Amber Quinn

2022-06-22

Suppose a line goes through the points $A=({x}_{1},{y}_{1})$ and $B=({x}_{2},{y}_{2})$. 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

$\begin{array}{}\text{(1)}& b=-\frac{det(A,B)}{\mathrm{\Delta}x}\end{array}$

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.

$b=-\frac{{x}_{1}{y}_{2}-{x}_{2}{y}_{1}}{{x}_{2}-{x}_{1}}$

where $b$ is the y-coordinate of the $y$-intercept. This can be written more suggestively as

$\begin{array}{}\text{(1)}& b=-\frac{det(A,B)}{\mathrm{\Delta}x}\end{array}$

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

Beginner2022-06-23Added 16 answers

Basically you are using the fact that the area enclosed by three collinear points is zero

$A({x}_{1},{y}_{1}),B({x}_{2},{y}_{x}),C(0,b)$ are the three points, then

$\left|\begin{array}{ccc}{x}_{1}& {y}_{1}& 1\\ {x}_{2}& {y}_{2}& 1\\ 0& b& 1\end{array}\right|=0$

Now expand along the bottom row to get the equation you have

$A({x}_{1},{y}_{1}),B({x}_{2},{y}_{x}),C(0,b)$ are the three points, then

$\left|\begin{array}{ccc}{x}_{1}& {y}_{1}& 1\\ {x}_{2}& {y}_{2}& 1\\ 0& b& 1\end{array}\right|=0$

Now expand along the bottom row to get the equation you have