Coordinate Geometry & Graphs

What you'll learn

Coordinate geometry combines algebra and geometry to solve problems involving points, lines and curves on the Cartesian plane. This topic appears in both Paper 2 and Paper 4 of CIE IGCSE Mathematics, testing your ability to find gradients, equations of straight lines, distances between points, midpoints, and transformations of graphs. Questions range from 2-mark gradient calculations to multi-step problems worth 6-8 marks involving perpendicular lines or applying multiple techniques.

Key terms and definitions

Gradient (m) — the measure of steepness of a line, calculated as change in y divided by change in x, or (y₂ - y₁)/(x₂ - x₁)

y-intercept (c) — the point where a line crosses the y-axis; the value of y when x = 0

Parallel lines — lines with equal gradients that never intersect

Perpendicular lines — lines that intersect at 90°; if one line has gradient m, the perpendicular line has gradient -1/m

Midpoint — the point exactly halfway between two coordinates, found using ((x₁ + x₂)/2, (y₁ + y₂)/2)

Distance formula — the length between two points: √[(x₂ - x₁)² + (y₂ - y₁)²], derived from Pythagoras' theorem

Linear equation — an equation representing a straight line, commonly written as y = mx + c

Asymptote — a line that a curve approaches but never touches

Core concepts

Finding the gradient between two points

The gradient measures how steep a line is. For any two points (x₁, y₁) and (x₂, y₂):

Gradient formula: m = (y₂ - y₁)/(x₂ - x₁)

Key examination points:

• Always subtract coordinates in the same order (second minus first for both x and y)
• Positive gradient: line slopes upward from left to right
• Negative gradient: line slopes downward from left to right
• Zero gradient: horizontal line (y = constant)
• Undefined gradient: vertical line (x = constant)

CIE papers frequently test whether you can identify parallel lines (same gradient) or calculate missing coordinates when the gradient is given.

Equations of straight lines

The most common form tested is y = mx + c, where m is the gradient and c is the y-intercept.

Finding the equation when given gradient and a point:

1. Substitute the gradient m into y = mx + c
2. Substitute the coordinates of the given point for x and y
3. Solve for c
4. Write the complete equation

Finding the equation when given two points:

1. Calculate the gradient using m = (y₂ - y₁)/(x₂ - x₁)
2. Use either point and follow the steps above

Alternative forms:

• ax + by + c = 0 (general form) — questions may ask you to rearrange into this format
• y - y₁ = m(x - x₁) (point-gradient form) — useful for quick working

Examiners test conversions between forms and whether you can identify the gradient and intercept from different formats.

Parallel and perpendicular lines

Parallel lines:

• Have identical gradients
• If line L₁ has equation y = 3x + 2, any line parallel to it has form y = 3x + k

Perpendicular lines:

• Their gradients multiply to give -1
• If one line has gradient m, the perpendicular has gradient -1/m
• Example: gradient 2 → perpendicular gradient is -1/2
• Example: gradient -3/4 → perpendicular gradient is 4/3

CIE IGCSE questions often combine these concepts, asking you to find the equation of a line perpendicular to a given line passing through a specific point. This is a 5-6 mark question requiring multiple steps.

Midpoint and distance between two points

Midpoint formula:

For points (x₁, y₁) and (x₂, y₂), the midpoint M is:

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Add the x-coordinates and divide by 2, then do the same for y-coordinates.

Distance formula:

The distance d between two points is:

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

This formula comes from applying Pythagoras' theorem to the right-angled triangle formed by the horizontal and vertical distances between points.

Exam questions may:

• Give the midpoint and one endpoint, asking for the other endpoint
• Ask for the perimeter or area of shapes drawn on coordinate grids
• Combine distance with other concepts like circles or proving geometric properties

Graphing linear inequalities

Linear inequalities create regions on a coordinate plane:

Method:

1. Replace the inequality sign with = and draw the boundary line
2. Use a solid line for ≤ or ≥ (points on the line are included)
3. Use a dashed line for < or > (points on the line are excluded)
4. Shade the required region:
   • For y > mx + c, shade above the line
   • For y < mx + c, shade below the line
   • For x > a, shade to the right of the vertical line
   • For x < a, shade to the left

When multiple inequalities are given, the solution region is where all shaded areas overlap. Label this region clearly as examiners specifically look for this in marking.

Transformations of graphs

CIE IGCSE tests transformations of standard functions. Understanding the effect of changes to function notation is essential:

f(x) + a — vertical translation upward by a units

f(x) - a — vertical translation downward by a units

f(x + a) — horizontal translation left by a units

f(x - a) — horizontal translation right by a units

-f(x) — reflection in the x-axis

f(-x) — reflection in the y-axis

af(x) (where a > 1) — vertical stretch, scale factor a

f(ax) (where a > 1) — horizontal compression, scale factor 1/a

Common functions you must recognise:

• y = x² — parabola (U-shape)
• y = x³ — cubic curve
• y = 1/x — reciprocal graph (two branches with asymptotes at x = 0 and y = 0)
• y = aˣ — exponential growth/decay
• y = sin x, y = cos x, y = tan x — trigonometric graphs

Paper 4 questions may show a transformed graph and ask you to write its equation, or give an equation and ask you to sketch the transformation.

Interpreting real-life graphs

Coordinate geometry applies to distance-time graphs, speed-time graphs, and other contextual situations:

Distance-time graphs:

• Gradient = speed
• Horizontal line = stationary
• Steeper gradient = higher speed

Speed-time graphs:

• Gradient = acceleration
• Area under graph = distance travelled
• Horizontal line = constant speed

Examiners test your ability to extract information, calculate gradients in context, and interpret what different sections of graphs represent.

Worked examples

Example 1: Points A(-2, 3) and B(4, 7) are vertices of a triangle. (a) Find the gradient of AB. [2] (b) Find the equation of the line AB in the form y = mx + c. [3] (c) A line perpendicular to AB passes through point C(3, 5). Find its equation. [3]

Solution:

(a) m = (y₂ - y₁)/(x₂ - x₁) = (7 - 3)/(4 - (-2)) = 4/6 = 2/3 [2 marks]

(b) Using y = mx + c with m = 2/3:

Substitute point A(-2, 3): 3 = (2/3)(-2) + c 3 = -4/3 + c c = 3 + 4/3 = 13/3

Equation: y = (2/3)x + 13/3 or y = (2x + 13)/3 [3 marks: 1 for method, 1 for substitution, 1 for correct equation]

(c) Perpendicular gradient = -1 ÷ (2/3) = -3/2

Using point C(3, 5): 5 = (-3/2)(3) + c 5 = -9/2 + c c = 5 + 9/2 = 19/2

Equation: y = -3x/2 + 19/2 or 2y = -3x + 19 [3 marks: 1 for perpendicular gradient, 2 for correct equation]

Example 2: The line L passes through points P(1, -2) and Q(5, 6). (a) Find the midpoint of PQ. [2] (b) Calculate the length of PQ. [2]

Solution:

(a) Midpoint = ((1 + 5)/2, (-2 + 6)/2) = (6/2, 4/2) = (3, 2) [2 marks]

(b) Distance = √[(5 - 1)² + (6 - (-2))²] = √[4² + 8²] = √[16 + 64] = √80 = 4√5 or 8.94 units [2 marks: 1 for correct substitution, 1 for answer]

Example 3: Sketch the region satisfied by all three inequalities: y ≥ 1, x + y ≤ 5, y < 2x. [4]

Solution:

1. y ≥ 1: horizontal line at y = 1 (solid), shade above
2. x + y ≤ 5 or y ≤ -x + 5: line through (0, 5) and (5, 0) (solid), shade below
3. y < 2x: line through origin with gradient 2 (dashed), shade below

The required region is where all three conditions overlap (bounded triangular region). [4 marks: 1 for each line correctly drawn, 1 for correct region identified]

Common mistakes and how to avoid them

• Mixing up coordinates when calculating gradient — always write out (x₁, y₁) and (x₂, y₂) clearly and subtract in the same order. The formula is (y₂ - y₁)/(x₂ - x₁), not the other way round.

• Finding the perpendicular gradient incorrectly — remember it's the negative reciprocal: flip the fraction AND change the sign. If the gradient is 2, the perpendicular gradient is -1/2, not 1/2 or -2.

• Forgetting to find c after calculating gradient — finding m = 3 is not the equation. You must substitute a point to find the y-intercept and write the complete equation y = 3x + c.

• Using solid lines for strict inequalities — y > 2x should be a dashed line because points on the line itself are not included. Use solid lines only for ≥ or ≤.

• Confusing horizontal and vertical translations in transformations — f(x + 3) moves the graph LEFT by 3 (opposite to what you might expect), while f(x) + 3 moves it UP by 3.

• Misreading coordinates from graphs — always check the scale carefully. Each square might represent 2 units, not 1. Read coordinates precisely, especially when finding the y-intercept.

Exam technique for Coordinate Geometry & Graphs

• Show your method clearly — gradient questions award method marks even if arithmetic errors occur. Write the formula, substitute values, then calculate. Don't just write an answer.

• Simplify final answers appropriately — equations can be left as y = (2/3)x + 5 or rearranged to 3y = 2x + 15 depending on the question requirement. Check command words like "Give your answer in the form ax + by + c = 0."

• Sketch vs. plot distinctions matter — "sketch" requires correct shape and key features (intercepts, asymptotes) but not precise plotting. "Plot" or "draw accurately" requires graph paper and exact coordinates.

• Use a ruler for straight-line graphs — examiners can penalise freehand lines. Even in sketches of transformations, straight lines should be ruled.

Quick revision summary

Gradient formula: m = (y₂ - y₁)/(x₂ - x₁). Equation of straight line: y = mx + c. Parallel lines have equal gradients; perpendicular gradients multiply to -1. Midpoint: average the coordinates. Distance: use Pythagoras √[(x₂ - x₁)² + (y₂ - y₁)²]. For inequalities, use dashed lines for < or >, solid for ≤ or ≥. Transformations: f(x) + a moves up, f(x + a) moves left, -f(x) reflects in x-axis. Always show working for method marks.