Coordinate geometry turns a diagram into a system of numbers without losing the geometry. A line can be described by two points, a gradient and an equation. A translation can be described by one vector. An enlargement can be described by a scale factor and centre. The strength of the coordinate method is that spatial relationships become calculable and checkable.
This Secondary 3 Mathematics Learning Guide develops gradient, distance, straight-line equations, coordinate-based geometry, translations and scale transformations. It also explains how coordinate methods connect to vectors, similarity and graph interpretation rather than treating them as separate chapters.
The official 2027 SEC G3 Mathematics syllabus K310 includes gradient from coordinates, length of a line segment, equations of straight lines in the form y = mx + c, geometric problems using coordinates, enlargement and reduction of plane figures, scale drawings, and translation by a vector. Other transformation language may be used by schools as supporting geometry; this guide keeps the core worked route tied to those stated ideas.
Use the Secondary Mathematics Hub for the wider route. Inside this article, go to diagnostic · gradients and lines · distance and midpoint · transformations · geometry problems · practice · answers.
Coordinates Are Locations, Not Vectors
The point A = (3, 5) describes a location relative to the origin. The vector from A to B describes a displacement between locations. If B = (8, 2), then AB = (5, −3).
This distinction matters because points are not added or subtracted arbitrarily. Coordinate differences have geometric meaning: they describe horizontal and vertical change.
A Six-Question Diagnostic
For A = (1, 2) and B = (5, 10), find the horizontal change, vertical change, gradient and length AB. Find the midpoint. Finally, write the equation of the line through A with gradient 2.
The answers are 4; 8; gradient 2; length √80 = 4√5; midpoint (3, 6); and y = 2x. The line equation follows because A = (1,2) lies on y=2x. These questions test difference, ratio, Pythagoras, averaging and line equations separately.
Gradient Measures Rate of Vertical Change Against Horizontal Change
For two distinct points (x₁,y₁) and (x₂,y₂) with x₂ ≠ x₁, gradient m = (y₂−y₁)/(x₂−x₁). The numerator and denominator must follow the same point order.
A positive gradient rises from left to right. A negative gradient falls. A zero gradient is horizontal. A vertical line has zero horizontal change, so the usual gradient ratio would require division by zero; its gradient is therefore undefined.
Worked Example 1: Gradient From Two Points
Find the gradient of the line through P = (−3, 4) and Q = (5, −8). Vertical change is −8−4 = −12. Horizontal change is 5−(−3)=8.
Therefore m = −12/8 = −3/2. Reversing both point orders gives 12/(−8), the same result. Reversing only one order would produce the wrong sign.
The Equation y = mx + c
In y = mx + c, m is the gradient and c is the y-intercept. The y-intercept is the y-value when x = 0.
For y = 3x − 7, the gradient is 3 and the line crosses the y-axis at (0, −7). Every point on the line satisfies the equation.
Worked Example 2: Find a Line Equation
A line has gradient 4 and passes through (2, 11). Find its equation. Start with y = 4x + c. Substitute the known point: 11 = 8 + c, so c = 3.
The equation is y = 4x + 3. Check the point: 4(2)+3=11.
Worked Example 3: Line Through Two Points
Find the equation through A=(1,5) and B=(4,14). Gradient = (14−5)/(4−1)=3. Use y=3x+c and point A: 5=3+c, so c=2.
Therefore y=3x+2. Substituting x=4 gives 14, so B is also verified.
Parallel and Perpendicular Lines as a Useful Extension
Parallel nonvertical lines have equal gradients. In the usual Cartesian setting, perpendicular nonvertical lines have gradients whose product is −1 when both gradients are defined. Use this relationship according to the geometry depth expected by your school.
For example, a line of gradient 2 is perpendicular to one of gradient −1/2. A horizontal line is perpendicular to a vertical line, which must be handled separately because the vertical gradient is undefined.
Distance Is Pythagoras Applied to Coordinate Change
Between A=(x₁,y₁) and B=(x₂,y₂), horizontal change is x₂−x₁ and vertical change is y₂−y₁. These form perpendicular legs of a right triangle.
Therefore AB = √[(x₂−x₁)² + (y₂−y₁)²]. Squaring means the final distance is independent of which point order is chosen.
Worked Example 4: Length of a Segment
A=(−2,3), B=(4,11). Find AB. Horizontal change is 6 and vertical change is 8. Hence AB=√(36+64)=10.
This matches the magnitude of vector AB=(6,8), showing the connection to Vectors and Geometric Relationships.
Midpoint as an Average Position
The midpoint of A=(x₁,y₁) and B=(x₂,y₂) is ((x₁+x₂)/2, (y₁+y₂)/2). Each coordinate lies halfway between the corresponding endpoint coordinates.
For A=(−4,7), B=(10,−1), midpoint = (3,3). This also follows from vector averaging of the two position vectors.
Worked Example 5: Prove a Triangle Is Isosceles
A=(0,0), B=(6,0), C=(3,4). Show triangle ABC is isosceles. AC=√(3²+4²)=5. BC=√[(3−6)²+(4−0)²]=5.
Since AC=BC, the triangle is isosceles. The coordinate method turns a geometric claim into two distance calculations.
Worked Example 6: Prove a Right Angle
A=(0,0), B=(4,2), C=(−1,2). Show angle BAC is 90°. Gradient AB = 2/4=1/2. Gradient AC = 2/(−1)=−2.
The product is −1, so the lines are perpendicular under the stated gradient criterion. Therefore angle BAC is 90°.
Translation: Add the Same Vector to Every Point
A translation by vector (p,q) maps (x,y) to (x+p,y+q). Since every point moves by the same displacement, lengths, angles and orientation are preserved.
Triangle vertices A=(1,2), B=(4,2), C=(1,6) translated by (3,−1) become A′=(4,1), B′=(7,1), C′=(4,5).
The side lengths remain the same because pairwise coordinate differences remain unchanged after the same vector is added to every point.
Worked Example 7: Recover a Translation Vector
Point P=(−2,5) maps to P′=(4,1). Find the translation vector. End minus start gives (4−(−2), 1−5)=(6,−4).
Any second point under the same translation should have exactly the same coordinate difference. That provides a direct consistency check.
Enlargement and Reduction
An enlargement with positive scale factor k multiplies every distance from the centre of enlargement by k along the same ray. A factor greater than 1 enlarges; a factor between 0 and 1 reduces.
If the centre is the origin, (x,y) maps simply to (kx,ky). For example, scale factor 3 sends (2,−1) to (6,−3).
If the centre is not the origin, first form the displacement from the centre to the point, scale that displacement, then add the centre back.
Worked Example 8: Enlargement About a Non-Origin Centre
Centre C=(1,2), point P=(4,6), scale factor 2. Find P′. CP=(3,4). Doubling gives CP′=(6,8).
Add the centre back: P′=(1,2)+(6,8)=(7,10). The point lies on the same ray from C and twice as far away.
Scale Factor Changes Length, Area and Volume Differently
For similar plane figures, a length factor k gives area factor k². For similar solids, it gives volume factor k³. The coordinate transformation controls lengths; dimensional scaling determines the resulting area or volume change.
This connection is developed further in Similarity, Scale Factors and Mensuration.
Reflection and Rotation as Representation Extensions
Schools may use reflections and rotations to strengthen transformation reasoning. Around the origin, simple coordinate rules include reflection in the x-axis: (x,y)→(x,−y); reflection in the y-axis: (x,y)→(−x,y); and a 180° rotation about the origin: (x,y)→(−x,−y).
Treat these as representation extensions unless your current school scope explicitly requires them. The central skill is to preserve the stated transformation exactly and check the image against fixed points, distances or directions.
Coordinate Geometry as Proof
A coordinate proof converts a geometric property into one or more numerical conditions. Equal gradients can establish parallelism. Perpendicular gradients can establish a right angle. Equal lengths can establish an isosceles triangle. A common midpoint can establish that diagonals bisect each other.
The method is powerful because several independent checks are often available. A claimed rectangle, for example, can be tested for parallel opposite sides and one right angle, or by appropriate coordinate relationships.
Worked Example 9: Show a Quadrilateral Is a Parallelogram
A=(1,1), B=(5,2), C=(7,6), D=(3,5). Show ABCD is a parallelogram.
AB=(4,1). DC=C−D=(4,1). Also BC=(2,4), while AD=D−A=(2,4). Both pairs of opposite directed sides are equal.
Therefore ABCD is a parallelogram. The vector method is equivalent to a coordinate-difference proof.
Worked Example 10: Intersection of Two Lines
Find the intersection of y=2x+1 and y=−x+7. At the intersection both equations have the same x and y, so set 2x+1=−x+7.
Then 3x=6, x=2, and y=5. The intersection is (2,5). This connects coordinate geometry to simultaneous equations and graph intersections.
Four Common Errors
Gradient sign error: numerator order and denominator order are inconsistent. Repair by writing both differences in the same point order.
Distance without squaring: coordinate differences are simply added. Repair by drawing the right triangle and using Pythagoras.
Translation applied to only one coordinate: repair by reading the vector as horizontal and vertical displacement together.
Enlargement from the wrong centre: coordinates are multiplied directly even though the centre is not the origin. Repair by scaling the displacement from the stated centre.
Independent Practice
1. Find the gradient through (2,3) and (8,15).
2. Find the distance between (−1,4) and (5,12).
3. Find the midpoint of (−6,2) and (4,10).
4. Find the equation of the line with gradient 3 through (2,9).
5. Find the equation through (−1,1) and (3,9).
6. State whether y=4x−2 and y=4x+7 are parallel.
7. Translate (3,−5) by vector (−4,7).
8. A maps to A′ by translation, where A=(2,6), A′=(−3,10). Find the vector.
9. Enlarge (4,−2) by scale factor 1/2 about the origin.
10. Centre C=(2,1), point P=(5,7), scale factor 2. Find P′.
11. Reflect (3,−4) in the x-axis as an extension.
12. Rotate (2,5) by 180° about the origin as an extension.
13. A=(0,0), B=(4,0), C=(4,3). Show angle ABC is 90° using gradients or vectors.
14. A=(1,2), B=(5,3), C=(7,7), D=(3,6). Show opposite sides are parallel.
15. Find the intersection of y=3x−4 and y=x+6.
Explained Answers
1. (15−3)/(8−2)=12/6=2.
2. Changes are 6 and 8, so distance=10.
3. Midpoint=(-1,6).
4. y=3x+c; 9=6+c, so y=3x+3.
5. Gradient=(9−1)/(3−(−1))=2. Using (−1,1), 1=−2+c, so y=2x+3.
6. Yes. Both gradients are 4 and the different intercepts show distinct parallel lines.
7. (−1,2).
8. A′−A=(−5,4).
9. (2,−1).
10. CP=(3,6); doubled=(6,12); P′=(8,13).
11. (3,4).
12. (−2,−5).
13. BA=(−4,0) is horizontal and BC=(0,3) is vertical, so they are perpendicular.
14. AB=(4,1), DC=C−D=(4,1); BC=(2,4), AD=D−A=(2,4). Opposite sides are equal and parallel.
15. 3x−4=x+6 gives x=5, y=11. Intersection=(5,11).
A Reliable Coordinate Workflow
Label the points. Decide whether the task needs difference, ratio, distance, midpoint or equation. Preserve point order in coordinate differences. For transformations, state centre, vector or scale factor before moving points. After calculating, return to the geometry: does the direction, length, parallelism or image location make sense?
Teacher and Parent Prompts
Ask “What geometric fact are these coordinates supposed to prove?” before starting calculations. When a line equation is wrong, ask whether the gradient or the intercept failed. For transformations, ask the learner to describe the movement in words before applying the coordinate rule.
For extension, give the transformed image and ask the student to recover the translation vector or centre-related displacement. Reverse problems test whether the representation is understood rather than copied.
Continue the Secondary 3 Learning Route
Continue with Vectors and Geometric Relationships for directed displacement, Trigonometry, Bearings and Navigation for angle-and-distance modelling, and Probability and Statistical Reasoning for data and uncertainty.
Coordinate geometry is successful when the algebra still describes the picture and the picture still verifies the algebra. Return to the Secondary Mathematics Hub for the full learning route.
