We have seen many theorems proven using two column, paragraph, flowchart, and algebraic proofs. Now we would like to prove some theorems using diagrams on the coordinate plane.

For these proofs, we can follow the steps below:

Create a diagram on the coordinate plane representing the given information.
Label all the key points, choosing coordinates will lead to easy algebraic work.
Use the coordinates of key points to determine other properties of the diagram.
Use the properties of the diagram to prove the proof goal of the theorem.

Trapezoid O P Q R plotted on a first quadrant coordinate plane without numbers. Vertex O is at (0, 0), vertex P at (2 b, 2 c), vertex Q at (2 d, 2 c), and vertex R at (2 a, 0). A point M is on side O P, and point N on side Q R. A segment is drawn from M to N.

For example:

The points P and Q have the same y-coordinate, so \overline{PQ} is parallel to the x-axis and \overline{OR}.

The vertices of trapezoid OPQR are all even, so finding the coordinates of midpoints will not involve fractions.

We can find the coordinates for M and N, then show that \overline{MN} is parallel to \overline{OR}.

Some tips for creating a diagram that leads to easier algebraic work are:

Use the origin, \left(0,0\right), for one of the key points.
If a diagram has midpoints, use even coordinates like \left(2a, 2b\right) to avoid fractions.
When sides are parallel, have one of the sides on the x-axis.
When two sides are perpendicular, have one on each axis.
If there is an important line of symmetry, make it overlap with one of the axes.
When possible, make the coordinates of one point be in terms of the variables of another point.

Since a proof needs to work for all examples, we must use variables in our coordinates rather than a specific numeric example.

Worked examples

Example 1

Draw and label the vertices of a square in the coordinate plane that would be helpful for proofs.

Solution

A square has congruent sides, two pairs of parallel sides, and its adjacent sides are perpendicular.

Some things we can do to make the diagram nicer for a coordinate proof are:

Have one side on the x-axis and one on the y-axis to ensure those sides meet at a right angle.
Have the other two sides be a vertical and horizontal line to get two pairs of parallel sides.
Let a represent the side length (or 2a if we expect to find midpoints).
Have one vertex at the origin.

With one vertex at the origin and a side length of a, we can determine that the vertices will be:

\left( 0,0 \right),\left( a,0 \right),\left( 0,a \right),\left( a,a \right)

Choosing the sides of the square to be horizontal and vertical line segments makes it very easy to choose vertices with a certain distance between them.

Example 2

Use coordinate geometry to prove that the midpoint of the hypotenuse of a right triangle is equidisant from the three vertices.

Approach

In the proof, we want to find the midpoint of a right triangle. Some of the tips that will help us choose our vertices are:

Use the origin for one of the key points.
When two sides are perpendicular, have one on each axis.
If a diagram has midpoints, use even coordinates.

Solution

Triangle O B A plotted on a first quadrant coordinate plane without numbers. Vertex O is at (0, 0), vertex B at (0, 2 b), and vertex A at (2 a, 0). A point M is the midpoint of side A B, and a segment is drawn from M to O.

Using our tips, we can choose the two perpendicular sides of the right triangle to be on the axes, with the vertex at the origin.

Since we want to find a midpoint, we also choose to have even coordinates, making sure to use variables so that the proof will work for all examples.

Given: A\left(2a,0\right), B\left(0,2b\right), O\left(0,0\right) and M is the midpoint of \overline{AB}.

Proof goal: BM=AM=OM

Use the midpoint formula to find the coordinates of point M

\displaystyle M	\displaystyle =	\displaystyle \left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)
	\displaystyle =	\displaystyle \left(\dfrac{0+2a}{2},\dfrac{2b+0}{2}\right)
	\displaystyle =	\displaystyle \left(\dfrac{2a}{2},\dfrac{2b}{2}\right)
	\displaystyle =	\displaystyle \left(a,b\right)

Use the distance formula to find the lengths BM, AM, and OM

\displaystyle BM	\displaystyle =	\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}
	\displaystyle =	\displaystyle \sqrt{\left(a-0\right)^2+\left(2b-b\right)^2}
	\displaystyle =	\displaystyle \sqrt{a^2+b^2}
\displaystyle AM	\displaystyle =	\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}
	\displaystyle =	\displaystyle \sqrt{\left(2a-a\right)^2+\left(0-b\right)^2}
	\displaystyle =	\displaystyle \sqrt{a^2+b^2}
\displaystyle OM	\displaystyle =	\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}
	\displaystyle =	\displaystyle \sqrt{\left(a-0\right)^2+\left(b-0\right)^2}
	\displaystyle =	\displaystyle \sqrt{a^2+b^2}

BM=AM=OM=\sqrt{a^2+b^2}

Therefore, BM=AM=OM, so the midpoint of the hypotenuse of a right triangle is equidisant from the three vertices.

Outcomes

MA.912.GR.3.2

Given a mathematical context, use coordinate geometry to classify or justify definitions, properties and theorems involving circles, triangles or quadrilaterals.

10.07 Proving theorems with coordinate proofs

Concept summary

Worked examples

Example 1

Solution

Example 2

Approach

Solution

Outcomes

MA.912.GR.3.2

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