Showing Three Points Are Collinear

Three points are collinear if they lie on one straight line. The test is: the gradient between two pairs is equal, and they share a common point. Both parts matter, equal gradients alone would only prove parallel.

Show that the points $A(1, 2)$, $B(3, 6)$ and $C(5, 10)$ are collinear. [3]

The working

Step 1, gradient of $AB$: $m_{AB} = \frac{6 - 2}{3 - 1} = \frac{4}{2} = 2 \quad \text{(M1)}$

Step 2, gradient of $BC$: $m_{BC} = \frac{10 - 6}{5 - 3} = \frac{4}{2} = 2 \quad \text{(A1)}$

Step 3, conclude. Since $m_{AB} = m_{BC} = 2$ and the two segments share the point $B$, the points $A$, $B$ and $C$ lie on the same straight line, they are collinear. $\blacksquare$ (A1)

Where the marks are won and lost

• Compute two gradients from the three points ($AB$ and $BC$, or $AB$ and $AC$). One gradient proves nothing.
• The conclusion must mention the shared point. Equal gradients only show the lines are parallel; the common point $B$ upgrades that to collinear. Stating this earns the final mark.
• Keep the coordinate order consistent in each gradient ($\frac{y_2 - y_1}{x_2 - x_1}$).

Common mistakes

• Computing only one gradient.
• Concluding “collinear” from equal gradients without noting the shared point.
• Mixing up the order of coordinates and getting a sign wrong.

Full method: Gradient, Midpoint & Length notes. Topic home: Straight-Line Graphs pillar.