“Who DARES to challenge the Mathematical Ninja?” he bellowed.

“It is I,” said the challenger. “@srcav, but Cav in real life.”

“Oh!” said the Mathematical Ninja. “Hello there, old chap, I was expecting someone else. Come in, I’ll put the kettle on.”

“Much obliged,” said Cav.

The challenge, which Cav repeated while dunking a ninjabread man into his cocoa, was this: “Why do you prefer $y-y_1 = m(x-x_1)$ over $y=mx+c$?”

“There are many reasons,” said the Mathematical Ninja. “Allow me to enumerate them.” He had offered Cav a ninjabread man, but not taken one himself; ninjabread men are notoriously sticky on the teeth, and would allow him to dominate the conversation.

The Special Case argument

“The simplest argument is this: $y=mx+c$ (henceforth known as the baby formula) is a special case of $y-y_1 = m(x-x_1)$, the grown-up version. If you know the $y$-intercept, $Y$, you can substitute $(x_1,y_1) = (0, Y)$ into it to get $y - Y = mx$, which is one step away from $y=mx+c$.”

The Unrestricted Given Point argument

“It baffles me that you’d - or rather, one - would introduce an unnecessary constant at GCSE, unless one were trying to torture students with more algebra than they need to do. If you’ve got a point that isn’t on the $y$-axis, you need to juggle things around, remember the difference between a constant and a variable, remember to write the equation down - none of which you need to do wit the proper version. You just substitute and rearrange.”

The Does-What-It-Says-On-The-Tin argument

“The grown-up formula, in its original form, tells you exactly what it’s doing: the difference from a given point in the $y$-direction is the same as the gradient multiplied by the difference in the $x$-direction. That’s pretty much the definition of a straight line. The baby formula does - technically - the same thing from the point where the line intersects the $y$-axis, but it’s nothing like as explicit. If either of the two promotes understanding, it’s the proper one.”

The fractional gradient argument

“When you’ve got a rational (but non-integer) gradient, especially at A-level when $ax + by + c = 0$ is the preferred way of reporting lines, the grown-up formula is much easier to manipulate. Let’s say you have a line with a gradient of $\frac{3}{2}$ that passes through $(3,5)$. The baby way:

$y = \frac{3}{2}x + c$, although I’ve yet to meet a student who wouldn’t write it as $1.5$. $5 = \frac{3}{2}\times 3 + c$ $c = 5 - \frac{9}{2} = \frac{1}{2}$ so $y = \frac{3}{2}x + \frac{1}{2}$. Better double that: $2y = 3x + 1$. Now rearrange: $0 = 3x - 2y + 1$. It’s possible, but what a mess!

$(y - 5) = \frac{3}{2}(x - 3)$ $2y - 10 = 3x - 9$ $0 = 3x - 2y + 1$

Would you seriously tell me the first way is better? Of course not, your mouth is stuck together.”

The only known counter-argument: the differential equation argument

“In the interest of balance, there is one possible counter-argument: if you have $\frac{\d y}{\d x} = m$ as a differential equation, you might separate it: $\int \d y = \int m \d x$, so $y = mx + c$. But I don’t do it that way.”

Of course you don’t, said Cav’s eyebrows.

“I’d do it as $y - y_0 = m(x - x_0)$, allowing me to substitute in what I’m given directly. That’s right: I use two constants rather than one.”

Cav almost unstuck his teeth in shock. Almost, but not quite.