You already plot points and work with straight-line graphs \(y=mx+c\). Additional Mathematics treats the line segment joining two given points as an object in its own right: its length, its midpoint and its gradient all come directly from the coordinates.
Line segment joining \(A(x_1,\,y_1)\) and \(B(x_2,\,y_2)\)
| Quantity | Formula | How it works |
|---|---|---|
| Distance \(AB\) | \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) | Pythagoras on the horizontal difference and the vertical difference. |
| Midpoint of \(AB\) | \(\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)\) | Average the \(x\)-coordinates; average the \(y\)-coordinates. |
| Gradient of \(AB\) | \(m=\dfrac{y_2-y_1}{x_2-x_1}=\tan\theta\) | \(\theta\) is the angle the line makes with the positive \(x\)-axis. |
The horizontal difference \(x_2-x_1\) and the vertical difference \(y_2-y_1\) form the two legs of a right-angled triangle: the distance is the hypotenuse, the gradient is the ratio of the legs, and the midpoint \(M\) sits halfway along each leg.
Collinear points
Points \(A\), \(B\) and \(C\) are collinear when they lie on one straight line: \(m_{AB}=m_{BC}=m_{AC}\). To show collinearity it is enough to show \(m_{AB}=m_{BC}\) — the two gradients share the point \(B\), so equal gradients force all three points onto the same line.
Keep the subtraction order consistent in every formula: \(\dfrac{y_2-y_1}{x_2-x_1}\) is correct, \(\dfrac{y_2-y_1}{x_1-x_2}\) flips the sign of the gradient.
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