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 are 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:
\(\ds \mathbf v_1 \cdot \mathbf v_2\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x_1 x_2 + y_1 y_2\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \frac {y_1} {x_1} + \frac {x_2} {y_2}\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \frac {x_1} {y_1}\) | \(=\) | \(\ds -\frac 1 {\paren {\dfrac {y_2} {x_2} } }\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds m_1\) | \(=\) | \(\ds -\frac 1 {m_2}\) |
$\blacksquare$