Distance Formula and Midpoint Formula: Algebraic Proofs in Geometry

Drawing geometric figures on a coordinate plane allows to prove simple geometric theorems algebraically. In this lesson, several theorems will be proven algebraically.

Catch-Up and Review

Here are a few recommended readings before getting started.

Try your knowledge on these topics.

a State the coordinates of the points plotted on the plane.

b Match each concept with its corresponding diagram.

c True or false?
A parallelogram has exactly two pairs of parallel sides.

In a parallelogram, adjacent angles are congruent.

The four sides of a parallelogram are congruent.

In a parallelogram, opposite sides and opposite angles are congruent.

The diagonals of a parallelogram bisect each other.

Adjacent angles in a parallelogram are supplementary.

d Use the Pythagorean Theorem to find the missing side length.

Challenge

Investigating the Properties of a Parallelogram

The figure below looks like a parallelogram.

Without using measuring tools such as a ruler or a protractor, how can it be proven that the quadrilateral above is actually a parallelogram?

Discussion

Distance Formula

In certain situations, it is required to find the length of a line segment or a side of a polygon. To find those lengths, it is recommended to calculate the distance between the endpoints of the segment, or between two vertices of a polygon. If those points are plotted on a coordinate plane, the Distance Formula can be used.

In this lesson, the Distance Formula will be used to prove properties of geometric figures.

Pop Quiz

Practice Finding the Distance Between Two Coordinate Pairs

Use the Distance Formula to calculate the distance between the points plotted in the coordinate plane. If needed, round the answer to two decimal places.

Discussion

Definition of a Midpoint

Sometimes the length of a segment is not what is needed. Instead, the coordinates of the point that lies exactly in the middle of a segment are needed.

If the points are plotted on a coordinate plane, the Midpoint Formula can be used to find the coordinates of the midpoint.

Rule

Midpoint Formula

The midpoint $M$ between two points $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ on a coordinate plane can be determined by the following formula.

$M (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$

The formula above is called the Midpoint Formula.

Proof

For simplicity, the points

A (x_{1}, y_{1})

and

B (x_{2}, y_{2})

will be arbitrarily plotted in Quadrant I. Also, consider the line segment that connects these points. The midpoint

M

between

A

and

B

is the midpoint of this segment. Note that the position of the points in the plane does not affect the proof.

Consider the horizontal distance

Δ x

and the vertical distance

Δ y

between

A

and

B

. Since

M

is the midpoint,

M

splits each distance,

Δ x

and

Δ y

, in half. Therefore, the horizontal and vertical distances from each endpoint to the midpoint are

\frac{Δ x}{2}

and

\frac{Δ y}{2} .

Let

x_{m}

and

y_{m}

be the coordinates of

M .

Now, focus on the

x -

coordinates. The difference between the corresponding

x -

coordinates gives the horizontal distances between the midpoint and the endpoints.

x_{m} - x_{1} and x_{2} - x_{m}

The graph above shows that these distances are both equal to

\frac{Δ x}{2} .

Therefore, by the Transitive Property of Equality, they are equal.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x_{m} - x_{1} = \frac{Δ x}{2} \frac{1}{2} x_{2} - x_{m} = \frac{Δ x}{2} \frac{1}{2} ⇓ x_{m} - x_{1} = x_{2} - x_{m}

This equation can be solved to find

x_{m},

the

x -

coordinate of the midpoint

M .

x_{m} - x_{1} = x_{2} - x_{m}

Solve for

x_{m}

AddEqn

$LHS + x_{m} = RHS + x_{m}$

2 x_{m} - x_{1} = x_{2}

AddEqn

$LHS + x_{1} = RHS + x_{1}$

2 x_{m} = x_{2} + x_{1}

CommutativePropAdd

\CommutativePropAdd

2 x_{m} = x_{1} + x_{2}

DivEqn

$LHS / 2 = RHS / 2$

x_{m} = \frac{x _{1} + x _{2}}{2}

The

x -

coordinate of

M

x_{m} = \frac{x _{1} + x _{2}}{2} .

In the same way, it can be shown that the

y -

coordinate of

M

y_{m} = \frac{y _{1} + y _{2}}{2} .

With this information, the coordinates of

M

can be expressed in terms of the coordinates of

A

and

B .

$M (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$

The Midpoint Formula can also be used to prove some properties of geometric figures.

Pop Quiz

Practice Finding Midpoints Using the Midpoint Formula

Use the Midpoint Formula to calculate the coordinates of the midpoint $M$ between the points plotted on the coordinate plane.

Example

Proving the Properties of a Parallelogram

Paulina has joined the bandwagon of making picture frames. She wants to prove that the diagonals of the parallelogram shown below bisect each other.

Help Paulina become an awesome frame maker by proving the statement.

Answer

See solution.

Hint

Use the Midpoint Formula to show that the diagonals intersect at their midpoint.

Solution

The diagonals bisect each other if and only if they intersect at their midpoint. Start by drawing the diagonals

\overline{B D}

and

\overline{A C} .

Then, identify the coordinates of their point of intersection.

It is seen above that the point of intersection of

\overline{B D}

and

\overline{A C}

(0.5, - 0.5) .

If this point is the midpoint of each diagonal, then the diagonals bisect each other. To prove that

(0.5, - 0.5)

is the midpoint of

\overline{B D},

the coordinates of the endpoints

B (- 2, 2)

and

D (3, - 3)

can be substituted into the Midpoint Formula.

M (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

SubstitutePoints

Substitute $(- 2, 2)$ & $(3, - 3)$

M (\frac{- 2 + 3}{2}, \frac{2 + ( - 3 )}{2})

Evaluate

AddTerms

Add terms

M (\frac{1}{2}, \frac{2 + ( - 3 )}{2})

AddNeg

$a + (- b) = a - b$

M (\frac{1}{2}, \frac{- 1}{2})

MoveNegNumToFrac

Put minus sign in front of fraction

M (\frac{1}{2}, - \frac{1}{2})

FracToDiv

$\frac{a}{b} = a \div b$

M (0.5, - 0.5) ✓

The midpoint of the diagonal

\overline{B D}

is the point of intersection of the diagonals. By following the same procedure, it can be shown that the midpoint of

\overline{A C}

is also

(0.5, - 0, 5) .

$M (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
Diagonal	Endpoints	Substitute	Simplify
$\overline{B D}$	$B (- 2, 2)$ $&$ $D (3, - 3)$	$M (\frac{- 2 + 3}{2}, \frac{2 + ( - 3 )}{2})$	$M (0.5, - 0.5)$ $✓$
$\overline{A C}$	$A (- 3, 0)$ $&$ $D (4, - 1)$	$M (\frac{- 3 + 4}{2}, \frac{0 + ( - 1 )}{2})$	$M (0.5, - 0.5)$ $✓$

The midpoint of both diagonals is the same as their point of intersection. Therefore, the diagonals bisect each other.

Example

Proving the Triangle Midsegment Theorem by Rigid Motions

Paulina's best friend is obsessed with triangles. He has requested for Paulina to make him a triangular picture frame. Suppose Paulina can prove the Triangle Midsegment Theorem for the triangle below. She will better understand how to work with triangles and therefore make a better frame!

Help Paulina improve her frame making skills!

Answer

See solution.

Hint

Translate the triangle so that it has one vertex at the origin and a consecutive vertex on the positive $x -$ axis. Then, consider the midpoints of $\overline{A B}$ and $\overline{B C} .$

Solution

Start by recalling the Triangle Midsegment Theorem.

Triangle Midsegment Theorem
The segment that connects the midpoints of two sides of a triangle — a midsegment — is parallel to the third side of the triangle and half its length.

To make the proof easier,

△ A B C

will be translated so that it has one vertex at the origin and a consecutive vertex on the positive

x -

axis. For simplicity, vertex

A

will be located at the origin and

C

on the positive

x -

axis. This is done by translating the triangle

1

unit to the left and

2

units down.

The vertices after the translation are

A (0, 0),

B (2, 2.5),

and

C (6, 0) .

Now the Triangle Midsegment Theorem can be proven for this triangle. Let

M_{1}

and

M_{2}

be the midpoints of

\overline{A B}

and

\overline{B C},

respectively.

The coordinates of the midpoints of

\overline{A B}

and

\overline{B C}

can be found by using the Midpoint Formula.

$M (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
Side	Endpoints	Substitute	Simplify
$\overline{A B}$	$A (0, 0)$ and $B (2, 2.5)$	$M_{1} (\frac{0 + 2}{2}, \frac{0 + 2 . 5}{2})$	$M_{1} (1, 1.25)$
$\overline{B C}$	$B (2, 2.5)$ and $C (6, 0)$	$M_{2} (\frac{2 + 6}{2}, \frac{2 . 5 + 0}{2})$	$M_{2} (4, 1.25)$

It can be seen above that both

M_{1}

and

M_{2}

have the same

y -

coordinate

1.25 .

This means that

\overline{M_{1} M_{2}}

is a horizontal segment. Since

\overline{A C}

is on the

x -

axis, it can be said that it is also a horizontal segment. Therefore, the midsegment is parallel to

\overline{A C} .

\overline{M_{1} M_{2}} ∥ \overline{A C}

Finally, it needs to be proven that

M_{1} M_{2}

is half

A C .

To do this, the Distance Formula will be used.

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Segment	Endpoints	Substitute	Simplify
$\overline{M_{1} M_{2}}$	$M_{1} (1, 1.25)$ and $M_{2} (4, 1.25)$	$M_{1} M_{2} = (4 - 1)^{2} + (1.25 - 1.25)^{2}$	$M_{1} M_{2} = 3$
$\overline{B C}$	$A (0, 0)$ and $C (6, 0)$	$A C = (6 - 0)^{2} + (0 - 0)^{2}$	$A C = 6$

Since

3

is half

6,

the length of

\overline{M_{1} M_{2}}

is half the length of

\overline{A C} .

M_{1} M_{2} = \frac{1}{2} A C

By following the same procedure, it can be proven that the other two midsegments of

△ A B C

are parallel and half the length of the third side. The Triangle Midsegment Theorem has been proven. Paulina has just leveled up and her friend will get a great frame!

Closure

Proving the Properties of a Parallelogram

The topics covered in this lesson can now be applied to the challenge. The figure below looks like a parallelogram; without using measuring tools, how can it be proven that the quadrilateral is a parallelogram?

Answer

See solution.

Hint

Place the quadrilateral on a coordinate plane. Then, translate it so that it has a vertex at the origin and a consecutive vertex on the $x -$ axis.

Solution

As advised, the quadrilateral will be placed on a coordinate plane. Then, it will be translated so that a vertex ends up at the origin and a consecutive vertex on the

x -

axis. For simplicity, the vertices have been labeled.

It is sufficient to prove that

A B C D

has one pair of congruent parallel sides to prove that

A B C D

is a parallelogram. Note that the

y -

coordinate of

B

and

C

2 .

Therefore,

\overline{B C}

is parallel to the

x -

axis. Also,

A

and

D

lie on the

x -

axis, so

\overline{B C}

and

\overline{A D}

are parallel.

To prove that the sides have the same length, the Distance Formula will be used.

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Segment	Points	Substitute	Simplify
$\overline{A D}$	$A (0, 0)$ $&$ $D (3, 0)$	$A D =$ $(3 - 0)^{2} + (0 - 0)^{2}$	$A D =$ $3$
$\overline{B C}$	$B (2, 2)$ $&$ $C (5, 2)$	$B C =$ $(5 - 2)^{2} + (2 - 2)^{2}$	$B C =$ $3$

The length of the opposite sides $\overline{A D}$ and $\overline{B C}$ is $3 .$ This means that $\overline{A D}$ and $\overline{B C}$ are congruent.

The quadrilateral has been proven to have a pair of congruent parallel sides. Therefore, it is a parallelogram.

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Catch-Up and Review

Proof

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Hint

Solution

Answer

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Solution

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Solution