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.

Given two points $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ on a coordinate plane, their distance $d$ is given by the following formula.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Proof

Start by plotting $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ on the coordinate plane. Both points can be arbitrarily plotted in Quadrant I for simplicity. Note that the position of the points in the plane does not affect the proof. Assume that $x_{2}$ is greater than $x_{1}$ and that $y_{2}$ is greater than $y_{1} .$

Next, draw a right triangle. The hypotenuse of this triangle will be the segment that connects points

A

and

B .

The difference between the

x -

coordinates of the points is the length of one of the legs of the triangle. Furthermore, the length of the other leg is given by the difference between the

y -

coordinates. Therefore, the lengths of the legs are

x_{2} - x_{1}

and

y_{2} - y_{1} .

Now, consider the Pythagorean Equation.

a^{2} + b^{2} = c^{2}

Here,

a

and

b

are the lengths of the legs, and

c

the length of the hypotenuse of a right triangle. Substitute the expressions for the legs for

a

and

b

to find the hypotenuse's length. Then, the equation can be solved for

c .

a^{2} + b^{2} = c^{2}

SubstituteII

$a = x_{2} - x_{1}$ , $b = y_{2} - y_{1}$

(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} = c^{2}

▼

Solve for

c

SqrtEqn

$LHS = RHS$

(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} = c

RearrangeEqn

Rearrange equation

c = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Note that, when solving for

c,

only the principal root was considered. The reason is that

c

represents the length of a side and therefore must be positive. Keeping in mind that

c

is the distance between

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

and

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

then

c = d .

By the Transitive Property of Equality, the Distance Formula is obtained.

{c = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} c = d ⇓ d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

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.

Example

Classifying a Parallelogram Using Its Coordinates

Ramsha and Davontay are learning to make picture frames so they want to determine whether the quadrilateral $A B C D$ shown below is a rectangle.

Ramsha says that

A B C D

is a rectangle. Davontay says it is not. Use the Distance Formula to decide who is correct.

Hint

Other than just using the Distance Formula, try the Converse of the Pythagorean Theorem to determine whether the angles of the quadrilateral are right angles.

Solution

If the four angles of the quadrilateral are right angles, then

A B C D

is a rectangle. First, it will be determined whether

∠ B

is a right angle. To do so, start by drawing the diagonal

\overline{A C}

and considering

△ A B C .

By the Converse of the Pythagorean Theorem, if

A C

squared is equal to the sum of the squares of

A B

and

B C,

then

△ A B C

is a right triangle. In this case, then

∠ B

is a right angle. To find

A C,

the coordinates of

A (- 3, 0)

and

C (4,, - 1)

will be substituted into the Distance Formula.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(- 3, 0)$ & $(4, - 1)$

d = (4 - (- 3))^{2} + (- 1 - 0)^{2}

▼

Evaluate right-hand side

SubNeg

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

d = 7^{2} + (- 1 - 0)^{2}

SubTerm

Subtract term

d = 7^{2} + (- 1)^{2}

CalcPow

Calculate power

d = 49 + 1

AddTerms

Add terms

d = 50

SplitIntoFactors

Split into factors

d = 25 (2)

SqrtProd

$a \cdot b = a \cdot b$

d = 252

CalcRoot

Calculate root

d = 52

The distance

d

between

A

and

C,

and therefore the length of the diagonal

\overline{A C},

52 .

The same procedure can be used to find

A B

and

B C .

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Segment	Points	Substitute	Simplify
$\overline{A C}$	$A (- 3, 0)$ $&$ $C (4, - 1)$	$A C =$ $(4 - (- 3))^{2} + (- 1 - 0)^{2}$	$A C =$ $52$
$\overline{A B}$	$A (- 3, 0)$ $&$ $B (- 2, 2)$	$A B =$ $(- 2 - (- 3))^{2} + (2 - 0)^{2}$	$A B =$ $5$
$\overline{B C}$	$B (- 2, 2)$ $&$ $C (4, - 1)$	$B C =$ $(4 - (- 2))^{2} + (- 1 - 2)^{2}$	$B C =$ $35$

Finally, as previously mentioned, it needs to be seen whether

A C

squared is equal to the sum of the squares of

A B

and

B C .

A C^{2} = ? A B^{2} + B C^{2}

SubstituteValues

Substitute values

(52)^{2} = ? (5)^{2} + (35)^{2}

▼

Evaluate

PowProdII

$(a b)^{m} = a^{m} b^{m}$

25 (2)^{2} = ? (5)^{2} + 9 (5)^{2}

PowSqrt

$(a)^{2} = a$

25 (2) = ? 5 + 9 (5)

Multiply

50 = ? 5 + 45

AddTerms

Add terms

50 = 50 ✓

Since a true statement was obtained, it can be said that

A C^{2} = A B^{2} + B C^{2} .

By the Converse of the Pythagorean Theorem,

△ A B C

is a right triangle. Therefore,

∠ B

is a right angle.

By following the same procedure, it can be shown that $∠ A,$ $∠ C,$ and $∠ D$ are also right angles.

Since each angle of $A B C D$ is a right angle, the quadrilateral is a rectangle. Therefore, Ramsha is correct. She's on her way toward making some nice frames.

Example

Determining If a Point Is On a Circle's Circumference

Consider the circle centered at the origin and containing the point

Q (0, 2) .

Maya has been asked to determine whether the point

P (1, 3)

lies on the above circle. Use the Distance Formula to help Maya find the answer.

Hint

Start by finding the radius of the circle.

Solution

The radius of a circle is constant. Therefore, if

P (1, 3)

lies on the circle, then its distance from the origin would also equal the radius of the circle. Point

Q (0, 2)

is given, so the radius can be solved by finding the distance between point

Q

and the origin, using the Distance Formula.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(0, 0)$ & $(0, 2)$

d = (0 - 0)^{2} + (2 - 0)^{2}

▼

Evaluate right-hand side

SubTerms

Subtract terms

d = 0^{2} + 2^{2}

CalcPow

Calculate power

d = 0 + 4

IdPropAdd

Identity Property of Addition

d = 4

CalcRoot

Calculate root

d = 2

The distance between the origin and

Q (0, 2),

and therefore the radius of the circle, is

2 .

Using the Distance Formula one more time, the distance between the origin and

P (1, 3)

will be calculated.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(0, 0)$ & $(1, 3)$

d = (1 - 0)^{2} + (3 - 0)^{2}

▼

Evaluate right-hand side

SubTerms

Subtract terms

d = 1^{2} + (3)^{2}

BaseOne

$1^{a} = 1$

d = 1 + (3)^{2}

PowSqrt

$(a)^{2} = a$

d = 1 + 3

AddTerms

Add terms

d = 4

CalcRoot

Calculate root

d = 2

The distance between

P (1, 3)

and the origin is the same as the radius of the circle,

2 .

Therefore, the point

P

lies on the circle.

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.

Concept

Midpoint

The midpoint of a line segment is the point that divides the segment into two segments of equal length.

The point

M

is the midpoint of the segment

\overline{A B}

since the distance from

A

M

is the same as the distance from

M

B .

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

Commutative Property of Addition

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.

Illustration

Using Rigid Motions to Prove a Theorem

Any figure on a coordinate plane — located at any position — can be transformed through a combination of rigid motions to have a vertex at the origin and a consecutive vertex on the

x -

axis. By this reasoning, when proving a theorem for a figure with a vertex at the origin and a consecutive vertex on the

x -

axis, that theorem is valid for any figure with the same shape and size.

Try to place a vertex of the polygon at the origin and a consecutive vertex on the positive

x - axis!

Extra

How to Use the Applet

The polygon can be translated by clicking inside it and dragging it.
The polygon can be rotated around the center of rotation by clicking and dragging any of its vertices.
The center of rotation can be moved wherever needed.

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.

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount }}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount }}
	{{ 'ml-lesson-time-estimation' \| message }}

{{ article.displayTitle }}

Catch-Up and Review

Proof

Hint

Solution

Hint

Solution

Proof

Answer

Hint

Solution

Extra

Answer

Hint

Solution

Answer

Hint

Solution