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Coordinate Proofs: Using Coordinates to Prove Geometry

In a coordinate proof we place a shape on the grid and use algebra to prove facts about it. The distance formula proves lengths, the midpoint formula proves bisecting, and slopes prove parallel (equal slopes) or perpendicular (slopes multiply to −1). For circles, a point (x, y) lies on the circle with centre (h, k) and radius r exactly when (x − h)² + (y − k)² = r²; completing the square turns a messy equation into this form.

🎬 Step-by-step story

  1. Two points A(−3, −2) and B(3, 6). Draw a right triangle between them: 6 across, 8 up. Pythagoras gives AB = √(6² + 8²) = 10. That is the distance formula.
  2. The midpoint M is the average of the x’s and the average of the y’s: M(0, 2). The slope is rise ÷ run = 8 ÷ 6 = 4/3.
  3. Proof time: is PQRS a rectangle? Opposite sides have equal slopes, so they are parallel. Next-door slopes multiply to −1, so the corners are right angles.
  4. A circle with centre O(0, 0) and radius 5 is x² + y² = 25. P(3, 4) is exactly 5 away, so it is on the circle. Q(4, −4) is about 5.66 away, so it is outside.
  5. Completing the square: x² + y² − 6x + 4y − 3 = 0 becomes (x − 3)² + (y + 2)² = 16. Watch the circle move to centre (3, −2) with radius 4.
  6. Free play: A and B are the ends of a diameter. Move P with the sliders. On the circle, angle APB is always 90° and the slopes of PA and PB multiply to −1.

Tip: drag the 3D scene to turn it. Use two fingers to zoom.

🤔 Common doubts, cleared

Why is the distance formula just Pythagoras?

Draw a horizontal line and a vertical line from the two points. They make a right triangle; the across and up distances are the legs and the segment is the hypotenuse.

Why do I average to get the midpoint?

The middle is halfway across and halfway up, and halfway between two numbers is their average.

Why do perpendicular slopes multiply to −1?

Turning a line by 90° swaps its rise and run and flips one sign: slope a/b becomes −b/a. Multiply them and you get −1.

How do I check a point is on a circle without drawing?

Put its x and y into the equation. If (x − h)² + (y − k)² equals r² exactly, it is on the circle; smaller means inside, bigger means outside.

Why do the signs flip when I read the centre?

(x − 3)² is zero when x = 3, so the centre's x is 3. (y + 2)² is zero when y = −2, so the centre's y is −2.

Does the 90° rule work for every point on the circle?

Yes, for every point except A and B themselves. Move P in free play: the slope product stays −1 on the circle and changes when P goes inside or outside.

The tools: distance, midpoint and slope

For points A(x₁, y₁) and B(x₂, y₂):

How to write a coordinate proof

  1. Place the figure on the grid. Put a corner at the origin and a side on an axis when you can; it keeps numbers simple.
  2. Decide what you must show: equal lengths (distance), parallel or perpendicular sides (slope), a bisected side (midpoint).
  3. Calculate each value clearly.
  4. Conclude with a sentence that links the numbers to the definition, e.g. "Both pairs of opposite sides have equal slopes, so PQRS is a parallelogram."

Useful tests: parallelogram = both pairs of opposite sides parallel (or diagonals bisect each other: same midpoint); rectangle = parallelogram with one right angle (or equal diagonals); rhombus = four equal sides; square = rectangle + rhombus; right triangle = two sides with slope product −1 (or Pythagoras).

General proofs use letters: put a rectangle at (0, 0), (a, 0), (a, b), (0, b). Both diagonals have length √(a² + b²), which proves that the diagonals of every rectangle are equal.

Circles on the coordinate plane

A circle is every point at distance r from a centre (h, k). Using the distance formula:

(x − h)² + (y − k)² = r²   (centre at the origin: x² + y² = r²)

Proving circle theorems with coordinates

Angle in a semicircle is 90°. Take the circle x² + y² = r² with diameter A(−r, 0), B(r, 0), and any other point P(x, y) on it. Slope PA = y/(x + r), slope PB = y/(x − r). Their product is y²/(x² − r²). Because x² + y² = r², we have y² = r² − x² = −(x² − r²), so the product is −1. So PA ⟂ PB and ∠APB = 90°.

The perpendicular from the centre bisects a chord. Take a chord between (x₁, y₁) and (x₂, y₂) on x² + y² = r². Show the slope from O to the chord's midpoint times the chord's slope is −1. (Try it in the practice.)

Try it: graph-paper proof

On graph paper, mark A(0, 0), B(6, 0), C(8, 4), D(2, 4). Predict: what kind of quadrilateral is it? Now prove it: find all four slopes and two side lengths. (Answer: AB and DC have slope 0, AD and BC have slope 2 → parallelogram; AB = 6 but AD = √20, so not a rhombus.) Then, in the 3D free play, set P's distance to 4 and to 6 and see that ∠APB becomes more or less than 90°.

Key formulas and definitions

Worked examples

1. Find the distance and midpoint of A(−3, −2) and B(3, 6).

AB = √((3 − (−3))² + (6 − (−2))²) = √(36 + 64) = √100 = 10. M = ((−3 + 3)/2, (−2 + 6)/2) = (0, 2).

2. Find the point that divides A(1, 2) to B(10, 14) in the ratio 1 : 2.

P = ((2 × 1 + 1 × 10)/3, (2 × 2 + 1 × 14)/3) = (12/3, 18/3) = (4, 6). Check: P is one third of the way from A.

3. Prove that P(−5, 1), Q(−3, −3), R(5, 1), S(3, 5) form a rectangle.

Slopes: PQ = (−3 − 1)/(−3 + 5) = −2; QR = (1 + 3)/(5 + 3) = ½; RS = (5 − 1)/(3 − 5) = −2; SP = (1 − 5)/(−5 − 3) = ½. Opposite sides have equal slopes → parallelogram. PQ × QR = −2 × ½ = −1 → right angle. A parallelogram with a right angle is a rectangle.

4. Does the point (−6, 8) lie on the circle centred at the origin through (10, 0)?

Radius = 10, so the circle is x² + y² = 100. For (−6, 8): 36 + 64 = 100. Yes, it lies on the circle.

5. Find the centre and radius of x² + y² − 6x + 4y − 3 = 0.

Group: (x² − 6x) + (y² + 4y) = 3. Add 9 and 4: (x − 3)² + (y + 2)² = 16. Centre (3, −2), radius 4.

6. Find the equation of the tangent to x² + y² = 25 at (3, 4).

Slope of radius = 4/3. The tangent is perpendicular, so slope = −3/4. y − 4 = −¾(x − 3) → 4y − 16 = −3x + 9 → 3x + 4y = 25.

7. Prove that the diagonals of any rectangle have equal length.

Place the rectangle at O(0, 0), A(a, 0), B(a, b), C(0, b). OB = √(a² + b²) and AC = √((0 − a)² + (b − 0)²) = √(a² + b²). They are equal for every a and b, so the diagonals of every rectangle are equal.

Common mistakes

Practice quiz

1. The slopes of two perpendicular lines are 3 and:
2. The midpoint of (2, −4) and (8, 6) is:
3. Centre and radius of (x + 1)² + (y − 5)² = 49:
4. To prove a quadrilateral is a parallelogram with slopes, you show:
5. The point (5, 12) on x² + y² = 169 is:

Practice: answer these yourself

Type or choose your answer, then press Check. Use a hint if you are stuck; the full solution appears after you answer.

Frequently asked questions

What is a coordinate proof?

It is a proof where you place a figure on the coordinate grid and use the distance, midpoint and slope formulas to show a geometric fact.

How do you prove a point lies on a circle?

Substitute the point into (x − h)² + (y − k)² = r². If both sides are equal, the point is on the circle.

How do you prove a quadrilateral is a rectangle with coordinates?

Show both pairs of opposite sides have equal slopes (parallelogram) and two next-door sides have slopes that multiply to −1 (right angle). Or show it is a parallelogram with equal diagonals.

Where this is taught

USA (Common Core, NGSS, AP)Grade 10Circles with and without coordinates

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