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10. Connecting Algebra and Geometry through Coordinates
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Geometry

10.05 Extension: Proving theorems in the coordinate plane

Lesson
Practice

Introduction

We have seen many theorems proven using two column, paragraph, flowchart, and algebraic proofs. Now, we will prove some theorems using diagrams on the coordinate plane. There are certain strategies that we can use to make the calculations in our proofs simpler, and we will explore those strategies in this lesson.

Proving theorems in the coordinate plane

To prove a theorem using the coordinate plane, we can follow the steps below:

  1. Represent the given information with a diagram on the coordinate plane.

  2. Label the coordinates of all the key points.

  3. Use the coordinates of the key points to determine other properties of the diagram.

  4. Use the properties of the diagram to prove the theorem.

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

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}.

Exploration

Position and label a figure on the coordinate plane that has the same properties as the given diagram.

  1. How did you decide where to put your first line?
  2. Explain how you know the two lines are parallel in the coordinate plane.
  3. Explain how you know one line is perpendicular to the other two in the coordinate plane.
  4. Can you simplify your diagram at all to make the coordinates easier to work with?

Using the axes to make constructions is helpful as it means that we do not require as many variables. For example, we can choose to label the x- and y-intercepts in this diagram to help us find the slopes of the lines.

Examples

Example 1

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

Worked Solution
Create a strategy

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.

Apply the idea

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.

Reflect and check

Using vertical and horizontal lines means we cannot prove lines perpendicular by showing the slopes are opposite reciprocals. We also cannot prove vertical lines are parallel by showing the slopes are the same. Both slopes are undefined, but that does not make them equal.

A different strategy would be to rotate the square to make a diamond shape.

Example 2

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

Worked Solution
Create a strategy

In the proof, we want to find the midpoint of the hypotenuse 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.

Apply the idea
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

  1. 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)
  2. 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.

Example 3

The triangle midsegment theorem says "the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length."

a

Draw and label the vertices of a triangle in the coordinate plane that would be helpful for proving this theorem.

Worked Solution
Create a strategy

The theorem refers to all triangles, so we need to make sure the triangle we draw does not look like one particular type since different types of triangles have different properties.

The triangle in this image has a generic shape and can be adjusted to look like other triangle types if needed.

We will need to prove that the segment in the middle of this image (the segment joining midpoints of two sides of a triangle) is parallel to the third side and half the length of it.

When placing this triangle on the coordinate plane, we can use even side lengths to avoid working with fractions when calculating the midpoints of the sides in the proof. We can also place the third side of the triangle along the x-axis to make slope calculations simpler.

Apply the idea

We will begin by placing the third side of the triangle along the x-axis and naming the key points.

We have the origin as the endpoint, O, for one side, and we can use P\left(2b,2c\right) as the second endpoint to avoid fractional values. The midpoint of \overline{OP} is:

\displaystyle M\displaystyle =\displaystyle \left(\dfrac{0+2b}{2},\dfrac{0+2c}{2}\right)
\displaystyle =\displaystyle \left(b,c\right)

For the next side, P\left(2b,2c\right) will be one endpoint, and we will let Q\left(2a,0\right) be the other. We are choosing \left(2a, 0\right) because we have to use a different variable than b or c since we cannot assume any side lengths are the same. The cooridnates of that midpoint are:

\displaystyle N\displaystyle =\displaystyle \left(\dfrac{2a+2b}{2},\dfrac{0+2c}{2}\right)
\displaystyle =\displaystyle \left(a+b,c\right)

Putting this onto the coordinate plane, we get the diagram shown below:

Reflect and check

Keep in mind that we strategically placed this on the coordinate plane to make the calculations easier, but the theorem can be applied to any triangle. If we had put \overline{OQ} on the y-axis instead, it would not have affected the proof in part (b). We could have also placed point Q to the left of point P, and this would not affect the proof either.

b

Prove the triangle midsegment theorem using a coordinate plane.

Worked Solution
Create a strategy

Using the setup from part (a), we need to prove \overline{MN}\parallel \overline{OQ} and MN=\frac{1}{2}OQ. First, we will find the slopes of both line segments, then we will find the lengths.

Apply the idea

First, we will find the slope of \overline{MN}:

\displaystyle m_{MN}\displaystyle =\displaystyle \dfrac{y_2-y_1}{x_2-x_1}State the slope formula
\displaystyle =\displaystyle \dfrac{c-c}{a+b-b}Substitute the coordinates
\displaystyle =\displaystyle \dfrac{0}{a}Evaluate the subtraction
\displaystyle m_{MN}\displaystyle =\displaystyle 0Evaluate the division

Next, we will find the slope of \overline{OQ}:

\displaystyle m_{OQ}\displaystyle =\displaystyle \dfrac{y_2-y_1}{x_2-x_1}State the slope formula
\displaystyle =\displaystyle \dfrac{0-0}{2a-0}Substitute the coordinates
\displaystyle =\displaystyle \dfrac{0}{2a}Evaluate the subtraction
\displaystyle m_{OQ}\displaystyle =\displaystyle 0Evaluate the division

Therefore, m_{MN}=m_{OQ} which means \overline{MN} \parallel \overline{OQ}.

Now, we need to find the lengths of each segment. Since both of them are horizontal lines, we can use absolute values to determine the length.

\displaystyle MN\displaystyle =\displaystyle |a+b-b|
\displaystyle =\displaystyle a
\displaystyle OQ\displaystyle =\displaystyle |2a-0|
\displaystyle =\displaystyle 2a

From this, we can see that OQ=2(MN) or that \frac{1}{2}OQ=MN.

We have proved that the segment joining the midpoints of two sides of a triangle is parallel to the third side and half the length of the third side.

Reflect and check

There is more than one way to prove this theorem; we previously proved it using congruent triangles and parallelograms.

Idea summary

When proving theorems on the coordinate plane, following these tips for creating a diagram will lead to easier algebraic work:

  • Use the origin 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.

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