IGCSE Maths · Coordinate Geometry · Core & Extended

Coordinate Geometry: Lines, Gradients, Lengths and Midpoints

Coordinate geometry is where algebra and geometry meet — every line, length, and midpoint you’ll find here can be worked out with a formula, without ever needing to measure anything by hand. Gradient from a grid, drawing/finding line equations, and parallel lines are shared by both tiers. Calculating gradient from coordinates, length, midpoint, more general line equations, and perpendicular lines are Extended-only, tagged throughout.

1. Cartesian Coordinates

A point is described by an ordered pair (x, y): x tells you how far right (or left, if negative) of the origin, and y tells you how far up (or down).

Reading coordinates (3, 2) (−3, 7) The point (3, 2) is 3 units right and 2 units up from the origin. The point (−3, 7) is 3 units left and 7 units up.

2. Finding the Gradient of a Line

From a grid

Gradient = vertical change ÷ horizontal change (“rise over run”), read directly from the squares on a grid.

Reading gradient from a grid run = 4 rise = 6 Gradient = rise ÷ run = 6 ÷ 4 = 1.5

From two coordinates Extended only

m = (y₂ − y₁) / (x₂ − x₁)
Worked example Find the gradient of the line through (2, 3) and (6, 11).
m = (11 − 3) / (6 − 2) = 8/4 = 2
Try it — Gradient
  1. A line on a grid rises 8 squares for every 2 squares across. Find its gradient.
  2. Find the gradient of the line through (1, 4) and (5, 12). Ext.
  3. Find the gradient of the line through (−2, 5) and (3, −5). Ext.
Show answers
  1. 8 ÷ 2 = 4
  2. (12−4)/(5−1) = 8/4 = 2
  3. (−5−5)/(3−(−2)) = −10/5 = −2

3. Length of a Line Segment Extended only

This is Pythagoras’ theorem in disguise: the horizontal and vertical differences form the two shorter sides of a right-angled triangle, and the line segment itself is the hypotenuse.

d = √[(x₂ − x₁)² + (y₂ − y₁)²]
Worked example (1,2) (4,6) Find the length of the line segment joining (1, 2) and (4, 6).
d = √[(4−1)² + (6−2)²] = √[3² + 4²] = √[9 + 16] = √25 = 5
Try it — Length of a Line Segment Extended
  1. Find the length of the line segment joining (0, 0) and (6, 8).
  2. Find the length of the line segment joining (2, 1) and (7, 13).
  3. Find the length of the line segment joining (−3, 2) and (1, −1).
Show answers
  1. √(36+64) = √100 = 10
  2. √(25+144) = √169 = 13
  3. √(16+9) = √25 = 5

4. Midpoint of a Line Segment Extended only

The midpoint is exactly halfway between two points — average the x-coordinates, and separately average the y-coordinates.

Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2)
Worked example Find the midpoint of (2, 3) and (8, 11).
Midpoint = ((2+8)/2, (3+11)/2) = (10/2, 14/2) = (5, 7)
Try it — Midpoint Extended
  1. Find the midpoint of (0, 0) and (10, 6).
  2. Find the midpoint of (−4, 5) and (6, −3).
  3. The midpoint of a line segment is (3, 4), and one endpoint is (1, 2). Find the other endpoint.
Show answers
  1. (5, 3)
  2. (1, 1)
  3. If (1,2) and (x,y) average to (3,4): (1+x)/2=3 → x=5; (2+y)/2=4 → y=6. Other endpoint: (5, 6)

5. Drawing and Finding the Equation of a Straight Line

The standard form is y = mx + c, where m is the gradient and c is the y-intercept. Extended questions may also use ax + by = c or x = k (a vertical line) — rearrange into y = mx + c form first if needed.

Worked example — finding an equation from a gradient and a point A line has gradient 2 and passes through (1, 5). Find its equation.
y = 2x + c. Substitute (1, 5): 5 = 2(1) + c → c = 3
Answer: y = 2x + 3
Worked example — finding an equation from two points Find the equation of the line through (1, 4) and (3, 10).
Gradient: (10−4)/(3−1) = 6/2 = 3
Using (1,4): 4 = 3(1) + c → c = 1
Answer: y = 3x + 1
Worked example — rearranging ax + by = c Extended Find the gradient and y-intercept of 3x + 2y = 12.
Rearrange: 2y = −3x + 12 → y = −1.5x + 6
Gradient: −1.5, y-intercept: 6
Try it — Equations of Lines
  1. Find the equation of a line with gradient 4, passing through (2, 9).
  2. Find the equation of the line through (0, 1) and (2, 9).
  3. Find the gradient and y-intercept of 4x + 2y = 10. Ext.
Show answers
  1. 9=4(2)+c → c=1. y=4x+1
  2. Gradient=(9−1)/(2−0)=4. Using (0,1): c=1. y=4x+1
  3. 2y=−4x+10 → y=−2x+5. Gradient −2, y-intercept 5

6. Parallel Lines

Parallel lines have exactly the same gradient.

Worked example Find the equation of the line parallel to y = 4x − 1 that passes through (2, 5).
Parallel means the same gradient: m = 4
5 = 4(2) + c → c = −3
Answer: y = 4x − 3
Try it — Parallel Lines
  1. Find the equation of the line parallel to y = 3x + 2 that passes through (1, 8).
  2. Find the equation of the line parallel to y = −2x + 5 that passes through (0, −3).
Show answers
  1. m=3. 8=3(1)+c → c=5. y=3x+5
  2. m=−2. y=−2x−3 (already passes through (0,−3) directly, since c=−3)

7. Perpendicular Lines Extended only

Perpendicular lines meet at a right angle. Their gradients are negative reciprocals of each other — multiply the two gradients together and you always get −1.

m₁ × m₂ = −1
Worked example Find the equation of the line perpendicular to y = 2x + 3 that passes through (4, 1).
Gradient of the given line: 2. Perpendicular gradient: −1/2
1 = −1/2(4) + c → 1 = −2 + c → c = 3
Answer: y = −½x + 3
Worked example — the perpendicular bisector Find the equation of the perpendicular bisector of the line segment joining (−3, 8) and (9, −2).
Midpoint: ((−3+9)/2, (8−2)/2) = (3, 3)
Gradient of the segment: (−2−8)/(9−(−3)) = −10/12 = −5/6
Perpendicular gradient: 6/5
Using the midpoint (3,3): 3 = (6/5)(3) + c → 3 = 18/5 + c → c = 3 − 3.6 = −0.6
Answer: y = (6/5)x − 0.6
Try it — Perpendicular Lines Extended
  1. Find the gradient of a line perpendicular to y = 5x − 1.
  2. Find the gradient of a line perpendicular to y = −¾x + 2.
  3. Find the equation of the line perpendicular to y = 3x − 2 that passes through (6, 1).
Show answers
  1. −1/5
  2. 4/3
  3. Perpendicular gradient = −1/3. 1 = −1/3(6) + c → 1 = −2 + c → c = 3. y = −⅓x + 3
Common mistakes to watch for
  • Mixing up the order of subtraction in the gradient, length, or midpoint formulas — as long as you’re consistent (always point 2 minus point 1), the answer works out correctly.
  • Forgetting to square both differences before adding, in the length formula.
  • Finding the reciprocal of a gradient for a perpendicular line, but forgetting to also flip the sign.
  • Rearranging ax + by = c into y = mx + c incorrectly — divide every term by b, including the constant.
  • Assuming two lines with the same y-intercept must be parallel — parallel depends only on gradient, not on the intercept.

8. Quick-Fire Challenge Round

True or False?
  1. Parallel lines always have the same gradient.
  2. Perpendicular gradients always add up to −1.
  3. The midpoint formula averages the x-coordinates and the y-coordinates separately.
  4. A vertical line has a gradient of 0.
Show answers
  1. True.
  2. False — perpendicular gradients multiply to give −1, they don’t add to −1.
  3. True. Ext.
  4. False — a vertical line has an undefined gradient; a horizontal line has gradient 0.
Spot the Error
  1. “Gradient of the line through (2,5) and (6,13): (6−2)/(13−5) = 4/8 = 0.5.”
  2. “Length between (0,0) and (3,4): √(3²+4²) = √(9+4) = √13.”
  3. “A line perpendicular to y=4x+1 has gradient 4.”
Show answers
  1. The formula is (y₂−y₁)/(x₂−x₁), not the other way round: (13−5)/(6−2) = 8/4 = 2.
  2. 3² + 4² = 9 + 16 = 25, not 9 + 4 = 13 — both terms must be squared before adding: √25 = 5.
  3. Perpendicular gradients are negative reciprocals, not the same value: the perpendicular gradient is −1/4, not 4.

9. Mixed Practice — Bringing It All Together

Core & Extended
  1. Find the equation of a line with gradient 3, passing through (0, 4).
  2. Find the equation of the line through (1, 2) and (4, 11).
  3. Find the equation of the line parallel to y = −x + 6 that passes through (2, 1).
Show answers
  1. y = 3x + 4
  2. Gradient=(11−2)/(4−1)=3. Using (1,2): c=−1. y=3x−1
  3. m=−1. 1=−1(2)+c → c=3. y=−x+3
Extended-level mixed set Extended only
  1. Find the length and midpoint of the line segment joining (1, 1) and (7, 9).
  2. Find the gradient and y-intercept of 2x + 4y = 20.
  3. Find the equation of the line perpendicular to y = ½x − 3 that passes through (2, 6).
Show answers
  1. Length: √(36+64)=√100=10. Midpoint: (4,5)
  2. 4y=−2x+20 → y=−0.5x+5. Gradient −0.5, y-intercept 5
  3. Perpendicular gradient = −2. 6=−2(2)+c → 6=−4+c → c=10. y=−2x+10

Practice at Home

A large set again — take it steadily.

Part A — Gradient
  1. A line on a grid rises 9 squares for every 3 squares across. Find its gradient.
  2. Find the gradient of the line through (3, 2) and (7, 14). Ext.
  3. Find the gradient of the line through (−1, 6) and (4, −4). Ext.
Part B — Length and midpoint Extended
  1. Find the length of the line segment joining (0, 0) and (5, 12).
  2. Find the length of the line segment joining (2, 3) and (10, 9).
  3. Find the midpoint of (1, 5) and (9, 1).
  4. The midpoint of a segment is (4, −1), and one endpoint is (0, 3). Find the other endpoint.
Part C — Equations of lines
  1. Find the equation of a line with gradient 5, passing through (1, 3).
  2. Find the equation of the line through (2, 3) and (5, 15).
  3. Find the gradient and y-intercept of 5x + y = 8. Ext.
Part D — Parallel and perpendicular lines
  1. Find the equation of the line parallel to y = 6x − 4 that passes through (1, 5).
  2. Find the gradient of a line perpendicular to y = 2x + 7. Ext.
  3. Find the equation of the line perpendicular to y = −4x + 1 that passes through (8, 3). Ext.
Part E — Applied / challenge
  1. A triangle has vertices A(0,0), B(6,0), and C(3,4). Find the length of each side, and state whether the triangle is isosceles.
  2. Find the equation of the perpendicular bisector of the line segment joining (2, 6) and (8, −2).
  3. Three points A(1,2), B(3,6), and C(5,10) are claimed to lie on the same straight line. Use the gradient formula to check whether this is true.

Key Vocabulary Recap

Gradient — steepness of a line; m = (y₂−y₁)/(x₂−x₁)
y-intercept — where a line crosses the y-axis
Midpoint — the point exactly halfway between two points
Parallel lines — lines with equal gradients
Perpendicular lines — lines whose gradients multiply to give −1 Ext.

Syllabus Reference

This note covers Cambridge IGCSE Mathematics (0580) sections C3.1, C3.2, C3.3, C3.5 and C3.6 (Core subject content) and sections E3.1–E3.7 (Extended subject content) — Coordinate geometry in full. Gradient from a grid, drawing/finding equations of lines (y=mx+c), and parallel lines are shared by both tiers. Gradient from coordinates, length of a line segment, midpoint, more general line equation forms, and perpendicular lines are Extended-only, marked above — the corresponding Core sections C3.4 and C3.7 are explicitly “Extended content only” in the syllabus.

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