Every so often, I see a tweet so marvellous I can't believe it's true. Then I bookmark it and forget about it for months, until I don't know what to write next.

An example is @robjlow's message from June:

a) Isn't that lovely? and b) Hang on, is it really true?

### Let's algebra

Working out the determinant using my favourite diagonal stripes method gives $x(y_1 - y_2) + y(x_2 - x_1) + (x_1 y_2 - x_2 y_1) = 0$.

Does that look plausible? Certainly it's a straight line, and certainly replacing $x$ and $y$ with either $x_1$ and $y_1$ or $x_2$ and $y_2$ makes it a true statement.

Maybe we can rearrange it into a nicer form? $y(x_1 - x_2) = x(y_1 - y_2) + (x_1 y_2 - x_2 y_1)$, or $y = \frac{y_1 - y_2}{x_1 - x_2} x + \frac{x_1 y_2 - x_2 y_1}{x_1 - x_2}$. The gradient looks right and the $y$-intercept looks plausible.

But wait, we can convince ourselves more elegantly!

### Let's **algebra**!

If a point lies on the line between $\br{x_1, y_1}$ and $\br{x_2, y_2}$, it can be written as (abusing notation slightly) $k \br{x_1, y_1} + \br{1-k} \br{x_2, y_2}$ - with the point where $k=0$ corresponding to $\br{x_1, y_1}$ and $k=1$ to $\br{x_2, y_2}$.

That is to say, any point that lies on the line between the two points is a linear combination of them - which means the top row of the matrix can be written as a linear combination of the other two. The rows of the matrix are not linearly independent, so its determinant is zero!

Similarly, you can show that *only* points on the line can be written in this form, so the determinant defines the line. How lovely!

- Thanks to Rob for the original tweet!

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.