Proof of Perpendicular Lines, $m_{1}m_{2}=-1$

It is noted that $m_{1}=-\frac{b}{a}$ and $m_{2}=\frac{c}{a}$ . Hence, $m_{1}m_{2}=-\frac{bc}{a^{2}}$ .

According to the diagram, $\theta_{1} + \theta_{2} = 90\deg$ . It can also be seen that $\tan \theta_{1}=\frac{b}{a}$ and $\tan \theta_{2}=\frac{c}{a}$ .

As such, taking $\tan \left(\theta_{1} + \theta_{2}\right) = \frac{\tan \theta_{1} + \tan \theta_{2}}{1 - \tan \theta_{1} \tan \theta_{2}} = \frac{b/a + c/a}{1 - b/a\left(c/a\right)}$ . Since $\tan \left(\theta_{1} + \theta_{2}\right) = \tan 90\deg = \text{undefined}$ , so from $1 - \frac{bc}{a^{2}}$ of the denominator, we get $1 = \frac{bc}{a^{2}}$ such that the denominator is zero.

Substituting $1 = \frac{bc}{a^{2}}$ back into $m_{1}m_{2}=-\frac{bc}{a^{2}}$ , we get $m_{1}m_{2}=-1$ . Hence, it is shown that $m_{1}m_{2}=-1$ .

#Geometry

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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
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Proof of Perpendicular Lines, m1m2=−1m_{1}m_{2}=-1m1​m2​=−1

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Proof of Perpendicular Lines, $m_{1}m_{2}=-1$