Collinearity

Video lesson · from 12:522 worked examples

One lesson video covers all of The Straight Line, so it opens at 12:52 for this topic — not from the beginning.

Points which lie on the same straight line are said to be collinear.

m_AB ≠ m_BC

B is common

m_AB = m_CD

No common point

m_AB = m_BC

B is common

To prove points A, B, and C are collinear, you must show that:

Write the conclusion out. Conclusions here are very often left unstated, and that is where the marks go. Equal gradients on their own do not finish it — state that the gradients are equal, that $B$ is common to both, and therefore the points are collinear.
The common point is half the proof: two separate lines can have equal gradients and never meet. Naming the shared point is what rules that out.
Do not stop at “parallel”: equal gradients alone prove the lines are parallel, which is a different statement.
Keep gradients exact: leave them as fractions. Rounding to decimals can make two genuinely equal gradients look different.

Example 1

Show that the points $P(-6, -1)$ , $Q(0, 2)$ and $R(8,6)$ are collinear.

\begin{aligned} m_{PQ} &= \frac{2 - (-1)}{0 - (-6)} \\ &= \frac{3}{6} \\ &= \frac{1}{2} \end{aligned}

\begin{aligned} m_{QR} &= \frac{6 - 2}{8 - 0} \\ &= \frac{4}{8} \\ &= \frac{1}{2} \end{aligned}

Since $m_{PQ} = m_{QR}$ and $Q$ is a common point, the points $P, Q, \text{and } R$ are collinear.

Example 2

The points $A(1, -1)$ , $B(-1, k)$ and $C(5, 7)$ are collinear. Find the value of $k$ .

Since they are collinear, $m_{AB} = m_{BC}$ .

\begin{aligned} m_{BC} &= \frac{7 - k}{5 - (-1)} \\ &= \frac{7 - k}{6} \end{aligned}

\begin{aligned} m_{AC} &= \frac{7 - (-1)}{5 - 1} \\ &= \frac{8}{4} \\ &= 2 \end{aligned}

Since the line is the same, $m_{BC}$ must be 2.

\frac{7 - k}{6} = 2

7 - k = 12 \implies k = -5