IP EMaths Notes (Upper Sec, Year 3-4): 07) Coordinate Geometry

Study guide08 Nov 2025, 00:00 ZUpdated 17 Jul 2026

Gradients, midpoints, and distance formulas for quick analytic geometry in the coordinate plane.

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For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.

How this chapter applies

Eclat core: gradient, distance, midpoint, straight-line equations, and coordinate problem solving form the main route.
School-sensitive extension: perpendicular bisectors, loci, closest-point arguments, and vector links may be assessed at different depths.
2027 national comparison: K310 Topic G6 covers gradient, segment length, equations in the form y = mx + c, and geometric problems using coordinates.
Check your school: confirm whether midpoint, perpendicular-line, or locus proofs require formal justification.
Exam-track route: use the separate O-Level and SEC G3 Mathematics notes for K310 topic ownership.

Q: What does IP EMaths Notes (Upper Sec, Year 3-4): 07) Coordinate Geometry cover?
A: Gradients, midpoints, and distance formulas for quick analytic geometry in the coordinate plane.

The core idea is simple: Coordinates turn geometry into formulas for gradient, midpoint, distance, and line equations.

Use it as a working check: Find the feature you need first: gradient for direction, midpoint for halfway points, distance for length, and negative reciprocal gradients for perpendicular lines.

Then go one layer deeper: Use the perpendicular bisector example as a template: calculate midpoint, calculate gradient, switch to the perpendicular gradient, then write the line through the midpoint.

Coordinate geometry connects algebra with diagrams. These formulas appear in locus questions, perpendicular bisectors, and vector tie-ins.

Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.

New to the Integrated Programme? Start with What is IP? | Browse all free IP notes.

Core formulas

Gradient between $(x_1, y_1)$

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Marcus Pang·Managing Director (Maths)

Question cue	First setup	Calculation move	Common trap
"Find the midpoint of $AB$ "	Average the x-coordinates and y-coordinates separately.	$M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$ .	Averaging one coordinate but copying the other.
"Find the length of $AB$ "	Find horizontal change and vertical change.	Use Pythagoras: $d=\sqrt{(\Delta x)^2+(\Delta y)^2}$
"Given midpoint and one endpoint, find the other endpoint"	Use the midpoint as the average of the two endpoints.	Set $\dfrac{x_1+x_2}{2}=x_M$

Line type	How to recognise it	Equation form	Perpendicular line
Horizontal	The two points have the same y-coordinate.	$y = k$	Vertical line $x = a$ .
Vertical	The two points have the same x-coordinate.	$x = a$	Horizontal line $y = k$ .
Neither	Both x and y coordinates change.	Use $y = mx + c$ or point-slope form.	Use the negative reciprocal gradient.

Question cue	First move	Check before final answer
"Find the gradient" or "parallel line"	Use the gradient formula, then reuse the same gradient for a parallel line.	A vertical line has no finite gradient, so do not force it into $y = mx + c$ .
"Find the midpoint"	Average the two x-coordinates and the two y-coordinates separately.	The midpoint should lie halfway between both coordinates, not just halfway in distance.
"Find the length"	Use the distance formula or Pythagoras on the horizontal and vertical changes.	Distance is always non-negative; keep exact surd form unless rounding is requested.
"Find a perpendicular bisector"	Find the midpoint, find the original gradient, then use the negative reciprocal gradient.	The final line must pass through the midpoint, not through either endpoint unless the diagram says so.
"Closest point on a line"	Draw the perpendicular from the point to the line, then intersect that perpendicular with the given line.	The closest route meets the line at a right angle; a random point on the line is not enough.

Step	What to do	Common trap
1	Find the gradient of the given line.	Using the coefficient of $x$ before rearranging into $y = mx + c$ .
2	Use the perpendicular gradient for the line through the outside point.	Forgetting that horizontal and vertical lines need separate handling.
3	Solve the original line and the perpendicular line simultaneously.	Picking an easy point on the original line instead of the foot of the perpendicular.
4	If the question asks for distance, use the distance formula from the outside point to the foot.	Measuring to an endpoint or intercept instead of the closest point.

IP EMaths Notes (Upper Sec, Year 3-4): 07) Coordinate Geometry

How this chapter applies

Core formulas

Sources

Midpoint, distance, or endpoint checkpoint

Horizontal and vertical line checkpoint

Line-task checkpoint

Worked example - Equation of a perpendicular bisector

Closest-point checkpoint

Practice Quiz

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