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Form 4 · Chapter 7

Coordinate Geometry, SPM Additional Mathematics Form 4

Coordinate Geometry is Chapter 7 of Form 4 Add Math, where algebra meets the xxyy plane. You learn to find the point that divides a line segment in a given ratio, decide when two lines are parallel or perpendicular and write their equations, calculate the area of a triangle or polygon straight from its vertices, and find the equation of a locus, the path traced by a moving point that obeys a fixed rule.

What this chapter is

Coordinate geometry is the branch of mathematics that describes points, lines and shapes using their coordinates on the Cartesian plane. In SPM Additional Mathematics this is Chapter 7 of Form 4, and it sits in the Geometry learning area.

The whole chapter rests on one powerful idea: once every point has an address (x,y)(x, y), geometric questions, how far apart, how steep, how large, what path, become algebra you can compute exactly.

You have already met the basics of coordinates in earlier years. Add Math takes them further in four directions: dividing a line segment in any ratio, not just the midpoint; using gradients to test whether lines are parallel or perpendicular and to write their equations; finding the area of any triangle or polygon directly from its corner coordinates; and, new to most students, deriving the equation of a locus, the set of all points that satisfy a stated condition.

Add Math is an elective subject, and Coordinate Geometry is one of the chapters where careful, visual working pays off quickly. A rough sketch of the points, drawn to no particular scale, turns almost every question into something you can see.

Because the answers are exact numbers and equations rather than long proofs, the chapter is a reliable source of marks for students who label their points clearly and substitute one value at a time.

By the end of the chapter you should be able to find a dividing point of a segment in a given ratio; state and use the gradient conditions for parallel and perpendicular lines; produce the equation of a straight line from a point and a gradient, or from two points; work out the area of a triangle and a quadrilateral from coordinates; and set up and simplify the equation of a locus, recognising when that locus is a straight line and when it is a circle. These tools reappear in later work on circles, vectors and calculus, so time spent here is repaid many times over.

Content standards

Coordinate Geometry is organised into four content standards. These codes and titles come directly from the KSSM DSKP, knowing them tells you exactly what a question is testing.

CodeStandardWhat you learn
7.1Divisor of a Line SegmentRelate the position of a point that divides a segment to the ratio; derive and use the divisor (division-point) formula on the Cartesian plane; and solve problems involving a dividing point.
7.2Parallel Lines and Perpendicular LinesMake and verify conjectures about the gradients of parallel and of perpendicular lines, generalise them, and solve problems on equations of parallel and perpendicular lines.
7.3Areas of PolygonsDerive the formula for the area of a triangle from its vertices, use it for triangles and quadrilaterals, generalise it to any polygon, and solve area problems.
7.4Equations of LociRepresent a locus graphically, a point at constant distance from a fixed point, or at a constant ratio of distances from two fixed points, find its equation, and solve locus problems.

The four standards build on one another. The division-point and gradient work in 7.1 and 7.2 gives you the equation of a line; areas in 7.3 use the same coordinates a different way; and loci in 7.4 combine distance, gradient and equation-forming into a single skill.

Read a question first to see which standard it is testing, and the right formula usually follows at once.

Key ideas

The distance between two points. The straight-line distance between A(x1,y1)A(x_{1}, y_{1}) and B(x2,y2)B(x_{2}, y_{2}) comes straight from Pythagoras applied to the horizontal and vertical gaps.

It underpins length, loci and the check for special triangles, so it is worth being completely fluent with it.

Distance between two pointsMust memorise
AB=(x2x1)2+(y2y1)2AB=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}
Not on the SPM formulae page, memorise it. The squares make the order of subtraction irrelevant.

The midpoint of a segment. The midpoint is simply the average of the two xx-coordinates and the average of the two yy-coordinates.

It is the special case of the division-point formula when the ratio is 1:11:1.

Midpoint of ABMust memorise
M=(x1+x22, y1+y22)M=\left(\dfrac{x_{1}+x_{2}}{2},\ \dfrac{y_{1}+y_{2}}{2}\right)
Not given in the exam, the ratio 1 : 1 case of the divisor formula.

Dividing a segment in a ratio (Standard 7.1). If the point PP divides ABAB so that AP:PB=m:nAP:PB=m:n, its coordinates are a weighted average of AA and BB.

The weight mm goes with the far point BB and nn with the near point AA, mixing them up is the most common slip here.

Divisor of a line segment, AP : PB = m : nMust memorise
P=(nx1+mx2m+n, ny1+my2m+n)P=\left(\dfrac{n x_{1}+m x_{2}}{m+n},\ \dfrac{n y_{1}+m y_{2}}{m+n}\right)
Not on the formulae page. Set m = n = 1 and it reduces to the midpoint.

Gradient. The gradient measures steepness as the change in yy over the change in xx.

A positive gradient rises to the right, a negative one falls, a horizontal line has gradient 00, and a vertical line has an undefined gradient because x2x1=0x_{2}-x_{1}=0.

Gradient of a line through two pointsMust memorise
m=y2y1x2x1m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}
Not given, keep the y-differences on top and the x-differences underneath, in the same order.

Parallel and perpendicular lines (Standard 7.2). Two non-vertical lines are parallel exactly when their gradients are equal, m1=m2m_{1}=m_{2}.

They are perpendicular exactly when the product of their gradients is 1-1, that is m1m2=1m_{1}m_{2}=-1; each gradient is the negative reciprocal of the other. These two conditions solve almost every 'parallel to' or 'perpendicular to' question.

Gradient conditionsMust memorise
Parallel: m1=m2Perpendicular: m1m2=1\text{Parallel: } m_{1}=m_{2}\qquad\text{Perpendicular: } m_{1}m_{2}=-1
Must be memorised. The perpendicular rule fails only for a horizontal–vertical pair, where one gradient is undefined.

The equation of a straight line. Given a point (x1,y1)(x_{1}, y_{1}) and a gradient mm, use the point–gradient form yy1=m(xx1)y-y_{1}=m(x-x_{1}); rearrange to y=mx+cy=mx+c to read off the gradient and yy-intercept.

If a line cuts the axes at (a,0)(a,0) and (0,b)(0,b), the intercept form xa+yb=1\frac{x}{a}+\frac{y}{b}=1 is often the quickest route.

Area of a polygon from coordinates (Standard 7.3). For a triangle with vertices (x1,y1)(x_{1},y_{1}), (x2,y2)(x_{2},y_{2}), (x3,y3)(x_{3},y_{3}), the area is half the modulus of a simple alternating sum.

Write the coordinates in order around the shape, apply the 'down-products minus up-products' pattern, and take half the absolute value. The same shoelace idea extends to quadrilaterals and larger polygons.

Area of a triangle from coordinatesMust memorise
Area=12x1(y2y3)+x2(y3y1)+x3(y1y2)\text{Area}=\tfrac{1}{2}\,\big|\,x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\,\big|
This coordinate (shoelace) formula is NOT the one on the exam page, that page gives only Area = ½ ab sin C, which needs a side and an angle.

Equations of loci (Standard 7.4). A locus is the set of all points P(x,y)P(x,y) obeying a rule.

Translate the rule into an equation using the distance formula, then simplify. When the distance from a fixed point is constant, the locus is a circle; when the ratio of distances from two fixed points is a fixed constant, it is generally a circle too (a straight line only when that ratio is exactly 1:11:1).

Squaring both sides early clears the roots and leaves a tidy equation.

How it is examined

Coordinate Geometry can appear in either written paper. Paper 1 (3472/1) lasts 2 hours and carries 80 marks: Section A has 12 questions worth 64 marks that you answer all of, and Section B has 3 questions worth 16 marks from which you answer 2.

Paper 2 (3472/2) lasts 2 hours 30 minutes and carries 100 marks across Section A (7 questions, 50 marks, answer all), Section B (4 questions, 30 marks, answer 3) and Section C (4 questions, 20 marks, answer 2).

Across both papers the items are limited-response subjective and structured questions, they are marked using analytic scoring, and you sit them with a non-programmable scientific calculator. Questions are set across a range of difficulty, the format uses a low-to-high difficulty ratio of 5 : 3 : 2, so expect both short substitution items and longer, multi-part problems that chain several skills together.

We do not predict how many marks any single chapter will carry, because that varies from year to year.

This chapter suits the show-your-working format especially well. A longer item often walks through a figure step by step: find a gradient, hence the equation of one line, hence the equation of a perpendicular line, hence a point of intersection, hence an area.

Because the scoring is analytic, each correct gradient, each equation and each substituted value can earn its own mark, so a small arithmetic slip early on need not cost the later marks, provided your method is clearly written and your later steps follow correctly from your own earlier answers.

Exam tip

None of this chapter's core formulae, distance, midpoint, division point, gradient, the parallel and perpendicular conditions, or the coordinate area formula, is printed on the SPM formulae page, so memorise them. Draw a quick sketch first, label every point with its coordinates, and write the formula in symbols before you substitute.

For areas, take the modulus so your answer is never negative; for loci, square both sides early to clear the square roots.

Common mistakes

Most marks lost in this chapter come from a handful of avoidable habits. Read these before every practice set until they become second nature.

  • Pairing the ratio weights with the wrong points. In AP:PB=m:nAP:PB=m:n, the weight mm multiplies the coordinates of the far point BB and nn multiplies the near point AA. Swapping them places PP on the wrong side of the segment.
  • Getting the perpendicular gradient wrong. Perpendicular means the negative reciprocal: if m1=23m_{1}=\tfrac{2}{3} then m2=32m_{2}=-\tfrac{3}{2}, because m1m2=1m_{1}m_{2}=-1. Flipping the sign without inverting the fraction (or the reverse) is a frequent error.
  • Forgetting the modulus in the area formula. The alternating sum can come out negative depending on the order in which you list the vertices; area is always positive, so take the absolute value of the whole expression before halving.
  • Listing polygon vertices out of order. The shoelace method only works if the corners are taken consistently around the shape (all clockwise or all anticlockwise). Jumping across the shape gives a wrong, often smaller, area.
  • Mixing up the coordinate order in the gradient. Keep the yy-differences on top and the xx-differences underneath, subtracted in the same order. Reversing one but not the other flips the sign of the gradient.
  • Leaving a locus equation un-simplified. After squaring both sides, expand fully, gather like terms, and simplify to standard form. An answer left as two square roots equal to each other rarely earns the final mark.

None of these mistakes is about ability, each one is a habit, and habits are fixable. Sketch the points first, write each formula before substituting, and check signs as you go; do that and your accuracy in this chapter climbs quickly.

How to study this chapter

Coordinate Geometry rewards a visual, ordered approach. Work through these steps in turn, then use the resources below to revise and test yourself.

Because none of the formulae are given in the exam, the first job is to hold them securely in memory, but memorise them alongside a sketch, not as isolated symbols. A gradient is a slope you can see; a locus is a path you can trace.

Short, frequent sessions that always begin with a quick diagram will fix both the formulae and the reasoning far better than one long cram.

  1. 1

    Memorise the six core formulae

    Distance, midpoint, division point, gradient, the parallel and perpendicular conditions, and the coordinate area formula, none is printed in the exam, so learn them all.

  2. 2

    Practise the division point

    Drill splitting a segment in ratios such as 2 : 3 and 3 : 1, and check each answer by confirming it lies the right fraction of the way from A to B.

  3. 3

    Build line equations both ways

    From a point and a gradient, and from two points, form the equation and rearrange it to y = mx + c; then find lines parallel and perpendicular to it.

  4. 4

    Master areas and loci

    Compute triangle and quadrilateral areas with the shoelace pattern (remembering the modulus), then set up locus equations and simplify them to standard form.

  5. 5

    Time yourself and review

    Attempt mixed, multi-part questions under a clock, then re-read your common-mistake list before the next session.

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Frequently asked questions

Are any coordinate geometry formulae given in the exam?

No, the distance, midpoint, division-point, gradient, parallel and perpendicular conditions, and the coordinate area formula are all not on the SPM formulae page, so you must memorise them. The page does list Area =12absinC=\tfrac{1}{2}ab\sin C, but that is the trigonometric form for a side and an included angle, not the coordinate (vertex) formula.

How do I find the gradient of a line perpendicular to another?

Take the negative reciprocal of the first gradient. If a line has gradient m1m_{1}, a line perpendicular to it has gradient m2=1m1m_{2}=-\dfrac{1}{m_{1}}, because m1m2=1m_{1}m_{2}=-1.

For example, a gradient of 23\tfrac{2}{3} becomes 32-\tfrac{3}{2}.

When is a locus a circle rather than a straight line?

When a point moves so its distance from a single fixed point stays constant, the locus is a circle. When the ratio of its distances from two fixed points is constant, it is generally a circle too, it is a straight line only in the special case where that ratio is exactly 1:11:1, giving the perpendicular bisector.

How do I use the area formula for a quadrilateral or larger polygon?

Use the same shoelace pattern: list the vertices in order around the shape, form the alternating sum of the coordinates, take the modulus, and halve it. Listing the corners consistently (all clockwise or all anticlockwise) is essential, or the answer will be wrong.

Source:SRC-DSKP-ENSRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

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