Amber Quinn

2022-06-22

Suppose a line goes through the points $A=\left({x}_{1},{y}_{1}\right)$ and $B=\left({x}_{2},{y}_{2}\right)$. One can easily check (as I did today while doodling) that
$b=-\frac{{x}_{1}{y}_{2}-{x}_{2}{y}_{1}}{{x}_{2}-{x}_{1}}$
where $b$ is the y-coordinate of the $y$-intercept. This can be written more suggestively as
$\begin{array}{}\text{(1)}& b=-\frac{det\left(A,B\right)}{\mathrm{\Delta }x}\end{array}$
The presence of det(A,B) suggests a geometrical interpretation, but I couldn't think of one. This reminds me of Cramer's Rule, but I couldn't make that connection explicit either.

Josie123

Basically you are using the fact that the area enclosed by three collinear points is zero
$A\left({x}_{1},{y}_{1}\right),B\left({x}_{2},{y}_{x}\right),C\left(0,b\right)$ are the three points, then
$|\begin{array}{ccc}{x}_{1}& {y}_{1}& 1\\ {x}_{2}& {y}_{2}& 1\\ 0& b& 1\end{array}|=0$
Now expand along the bottom row to get the equation you have

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