Solving Problems on Coordinate Axes Revision Notes | GCSE AQA Higher Maths

1. Why Coordinate Geometry Solves Shape Problems

Geometry questions often ask you to prove a shape is a square, find the area of a triangle, or show two lines are perpendicular. When those shapes live on a coordinate grid, you have a powerful toolkit: algebra. Instead of measuring with a ruler, you calculate.

Every geometrical problem on coordinate axes comes down to three core tools: gradient, midpoint, and distance. Master these three and you can classify any shape, find any missing vertex, and calculate any area the examiner throws at you.

2. The Three Core Formulas

Before working through problems, here are the three formulas you need. Each one comes directly from ideas you already know — the gradient formula from rise-over-run, the midpoint from averaging, and the distance formula from Pythagoras.

Notice how the distance formula is Pythagoras in disguise: the horizontal gap $(x_2 - x_1)$ and vertical gap $(y_2 - y_1)$ form the two shorter sides of a right-angled triangle, and $d$ is the hypotenuse.

3. Using Gradient to Identify Parallel and Perpendicular Lines

Two lines are parallel when they have the same gradient. Two lines are perpendicular when their gradients multiply to $-1$. This second rule is the one that trips people up, so let's make it concrete.

This rule is what lets you prove a shape contains a right angle without measuring anything. Calculate the gradients of two sides meeting at a vertex — if their product is $-1$, the angle is $90^\circ$.

Worked example: Proving Two Lines Are Perpendicular

Points $A(2, 1)$, $B(6, 3)$, and $C(5, 5)$ are given. Show that $AB$ is perpendicular to $BC$.

1. $m_{AB} = \dfrac{3 - 1}{6 - 2} = \dfrac{2}{4} = \dfrac{1}{2}$ — Apply $m = \frac{y_2 - y_1}{x_2 - x_1}$ with $A(2,1)$ and $B(6,3)$. [M1]
2. $m_{BC} = \dfrac{5 - 3}{5 - 6} = \dfrac{2}{-1} = -2$ — Apply the same formula with $B(6,3)$ and $C(5,5)$. [M1]
3. $m_{AB} \times m_{BC} = \dfrac{1}{2} \times (-2) = -1$ — The product of the gradients equals $-1$, confirming perpendicularity. [A1]
4. Since $m_{AB} \times m_{BC} = -1$, $AB$ is perpendicular to $BC$. — Always state the conclusion explicitly — 'show that' questions require a clear final statement.

Answer: $AB \perp BC$ because $m_{AB} \times m_{BC} = \frac{1}{2} \times (-2) = -1$.

4. Using Distance and Midpoint to Classify Shapes

Once you can calculate side lengths and midpoints, you can classify any quadrilateral or triangle. The strategy is always the same: work out what you need to prove, then choose the right formula to prove it.

• To prove a square: show all four sides are equal AND adjacent sides are perpendicular.
• To prove a rhombus: show all four sides are equal (perpendicularity is not required).
• To prove a rectangle: in a parallelogram, show the diagonals are equal in length; alternatively show adjacent sides are perpendicular.
• To prove an isosceles triangle: show exactly two sides are equal in length.
• To prove a right-angled triangle: show the perpendicularity condition holds at one vertex, or verify Pythagoras' theorem holds.

Worked example: Classifying a Quadrilateral Using Coordinates

The vertices of quadrilateral $PQRS$ are $P(0, 0)$, $Q(4, 0)$, $R(5, 3)$, $S(1, 3)$. Show that $PQRS$ is a parallelogram but not a rectangle.

1. $m_{PQ} = \dfrac{0 - 0}{4 - 0} = 0$ — $PQ$ is horizontal. [M1]
2. $m_{SR} = \dfrac{3 - 3}{5 - 1} = 0$ — $SR$ is also horizontal, so $PQ \parallel SR$. [A1]
3. $m_{PS} = \dfrac{3 - 0}{1 - 0} = 3$ — Apply the gradient formula with $P(0,0)$ and $S(1,3)$. [M1]
4. $m_{QR} = \dfrac{3 - 0}{5 - 4} = \dfrac{3}{1} = 3$ — $QR$ has the same gradient as $PS$, so $PS \parallel QR$. Both pairs of opposite sides are parallel, confirming a parallelogram. [A1]
5. $m_{PQ} \times m_{PS} = 0 \times 3 = 0 eq -1$ — A horizontal line has gradient $0$. For perpendicularity we need the product to equal $-1$, but $0 \times 3 = 0$. Adjacent sides are NOT perpendicular, so $PQRS$ is not a rectangle. [A1]
6. $PQRS$ is a parallelogram (two pairs of parallel sides) but not a rectangle (no right angles). — Always write a clear conclusion that directly addresses the question.

Answer: $PQRS$ is a parallelogram because $PQ \parallel SR$ and $PS \parallel QR$. It is not a rectangle because adjacent sides are not perpendicular ($m_{PQ} \times m_{PS} = 0 eq -1$).

5. Finding Areas of Shapes on a Coordinate Grid

Area questions on coordinate axes are common at GCSE. The most reliable method is to enclose the shape in a rectangle, then subtract the surrounding right-angled triangles. This avoids needing a special formula for every shape.

For triangles, you can also use the formula $\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height}$, provided you can identify a base and a perpendicular height from the coordinates. When the base is horizontal or vertical, this is straightforward.

Worked example: Finding the Area of a Triangle on a Coordinate Grid

Triangle $ABC$ has vertices $A(1, 1)$, $B(7, 1)$, $C(4, 5)$. Calculate the area of triangle $ABC$.

1. $AB$ runs from $(1, 1)$ to $(7, 1)$, so base $= 7 - 1 = 6$ — Both points share $y = 1$, so $AB$ is horizontal. The base length is the difference in $x$-coordinates. [M1]
2. $C$ is at $y = 5$; base $AB$ is at $y = 1$; height $= 5 - 1 = 4$ — The perpendicular height is the vertical distance from $C$ down to the line $AB$. [M1]
3. $\text{Area} = \dfrac{1}{2} \times 6 \times 4 = \dfrac{1}{2} \times 24 = 12$ — Substitute base $= 6$ and height $= 4$ into $\frac{1}{2} \times b \times h$. [A1]
4. $\text{Area} = 12 \text{ square units}$ — Coordinates are dimensionless unless the question specifies a scale, so write 'square units'. [A1]

Answer: Area $= 12$ square units

6. Multi-Step Coordinate Problems

Exam questions often chain several of these skills together. You might need to find a midpoint, then use that midpoint to calculate a gradient, then use the gradient to write an equation. The key is to break the problem into clear stages and show each one.

Worked example: Finding the Equation of a Perpendicular Bisector

Points $A(2, 6)$ and $B(8, 2)$ are given. Find the equation of the perpendicular bisector of $AB$.

1. $M = \left(\dfrac{2 + 8}{2},\ \dfrac{6 + 2}{2}\right) = \left(\dfrac{10}{2},\ \dfrac{8}{2}\right) = (5, 4)$ — The perpendicular bisector passes through the midpoint of $AB$. [M1]
2. $m_{AB} = \dfrac{2 - 6}{8 - 2} = \dfrac{-4}{6} = -\dfrac{2}{3}$ — We need this to find the perpendicular gradient. [M1]
3. $m_{\perp} = -\dfrac{1}{m_{AB}} = -\dfrac{1}{-\frac{2}{3}} = \dfrac{3}{2}$ — Negative reciprocal: flip $-\frac{2}{3}$ and change the sign to get $\frac{3}{2}$. [M1]
4. $4 = \dfrac{3}{2}(5) + c$ $4 = \dfrac{15}{2} + c$ $c = 4 - \dfrac{15}{2} = \dfrac{8}{2} - \dfrac{15}{2} = -\dfrac{7}{2}$ — Substitute $M(5, 4)$ and $m = \frac{3}{2}$ into $y = mx + c$ to find $c$. [M1]
5. $y = \dfrac{3}{2}x - \dfrac{7}{2}$ — This can also be written as $2y = 3x - 7$ or $3x - 2y - 7 = 0$. [A1]

Answer: $y = \dfrac{3}{2}x - \dfrac{7}{2}$ (equivalently $2y = 3x - 7$)

7. Deliberate Practice

These problems cover the full range of skills from this lesson. Work through each one before checking the solution — the process of attempting it first is what builds exam confidence.

8. Exam Technique and Common Mistakes

Coordinate geometry questions are reliable mark-earners if you follow a disciplined method. The most common reason students lose marks is not a lack of knowledge — it is rushing the substitution step and making arithmetic errors with negatives.

One final point on exam strategy: when a question says 'prove' or 'show that', every line of working must follow logically from the previous one. Examiners award marks for the method chain, not just the final statement. Write a conclusion sentence at the end that explicitly references the condition you have verified.