# Equations defining Projection in Plane/Cartesian

## Theorem

Let $M$ and $N$ be distinct lines in the plane, both of which pass through the origin.

Let the angle between $M$ and the $x$-axis be $\theta$.

Let the angle between $N$ and the $x$-axis be $\phi$.

Let $\pr_{M, N}$ be the projection on $M$ along $N$:

$\forall P \in \R^2: \map {\pr_{M, N} } P =$ the intersection of $M$ with the line through $P$ parallel to $N$.

Let $P = \tuple {x, y}$ be an arbitrary point in the plane.

Then:

$\map {\pr_{M, N} } P = \begin {cases} \tuple {0, y - x \tan \phi} & : \theta = \dfrac \pi 2 \\ \tuple {x, x \tan \theta} & : \phi = \dfrac \pi 2 \\ \tuple {\dfrac {x \tan \phi} {\tan \phi - \tan \theta} - \dfrac y {\tan \phi - \tan \theta}, \dfrac {y \tan \theta} {\tan \theta - \tan \phi} - \dfrac {x \tan \theta \tan \phi} {\tan \theta - \tan \phi} } & : \text {otherwise} \end {cases}$

## Proof

Let $P = \tuple {x_1, y_1} \in \R^2$ be arbitrary.

From Equation of Straight Line in Plane: Slope-Intercept Form, we can express $M$ and $N$ as follows:

 $\ds M: \,$ $\ds y$ $=$ $\ds m x$ $\ds N: \,$ $\ds y$ $=$ $\ds n x$

where:

 $\ds m$ $=$ $\ds \tan \theta$ $\ds n$ $=$ $\ds \tan \phi$

Let $\pr_{M, N}$ map $P = \tuple {x_1, y_1}$ onto $P' = \tuple {x_2, y_2}$.

From Equation of Straight Line in Plane: Point-Slope Form, $PP'$ can be expressed as:

$y - y_1 = n \paren {x - x_1}$

This holds also for $\tuple {x_2, y_2}$, and so:

 $\ds y_2 - y_1$ $=$ $\ds n \paren {x_2 - x_1}$ $\ds \leadsto \ \$ $\ds n x_2$ $=$ $\ds y_2 - y_1 + n x_1$ $\ds$ $=$ $\ds m x_2 - y_1 + n x_1$ as $\tuple {x_2, y_2} \in M$ $\ds \leadsto \ \$ $\ds \paren {n - m} x_2$ $=$ $\ds n x_1 - y_1$ $\ds \leadsto \ \$ $\ds x_2$ $=$ $\ds \dfrac {n x_1} {n - m} - \dfrac {y_1} {n - m}$

Similarly:

 $\ds y_2 - y_1$ $=$ $\ds n \paren {x_2 - x_1}$ $\ds \leadsto \ \$ $\ds \dfrac {y_2} n$ $=$ $\ds x_2 - x_1 + \dfrac {y_1} n$ $\ds$ $=$ $\ds \dfrac {y_2} m - x_1 + \dfrac {y_1} n$ as $\tuple {x_2, y_2} \in M$ $\ds \leadsto \ \$ $\ds \paren {\dfrac 1 n - \dfrac 1 m} y_2$ $=$ $\ds \dfrac {y_1} n - x_1$ $\ds \leadsto \ \$ $\ds y_2$ $=$ $\ds \dfrac {m y_1} {m - n} - \dfrac {m n x_1} {m - n}$

The result follows by substituting for $m$ and $n$.

$\Box$

### Special Cases

When either $M$ or $N$ are coincident with the $y$-axis, then either $\tan \theta$ or $\tan \phi$ is undefined.

Hence these cases are analysed as special cases.

Let $M$ be coincident with the $y$-axis.

By definition of the projection on $M$ along $N$:

$x_2 = 0$

while from the definition of tangent:

$\tan \phi = \dfrac {y_1 - y_2} {x_1}$

from which it follows that:

$y_2 = y_1 - x_1 \tan \phi$

Hence if $P = \tuple {x, y}$ is an arbitrary point in the plane, then:

$\map {\pr_{M, N} } P = \tuple {0, y - x \tan \phi}$

Let $N$ be coincident with the $y$-axis.

By definition of the projection on $M$ along $N$:

$x_2 = x_1$

while from the definition of tangent:

$\tan \theta = \dfrac {y_2} {x_2}$

from which it follows that:

$y_2 = x_1 \tan \theta$

Hence if $P = \tuple {x, y}$ is an arbitrary point in the plane, then:

$\map {\pr_{M, N} } P = \tuple {x, x \tan \theta}$

$\blacksquare$