Geometry — Coordinate Geometry and Proof

Coordinate geometry turns shapes into numbers. Once a point has coordinates, you can calculate a length, locate a midpoint, compare gradients, and prove a geometric fact without relying on a drawing that may not be to scale. Keep the algebra organised: write the formula first, substitute carefully, then state what the result means.

Coordinates, Slope, and Midpoint

A point is written as (x, y). For points A(x1, y1) and B(x2, y2), slope (also called gradient) measures vertical change per horizontal change:

m = (y₂ − y₁) / (x₂ − x₁)

A positive slope rises from left to right; a negative slope falls. A horizontal line has slope 0. A vertical line has no defined slope because its horizontal change is 0.

The midpoint is the average of each coordinate:

M = ((x₁ + x₂) / 2, (y₁ + y₂) / 2)

Worked example: slope and midpoint

Find the slope and midpoint of the segment joining A(−2, 1) and B(4, 7).

Slope:
m = (7 − 1) / (4 − (−2))
m = 6 / 6
m = 1

Midpoint:
M = ((−2 + 4) / 2, (1 + 7) / 2)
M = (2 / 2, 8 / 2)
M = (1, 4)

The segment rises 1 unit for every 1 unit it moves right, and its centre is M(1, 4).

Distance Between Two Points

The distance formula is Pythagoras’ theorem applied to the horizontal and vertical changes:

AB = √((x₂ − x₁)² + (y₂ − y₁)²)

Worked example: exact distance

Find the distance from P(−1, 2) to Q(5, 10).

PQ = √((5 − (−1))² + (10 − 2)²)
PQ = √(6² + 8²)
PQ = √(36 + 64)
PQ = √100
PQ = 10

Keep the exact square root until the end. Here it simplifies exactly to 10 units.

Equations of Lines

The slope-intercept form is y = mx + b, where m is slope and b is the y-intercept. When you know a slope and one point, point-slope form is often quicker:

y − y₁ = m(x − x₁)

Worked example: line through a point

Find the equation of the line with slope −3 through (2, 5).

y − 5 = −3(x − 2)
y − 5 = −3x + 6
y = −3x + 11

So the equation is y = −3x + 11. Check it: when x = 2, y = −6 + 11 = 5, so the point is on the line.

Parallel and Perpendicular Lines

Parallel non-vertical lines have equal slopes. Perpendicular non-vertical lines have slopes that are negative reciprocals, so their product is −1:

parallel:      m₁ = m₂
perpendicular: m₁m₂ = −1

A horizontal line is perpendicular to a vertical line. Treat that special case separately because a vertical line does not have a slope.

Worked example: a perpendicular line

Find the equation of the line perpendicular to y = (1/2)x − 2 that passes through (6, 1).

Given slope = 1/2
Perpendicular slope = −2

y − 1 = −2(x − 6)
y − 1 = −2x + 12
y = −2x + 13

The required equation is y = −2x + 13. The slope check is (1/2) × (−2) = −1.

A line and its perpendicular through a point A coordinate grid shows the line y equals one half x minus 2 and the perpendicular line y equals negative 2x plus 13. The lines meet at the marked point 6, 1. A small square marks their right angle. −8−42610 841−2 xy y = ½x − 2 y = −2x + 13 (6, 1)
The slopes 1/2 and −2 are negative reciprocals, so the two lines are perpendicular. The right-angle marker shows the geometric conclusion; the coordinates let you verify it algebraically.

Circles on the Coordinate Plane

A circle with centre (h, k) and radius r has equation:

(x − h)² + (y − k)² = r²

The signs inside the brackets are opposite the centre coordinates. For example, (x + 2)² means h = −2.

Worked example: identify a circle

State the centre and radius of (x − 3)² + (y + 1)² = 25, then check whether (6, 3) lies on it.

Centre = (3, −1)
r² = 25
r = 5

Check (6, 3):
(6 − 3)² + (3 + 1)² = 25
3² + 4² = 25
9 + 16 = 25
25 = 25

The point lies on the circle because it satisfies the equation.

Congruence and Similarity

Congruent figures have the same shape and the same size: corresponding sides and angles are equal. Similar figures have the same shape but may have different sizes: corresponding angles are equal and corresponding side lengths have one common scale factor.

  • Useful congruence tests for triangles: SSS, SAS, ASA, AAS, and RHS (right angle, hypotenuse, side).
  • Useful similarity tests: AA, SAS with proportional matching sides, and SSS with proportional matching sides.

Worked example: test similarity

Triangle A has side lengths 3, 4, 5. Triangle B has side lengths 6, 8, 10. Are they similar or congruent?

6 / 3 = 2
8 / 4 = 2
10 / 5 = 2

All corresponding side ratios are 2.
Therefore the triangles are similar by SSS.
They are not congruent because corresponding side lengths are not equal.

A Concise Coordinate Proof

Prove that the quadrilateral with vertices A(1, 1), B(5, 3), C(3, 7), and D(−1, 5) is a square. A reliable plan is to show adjacent sides are perpendicular and all four sides have equal length.

1. Find slopes of adjacent sides:
mAB = (3 − 1) / (5 − 1) = 2 / 4 = 1/2
mBC = (7 − 3) / (3 − 5) = 4 / (−2) = −2

(1/2)(−2) = −1, so AB is perpendicular to BC.

2. Find squared side lengths (avoids unnecessary square roots):
AB² = (5 − 1)² + (3 − 1)² = 4² + 2² = 20
BC² = (3 − 5)² + (7 − 3)² = (−2)² + 4² = 20
CD² = (−1 − 3)² + (5 − 7)² = (−4)² + (−2)² = 20
DA² = (1 − (−1))² + (1 − 5)² = 2² + (−4)² = 20

3. Conclude:
All four sides have equal length, and AB is perpendicular to BC.
Therefore ABCD is a square.

Using squared lengths is valid here: positive lengths are equal exactly when their squares are equal.

Common Mistakes

  • Mixing coordinate order: always subtract x-values from x-values and y-values from y-values.
  • Losing negative signs: write brackets, such as 4 − (−2), before simplifying.
  • Using a reciprocal but not changing the sign: the perpendicular slope to 2/3 is −3/2, not 3/2.
  • Reading the circle centre directly: in (x + 4)², the x-coordinate of the centre is −4.
  • Proving only one property: equal adjacent sides alone do not prove a square; include a right angle or another sufficient condition.

Practice Set

  1. Find the slope and midpoint of R(2, −3) and S(8, 9).
  2. Find the distance between (−2, 4) and (4, −4).
  3. Write the equation of the line through (−1, 6) with slope 4.
  4. Write the equation of the line through (3, 2) perpendicular to y = −(1/3)x + 5.
  5. State the centre and radius of (x + 5)² + (y − 2)² = 16.
  6. Triangles have side lengths 5, 7, 9 and 10, 14, 18. Are they similar? State why.

Answers

  1. m = (9 − (−3)) / (8 − 2) = 12/6 = 2; midpoint ((2 + 8)/2, (−3 + 9)/2) = (5, 3).
  2. √((4 − (−2))² + (−4 − 4)²) = √(6² + (−8)²) = √100 = 10.
  3. y − 6 = 4(x + 1), so y = 4x + 10.
  4. The perpendicular slope is 3. y − 2 = 3(x − 3), so y = 3x − 7.
  5. Centre (−5, 2); radius √16 = 4.
  6. Yes. 10/5 = 14/7 = 18/9 = 2, so they are similar by SSS.