# Image of Projection in Plane

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## Theorem

Let $M$ and $N$ be distinct lines in the plane.

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

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

Then $M$ is the image of $\pr_{M, N}$.

## Proof

Let $x \in \R^2$ be arbitrary.

By definition, the image of $x$ is the intersection of $M$ with the line through $x$ parallel to $N$.

Therefore $\map {\pr_{M, N} } x \in M$.

Hence:

- $\Img {\pr_{M, N} } \subseteq M$.

Now consider $y \in M$.

By Playfair's axiom there exists exactly one straight line $L$ parallel to $N$ passing through $y$.

Hence $y$ is the image of every point on $L$.

As $y$ is arbitrary, it follows that each point on $M$ is the image of some straight line in $\R^2$ under $\pr_{M, N}$.

That is:

- $M \subseteq \Img {\pr_{M, N} }$

The result follows by definition of set equality.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.5$