Coordinate geometry turns a diagram into numbers without losing the geometry. A line becomes an equation. Steepness becomes a gradient. A segment becomes a distance calculation. An intersection becomes a pair of values satisfying two relationships at once.
This Secondary 3 Mathematics Learning Guide develops gradients, lengths of line segments, straight-line equations in the form y=mx+c, intercepts, intersections, point testing and geometric problems using coordinates. It complements Coordinate Geometry and Transformations by giving the K310 straight-line system its own complete route.
Official scope: the current 2027 SEC G3 Mathematics syllabus listing identifies Mathematics as K310. K310 coordinate geometry includes finding the gradient of a straight line from two coordinates, finding the length of a line segment from its endpoints, interpreting and finding y=mx+c, and geometric problems involving coordinates.
Coordinate route: identify the points → calculate horizontal and vertical changes → choose gradient, distance or line equation → preserve signs → solve intersections if two conditions are present → substitute back into the original equations → interpret the result geometrically.
Coordinates Are Ordered Pairs
A point (x,y) records horizontal position first and vertical position second. Reversing the order changes the point.
For A(−3,5), the x-coordinate is −3 and the y-coordinate is 5. The signs matter because they determine which quadrant or axis region contains the point.
Gradient Measures Vertical Change per Horizontal Change
For points (x₁,y₁) and (x₂,y₂) on a non-vertical line, gradient m=(y₂−y₁)/(x₂−x₁).
Subtract coordinates in the same order on top and bottom. Reversing both orders gives the same gradient; reversing only one changes the sign incorrectly.
Worked Example 1: Find a Positive Gradient
A line passes through A(−3,5) and B(5,21). Find its gradient.
m=(21−5)/(5−(−3))=16/8=2.
The line rises 2 vertical units for every 1 horizontal unit to the right.
Worked Example 2: Negative Gradient
A line passes through P(2,11) and Q(8,−1). Find its gradient.
m=(−1−11)/(8−2)=−12/6=−2.
The negative sign means the line falls as x increases. It does not mean a distance is negative.
Vertical Lines Need Special Treatment
If two points have the same x-coordinate, then x₂−x₁=0. The gradient formula would require division by zero, so the gradient is undefined.
A vertical line through x=4 is written x=4, not in the form y=mx+c with a finite gradient.
Distance Comes From Pythagoras
The horizontal change and vertical change between two points form the legs of a right triangle. Therefore the segment length is √[(x₂−x₁)²+(y₂−y₁)²].
The squares remove sign from the length calculation. A segment does not become shorter because one coordinate difference is negative.
Worked Example 3: Exact and Approximate Segment Length
Find the distance between A(−2,3) and B(4,15).
Horizontal change=6. Vertical change=12.
AB=√(6²+12²)=√180=6√5≈13.4.
The exact form 6√5 preserves structure; the decimal form is useful when a numerical measurement is requested.
The Equation y=mx+c Separates Gradient and Intercept
In y=mx+c, m is the gradient and c is the y-intercept: the y-value when x=0.
A line can be found from a known gradient plus one point by substituting the point to solve for c.
Worked Example 4: Equation From Gradient and Point
Find the equation of the line with gradient 2 passing through (−2,3).
Write y=2x+c. Substitute the point: 3=2(−2)+c, so c=7.
The equation is y=2x+7.
Check by substituting x=−2: y=−4+7=3, which returns the given point.
Worked Example 5: Equation From Two Points
Find the equation of the line through (4,−1) and (10,11).
Gradient=(11−(−1))/(10−4)=12/6=2.
Use y=2x+c and point (4,−1): −1=8+c, so c=−9.
Therefore y=2x−9.
Check the second point: when x=10, y=20−9=11.
Intercepts Are Coordinate Events
At the y-axis, x=0. At the x-axis, y=0. These substitutions turn graphical intercepts into algebraic calculations.
Worked Example 6: Find Both Intercepts
For y=−3x+12, find the x- and y-intercepts.
At x=0, y=12, so y-intercept is (0,12).
At y=0, 0=−3x+12, so x=4. The x-intercept is (4,0).
These two points are enough to draw the straight line accurately on suitable axes.
Intersections Satisfy Two Equations at Once
If two lines intersect, the intersection point lies on both. Its coordinates therefore satisfy both line equations simultaneously.
This connects coordinate geometry directly to simultaneous equations.
Worked Example 7: Intersection of Two Lines
Find the intersection of y=2x+1 and y=−x+7.
At the intersection, 2x+1=−x+7. Hence 3x=6 and x=2.
Then y=2(2)+1=5. The intersection is (2,5).
Check in the second equation: −2+7=5.
A Point Can Be Tested Against a Line
To decide whether a point lies on a line, substitute its coordinates into the equation. If the equality is true, the point lies on the line.
Worked Example 8: Point Membership
Does P(3,7) lie on y=2x+1? Does Q(4,10)?
For P, 2(3)+1=7, so P lies on the line.
For Q, 2(4)+1=9≠10, so Q does not lie on the line.
This substitution check is often faster than drawing an accurate graph.
Coordinate Geometry Can Prove Length Relationships
A geometric problem can be translated into several segment lengths, gradients or line equations. The coordinates provide the data; the geometry determines what needs to be compared.
Worked Example 9: A Right Triangle From Coordinates
A=(0,0), B=(6,0), C=(6,8). Show that triangle ABC is right-angled and find its area.
AB=6 and BC=8 directly from the horizontal and vertical changes. AC=√(6²+8²)=10.
Since 6²+8²=10², the triangle is right-angled at B.
Area=1/2×6×8=24 square units.
The coordinate system has made two sides horizontal and vertical, but the Pythagorean check also verifies the side-length structure.
Worked Example 10: Recover a Missing Coordinate From Gradient
The line through A(2,5) and B(8,p) has gradient 3/2. Find p.
(p−5)/(8−2)=3/2.
(p−5)/6=3/2, so p−5=9 and p=14.
Check: vertical change=9 and horizontal change=6, giving gradient 9/6=3/2.
Worked Example 11: Recover an Unknown Intercept
A line y=−4x+c passes through (3,−5). Find c.
−5=−4(3)+c=−12+c, so c=7.
The line is y=−4x+7. Its y-intercept is therefore (0,7).
Worked Example 12: Geometry Through Two Line Conditions
A path is modelled by y=0.5x+2 and a boundary by y=−x+8. Find where they meet.
0.5x+2=−x+8. Hence 1.5x=6 and x=4.
Then y=0.5(4)+2=4. They meet at (4,4).
In a context, this coordinate should be interpreted using the units and axis meanings given by the model. The algebra alone does not tell us whether x is metres, minutes or another quantity.
Gradient Can Represent a Rate
When axes carry real quantities, gradient inherits a rate unit. On a distance-time graph, gradient may have units km/h. On a cost-versus-quantity graph, gradient may have units dollars per item.
The coordinate formula is unchanged, but interpretation depends on the axis labels. This connects to Distance-Time and Speed-Time Graphs.
A Reliable Coordinate-Geometry Checklist
- Write the points clearly as ordered pairs.
- For gradient, subtract y-values and x-values in the same order.
- For length, square both coordinate changes before adding.
- For y=mx+c, determine m first unless it is already given.
- Use a known point to solve for c.
- At the x-axis set y=0; at the y-axis set x=0.
- For an intersection, solve the two conditions simultaneously.
- Verify a point by substitution into the original equation.
- Return the final coordinate to the diagram or context.
Common Errors and Their First Repair
Mixed subtraction order in gradient: if y₂−y₁ is used on top, use x₂−x₁ below.
Forgetting brackets around a negative coordinate: 5−(−3)=8, not 2.
Using gradient formula for distance: gradient is a ratio; distance uses Pythagoras.
Assuming c from an arbitrary point: c is the y-value only when x=0.
Stopping after finding x at an intersection: substitute back to find y unless the question requests only x.
Accepting a line equation without checking a point: substitute one original coordinate pair.
Independent Practice
1. Find the gradient through (−1,4) and (5,16).
2. Find the gradient through (3,10) and (9,−2).
3. Find the length between (1,2) and (7,10), giving exact and 3 s.f. forms.
4. Find the equation of a line with gradient 3 through (2,−1).
5. Find the equation through (−2,7) and (4,−5).
6. Find both intercepts of y=4x−20.
7. Find the intersection of y=3x−2 and y=−2x+13.
8. Test whether (5,11) lies on y=2x+1.
9. Points A(0,0), B(9,0), C(9,12) form a triangle. Find all side lengths and area.
10. The line through (1,4) and (7,p) has gradient 5/3. Find p.
11. A line y=−2x+c passes through (6,−7). Find c.
12. Two models y=0.25x+5 and y=−0.5x+14 intersect. Find the intersection.
Explained Answers
1. (16−4)/(5−(−1))=12/6=2.
2. (−2−10)/(9−3)=−12/6=−2.
3. √(6²+8²)=10; exact and decimal forms coincide.
4. y=3x+c; −1=6+c, so y=3x−7.
5. Gradient=(−5−7)/(4−(−2))=−12/6=−2. Using (−2,7): 7=4+c, so y=−2x+3.
6. y-intercept (0,−20); x-intercept from 0=4x−20 is (5,0).
7. 3x−2=−2x+13 gives x=3 and y=7, so (3,7).
8. 2(5)+1=11, so yes.
9. AB=9, BC=12, AC=15. Area=1/2×9×12=54 square units.
10. (p−4)/6=5/3, so p−4=10 and p=14.
11. −7=−12+c, so c=5.
12. 0.25x+5=−0.5x+14 gives 0.75x=9, x=12 and y=8, so (12,8).
How to Know the Coordinate Skill Has Transferred
Change the job rather than only the numbers. Ask the learner to move from two points to an equation, from an equation to intercepts, from two equations to an intersection, or from coordinates to a geometric length. The student should select the relationship rather than wait for the chapter heading to announce it.
A strong solution also checks itself: a line equation should reproduce the original points; an intersection should satisfy both equations; a distance should be nonnegative and geometrically plausible.
Continue the Secondary 3 Learning Route
Continue with Direct and Inverse Proportion, Constants and Model Choice, Pythagoras, 3D Trigonometry, Elevation and Depression, and Composite Solids, Surface Area, Volume and Unit Conversion.
A coordinate-geometry answer is secure when the coordinate changes, equation, point checks and geometric interpretation all describe the same line or segment. Return to the Secondary Mathematics Hub.
