Slope of Orthogonal Curves

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Theorem

Let $C_1$ and $C_2$ be curves in a cartesian plane.

Let $C_1$ and $C_2$ intersect each other at $P$.

Let the slope of $C_1$ and $C_2$ at $P$ be $m_1$ and $m_2$.


Then $C_1$ and $C_2$ are orthogonal if and only if:

$m_1 = -\dfrac 1 {m_2}$


Proof

Let the slopes of $C_1$ and $C_2$ at $P$ be defined by the vectors $\mathbf v_1$ and $\mathbf v_2$ represented as column matrices:

$\mathbf v_1 = \begin{bmatrix} x_1 \\ y_1 \end{bmatrix} , \mathbf v_2 = \begin{bmatrix} x_2 \\ y_2 \end{bmatrix}$

By Non-Zero Vectors Orthogonal iff Perpendicular:

$\mathbf v_1 \cdot \mathbf v_2 = 0$ if and only if $C_1$ is orthogonal to $C_2$

where $\mathbf v_1 \cdot \mathbf v_2$ denotes the dot product of $C_1$ and $C_2$.

Thus:

\(\displaystyle \mathbf v_1 \cdot \mathbf v_2\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x_1 x_2 + y_1 y_2\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \frac {y_1} {x_1} + \frac {x_2} {y_2}\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \frac {x_1} {y_1}\) \(=\) \(\displaystyle -\frac 1 {\paren {\dfrac {y_2} {x_2} } }\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle m_1\) \(=\) \(\displaystyle -\frac 1 {m_2}\)

$\blacksquare$