverse.
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