Coordinate Proofs

Q: Where should I place the figure?

Put one vertex at the origin and one side along the $x$-axis whenever possible. Zeros in the coordinates make every distance, slope, and midpoint computation shorter. Just make sure the placement matches the figure's definition without forcing extra properties.

A coordinate proof proves a geometric claim with algebra. Instead of two columns of statements and reasons, you place the figure on the coordinate plane, label its vertices, and let the distance, slope, and midpoint formulas do the arguing.

The key habit is using variables like $a$ and $b$ for the coordinates rather than specific numbers. A computation done with variables covers every rectangle, every isosceles triangle, every case at once — that is what makes it a proof instead of an example.

Place the figure wisely

You get to choose where the figure sits, so choose the placement that makes the algebra easiest. Put one vertex at the origin and run one side along the $x$ -axis. For a rectangle, the natural placement is $(0, 0)$ , $(a, 0)$ , $(a, b)$ , and $(0, b)$ .

The placement must not accidentally add extra conditions. Vertices like $(a, a)$ force a square when you meant any rectangle; completely random vertices like $(d, e)$ fail to guarantee a rectangle at all. Use just enough structure to match the figure's definition — no more, no less.

Variables prove every case

If you verify a property on the rectangle with vertices $(0, 0)$ , $(6, 0)$ , $(6, 4)$ , and $(0, 4)$ , you have proven it for that one rectangle only. A $7$ -by- $3$ rectangle is untouched by your computation.

Write the vertices as $(0, 0)$ , $(a, 0)$ , $(a, b)$ , and $(0, b)$ instead, and the same computation runs for every possible pair of dimensions simultaneously. Whatever the algebra shows is now true for all rectangles — that is the whole point of a coordinate proof.

Pick the right formula

Match the formula to the claim. To show two segments are congruent, compute both lengths with the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ and show they are equal. To show two sides are parallel or perpendicular, compare slopes with $\displaystyle \frac{y_2 - y_1}{x_2 - x_1}$ . To show a segment is bisected, compute midpoints with $\displaystyle \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ and show they coincide.

End the proof with a sentence that connects the algebra back to the claim: "both diagonals have length $\sqrt{a^2 + b^2}$ , so the diagonals of a rectangle are congruent."

Worked examples

Example 1: the diagonals of a rectangle are congruent

Prove that the diagonals of any rectangle are congruent.

Place a general rectangle with variable coordinates

(0, 0),\; (a, 0),\; (a, b),\; (0, b)

Length of the diagonal from

(0, 0)

(a, b)

\sqrt{(a - 0)^2 + (b - 0)^2} = \sqrt{a^2 + b^2}

Length of the diagonal from

(a, 0)

(0, b)

\sqrt{(0 - a)^2 + (b - 0)^2} = \sqrt{a^2 + b^2}

Both diagonals have the same length, for every

a

and

b

\text{the diagonals are congruent}

Answer: Both diagonals measure $\sqrt{a^2 + b^2}$ , so the diagonals of any rectangle are congruent.

Example 2: prove a triangle is isosceles

Triangle $PQR$ has vertices $P(-3, 0)$ , $Q(3, 0)$ , and $R(0, 5)$ . Prove that $\triangle PQR$ is isosceles.

Isosceles means two congruent sides, so compute side lengths

\text{distance formula}

Length of

\overline{PR}

\sqrt{(0 - (-3))^2 + (5 - 0)^2} = \sqrt{34}

Length of

\overline{QR}

\sqrt{(0 - 3)^2 + (5 - 0)^2} = \sqrt{34}

Two sides have equal length

PR = QR

Answer: $PR = QR = \sqrt{34}$ , so $\triangle PQR$ is isosceles.

Example 3: diagonals of a parallelogram bisect each other

A parallelogram has vertices $(0, 0)$ , $(a, 0)$ , $(a + c, b)$ , and $(c, b)$ . Prove that its diagonals bisect each other.

Midpoint of the diagonal from

(0, 0)

(a + c, b)

\displaystyle \left(\frac{a + c}{2}, \frac{b}{2}\right)

Midpoint of the diagonal from

(a, 0)

(c, b)

\displaystyle \left(\frac{a + c}{2}, \frac{b}{2}\right)

The midpoints are the same point, so each diagonal passes through the other's midpoint

\text{the diagonals bisect each other}

Answer: Both diagonals share the midpoint $\displaystyle \left(\frac{a + c}{2}, \frac{b}{2}\right)$ , so they bisect each other.

Try one yourself

Common questions

Why can't I just use numbers for the coordinates?

Numbers prove the claim for one specific figure only. A computation on a $6$ -by- $4$ rectangle says nothing about a $7$ -by- $3$ one. Variable coordinates like $(a, 0)$ and $(0, b)$ run the computation for every case at once, which is what a proof requires.

Where should I place the figure?

Put one vertex at the origin and one side along the $x$ -axis whenever possible. Zeros in the coordinates make every distance, slope, and midpoint computation shorter. Just make sure the placement matches the figure's definition without forcing extra properties.

How do I know which formula to use?

Read the claim. Congruent segments call for the distance formula, parallel or perpendicular sides call for the slope formula $\displaystyle \frac{y_2 - y_1}{x_2 - x_1}$ , and bisecting calls for the midpoint formula. Some proofs need two of them.

Is a coordinate proof a real proof?

Yes. As long as the placement is general — variables, not specific numbers — the algebra covers every possible case, which is exactly what a two-column proof accomplishes with statements and reasons. It is an equally valid style.

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Coordinate Proofs

Place the figure wisely

Variables prove every case

Pick the right formula

Worked examples

Example 1: the diagonals of a rectangle are congruent

Example 2: prove a triangle is isosceles

Example 3: diagonals of a parallelogram bisect each other

Try one yourself

Common questions

Keep learning

Classifying Triangles & Quadrilaterals by Coordinates

The Distance & Midpoint Formulas on the Coordinate Plane

Calculating Slope from Two Points

Isosceles & Equilateral Triangle Theorems