Coordinate geometry turns shape into number without losing the geometry. A distance becomes a square-root expression. A direction becomes a gradient. A straight line becomes an equation. Parallelism and perpendicularity become relationships between gradients. Midpoints become averages of coordinates.
This forty-second Secondary 4 Mathematics Learning Guide develops coordinate geometry as a synthesis of algebra, graphs and Euclidean structure. It belongs to the Secondary Mathematics Hub and S1–S4 Capability Map.
It complements Linear Graphs, Gradient, Intercepts and Rate of Change and Coordinate Geometry and Transformations as Representation.
Segment length comes from Pythagoras
For A(x₁,y₁) and B(x₂,y₂), the horizontal change is x₂−x₁ and the vertical change is y₂−y₁. These form the perpendicular legs of a right triangle.
AB=√[(x₂−x₁)²+(y₂−y₁)²].
Worked Example 1 | Find a segment length
Find the distance between A(2,3) and B(8,11).
Horizontal change=6. Vertical change=8.
AB=√(6²+8²)=√100=10.
Midpoint is coordinate averaging
The midpoint of A(x₁,y₁) and B(x₂,y₂) is:
((x₁+x₂)/2,(y₁+y₂)/2).
Worked Example 2 | Midpoint
Find the midpoint of (−4,7) and (10,−1).
((−4+10)/2,(7−1)/2)=(3,3).
The midpoint is equally distant from both endpoints along the segment.
Gradient records direction and steepness
For two points on a non-vertical line:
m=(y₂−y₁)/(x₂−x₁).
A positive gradient rises left to right; a negative gradient falls. A horizontal line has gradient 0. A vertical line has undefined gradient.
Worked Example 3 | Gradient from endpoints
Find the gradient through P(−2,5) and Q(4,−7).
m=(−7−5)/(4−(−2))=−12/6=−2.
Straight-line equation from gradient and one point
Use y=mx+c. Substitute a known point to find c.
Worked Example 4 | Equation through one point
A line has gradient 3 and passes through (2,−1). Find its equation.
Use y=3x+c. Substitute:
−1=6+c, so c=−7.
y=3x−7.
Equation from two points
First calculate the gradient, then substitute either point into y=mx+c.
Worked Example 5 | Two-point line
Find the equation through (1,6) and (5,14).
m=(14−6)/(5−1)=2.
6=2(1)+c, so c=4.
y=2x+4.
Parallel lines preserve gradient
Distinct parallel non-vertical lines have equal gradients. Their intercepts differ.
Worked Example 6 | Parallel condition
Find the equation of the line through (3,5) parallel to y=−4x+1.
The required gradient is −4.
5=−4(3)+c, so c=17.
y=−4x+17.
Perpendicular lines use negative reciprocal gradients
For non-horizontal, non-vertical perpendicular lines:
m₁m₂=−1.
If one gradient is 3/2, a perpendicular gradient is −2/3.
Worked Example 7 | Perpendicular bisector idea
A segment joins A(0,2) and B(6,8). Find the equation of its perpendicular bisector.
Midpoint=(3,5).
Gradient AB=(8−2)/(6−0)=1.
Perpendicular gradient=−1.
Use y=−x+c through (3,5): 5=−3+c, so c=8.
y=−x+8.
Coordinate geometry can prove a quadrilateral property
Instead of relying on appearance, use gradients and lengths to prove relationships.
Worked Example 8 | Show opposite sides are parallel
Let A(0,0), B(4,2), C(7,8), D(3,6).
Gradient AB=2/4=1/2.
Gradient DC=(8−6)/(7−3)=2/4=1/2.
Therefore AB∥DC.
Gradient BC=(8−2)/(7−4)=6/3=2.
Gradient AD=(6−0)/(3−0)=2.
Therefore BC∥AD, so ABCD is a parallelogram.
Coordinate geometry can prove equal lengths
To show two segments are equal, compare their squared lengths. Squaring avoids unnecessary square roots.
Worked Example 9 | Equal side proof
A(1,1), B(5,4), C(8,0). Compare AB and BC.
AB²=(5−1)²+(4−1)²=16+9=25.
BC²=(8−5)²+(0−4)²=9+16=25.
AB=BC=5.
Thus triangle ABC is isosceles at B.
Intersections solve simultaneous geometric conditions
If two straight-line equations describe boundaries or paths, their intersection is the coordinate satisfying both equations.
Worked Example 10 | Intersection of two paths
Find the intersection of y=2x+1 and y=−x+7.
Set 2x+1=−x+7.
3x=6, so x=2 and y=5.
Intersection=(2,5).
Coordinate area can be decomposed geometrically
When vertices lie on horizontal or vertical lines, area can often be found by splitting into rectangles and triangles rather than introducing new formulae.
Worked Example 11 | Triangle area from coordinates
A(1,2), B(7,2), C(7,6). Find the area of triangle ABC.
AB=6 and BC=4. The sides are perpendicular because one is horizontal and one vertical.
Area=1/2×6×4=12 square units.
A line equation is a geometric constraint
Every point on y=2x+3 must satisfy that equation. If a proposed point does not, it is not on the line.
Worked Example 12 | Test a point
Does (4,11) lie on y=2x+3?
2(4)+3=11.
Yes. The coordinate satisfies the line equation.
A coordinate-geometry workflow
- Mark the relevant coordinates clearly.
- Decide whether the required relationship concerns length, midpoint, gradient, line equation or intersection.
- Use the smallest appropriate formula.
- Keep subtraction order consistent.
- Use exact values where practical.
- Translate the numerical result back into the geometric claim.
- Verify using a second property when possible.
Common failure modes
| Error | Cause | Repair |
|---|---|---|
| Distance formula drops one square | Pythagorean structure forgotten | Write horizontal²+vertical² |
| Midpoint subtracts coordinates | Midpoint and gradient mixed | Average coordinates |
| Gradient subtraction order inconsistent | Point order changes between numerator and denominator | Use the same point order |
| Parallel and perpendicular rules reversed | Relationship memorised without meaning | Parallel=same slope; perpendicular=negative reciprocal |
| Claims shape from diagram appearance | Visual evidence replaces proof | Use gradients and lengths |
| Finds intersection but gives only x | Coordinate pair incomplete | Substitute back to get y |
Independent practice
- Find the distance between (1,2) and (7,10).
- Find the midpoint of (−6,4) and (8,12).
- Find the gradient through (3,9) and (7,1).
- Find the equation of the line with gradient −3 through (2,5).
- Find the equation through (0,4) perpendicular to y=2x−1.
- Find the intersection of y=3x−2 and y=−x+6.
Explained answers
1. √[(7−1)²+(10−2)²]=√(36+64)=10.
2. ((−6+8)/2,(4+12)/2)=(1,8).
3. (1−9)/(7−3)=−8/4=−2.
4. 5=−3(2)+c, so c=11. y=−3x+11.
5. Perpendicular gradient=−1/2. Through (0,4): y=−x/2+4.
6. 3x−2=−x+6 gives x=2, y=4. (2,4).
Final thought
Coordinate geometry is powerful because the same object can be read in two languages. The diagram shows spatial structure; the coordinates turn that structure into calculations that can be checked and justified.
Use coordinates to make the geometry measurable, then return the numbers to the geometric claim they are meant to prove.
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