Slope of Orthogonal Curves

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:

$m_1 = -\dfrac 1 {m_2}$

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:

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

$\mathbf v_1 \cdot \mathbf v_2 = 0$ 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 \ \ $

$\ds m_1$

$=$

$\ds -\frac 1 {m_2}$

$\blacksquare$