Points of Intersection

Video lesson · from 54:361 worked example

One lesson video covers all of The Straight Line, so it opens at 54:36 for this topic — not from the beginning.

Theory

Many problems involve lines which intersect (cross each other). Once we have equations for the lines, there are three ways of calculating the point of intersection using simultaneous equations:

Elimination
Equating
Substitution

Use whichever method is most efficient for the problem you are tackling.

Line Type	Point Used	Gradient Used
Perpendicular Bisector of AB	Midpoint of AB	$m_{\perp}$ from $m_{AB}$
Altitude from C	Vertex C	$m_{\perp}$ from $m_{AB}$
Median from C	Vertex C and Midpoint of AB	$m$ between C and Midpoint of AB

⚠️ Common Examiner Traps

Substitution is usually the faster route: candidates reach for elimination out of habit where substitution is far quicker, and the extra steps invite errors. If one equation is already in the form $y = \ldots$ , substitute it.
Do not score out terms mid-line: scoring out terms in simultaneous equations produces lines of working that no longer follow from each other, and that costs marks even when the answer is right. Write each new equation out in full.
Answer as coordinates: finding $x$ is half the job. Substitute back for $y$ and give the point.
Check in the other equation: substituting your point into the equation you did not use catches most arithmetic slips in seconds.

Worked examples

Example 1

Triangle PQR has vertices $P(8, 3)$ , $Q(-1, 6)$ and $R(2, -3)$ .

Find the equation of altitude QS.
Find the equation of median RT.
Hence find the coordinates of M (the point of intersection of QS and RT).

a) Altitude QS

\begin{aligned} m_{PR} &= \frac{-3 - 3}{2 - 8} \\ &= \frac{-6}{-6} \\ &= 1 \end{aligned}

m_{\perp} = -1

y - 6 = -1(x - (-1))

y - 6 = -x - 1 \implies x + y = 5 \text{ (1)}

b) Median RT

Midpoint of PQ (T):

\begin{aligned} T &= \left(\frac{8 + (-1)}{2}, \frac{3 + 6}{2}\right) \\ &= \left(\frac{7}{2}, \frac{9}{2}\right) \end{aligned}

\begin{aligned} m_{RT} &= \frac{\frac{9}{2} - (-3)}{\frac{7}{2} - 2} \\ &= \frac{\frac{15}{2}}{\frac{3}{2}} \\ &= 5 \end{aligned}

y - (-3) = 5(x - 2)

y + 3 = 5x - 10

y = 5x - 13 \text{ (2)}

c) Point of Intersection M

Substitute (2) into (1):

x + (5x - 13) = 5

\begin{aligned} 6x - 13 &= 5 \\ 6x &= 18 \\ x &= 3 \end{aligned}

Substitute $x=3$ back into (2):

y = 5(3) - 13 = 15 - 13 = 2

The coordinates of M are $(3, 2)$ .