Intersection of Straight Line in Homogeneous Cartesian Coordinates with Axes

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Theorem

Let $\LL$ be a straight line embedded in a cartesian plane $\CC$.

Let $\LL$ be given in homogeneous Cartesian coordinates by the equation:

$l X + m Y + n Z = 0$

such that $l$ and $m$ are not both zero.


Then $\LL$ intersects:

the $x$-axis $Y = 0$ at the point $\tuple {-n, 0, l}$
the $y$-axis $X = 0$ at the point $\tuple {0, -n, m}$


When $l = 0$, $\LL$ is parallel to the $x$-axis with its point at infinity at $\tuple {-n, 0, 0}$

When $m = 0$, $\LL$ is parallel to the $y$-axis with its point at infinity at $\tuple {0, -n, 0}$.


Proof

The intersection of $\LL$ with the $x$-axis is the point $\tuple {X, Y, Z}$ satisfied by:

\(\ds l X + m Y + n Z\) \(=\) \(\ds 0\)
\(\ds Y\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds l X + n Z\) \(=\) \(\ds 0\)

which is satisfied by setting $X = -n$ and $Z = l$, while $Y = 0$.


The intersection of $\LL$ with the $y$-axis is the point $\tuple {X, Y, Z}$ satisfied by:

\(\ds l X + m Y + n Z\) \(=\) \(\ds 0\)
\(\ds X\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds m Y + n Z\) \(=\) \(\ds 0\)

which is satisfied by setting $Y = -n$ and $Z = m$, while $X = 0$.


When $l = 0$ we have:

\(\ds m Y + n Z\) \(=\) \(\ds 0\)
\(\ds Y\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds n Z\) \(=\) \(\ds 0\)


When $m = 0$ we have:

\(\ds l X + n Z\) \(=\) \(\ds 0\)
\(\ds X\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds n Z\) \(=\) \(\ds 0\)


The result follows.

$\blacksquare$


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