Unlock Geometric Theorems Using Coordinates, Distance Formula, and Circle Equation

This lesson will explore the use of coordinates to prove simple geometric theorems algebraically.

Catch-Up and Review

Here is some recommended reading before getting started.

Equation of a Circle
Slopes of Parallel Lines Theorem
Slopes of Perpendicular Lines Theorem
Midpoint Formula
Slope Formula
Distance Formula

Try your knowledge on these topics.

a Find the standard equation of the circle. Write the answer in the form

where

are the coordinates of the center, and

is the radius.

b Consider lines

and

on the coordinate plane.

Select all the statements that are true.

c Consider lines

and

on the coordinate plane. Select all the statements that are true.

Challenge

Investigating a Parallelogram in a Coordinate Plane

Without using the Pythagorean Theorem nor the Distance Formula, can it be determined whether the quadrilateral shown below is a rectangle?

Example

Proving Statements Using Different Methods

In mathematics, there are usually several ways of proving or disproving a statement. The following example will explore two ways of determining whether a point is on a circle.

Ramsha and Dylan are taking Geometry together. They were asked to determine whether the point

is on the circle that passes through

and whose center is at the origin. They decided to do it using different methods.

Use both methods to determine whether the point is on the circle.

Hint

The standard equation of the circle with center at and radius is The distance between two points with coordinates and is

Solution

Ramsha has decided to write the standard equation of the circle, and then check whether the coordinates of satisfy this equation. Conversely, Dylan has decided to use the Distance Formula. These two options will be explored one at a time.

Ramsha's Solution

As already said, Ramsha has decided first to write the standard equation of the circle and use it to see whether the point is on that circle. To do so, she will follow two steps.

Find the equation of the circle.
Verify whether the coordinates of the point satisfy the equation.

These steps will be done one at a time.

Finding the Equation of the Circle

The standard equation of a circle with center at

and radius

can be expressed.

Since the center is

it is known that

and

Furthermore, since

is on the circle, the distance between this point and the origin is the radius of the circle. This distance can be found by using the Distance Formula.

SubstituteValues

Substitute values

▼

Simplify right-hand side

SubTerms

Subtract terms

CalcPow

Calculate power

IdPropAdd

Identity Property of Addition

CalcRoot

Calculate root

The radius of the circle is

With this information, the standard equation of the circle can be written.

The circle with center at the origin and radius

is shown below.

Verifying If the Coordinates of the Point Satisfy the Equation

Finally, to determine whether point

is on the circle, Ramsha will substitute

and

in the obtained equation. If a true statement is obtained, then the point is on the circle. Otherwise, the point is not on the circle.

SubstituteII

▼

Evaluate

Subtract terms

Calculate power

Add terms

$4=4\ {\color{#009600}{\bm{\Large{\checkmark }}}}$

A true statement was obtained. Therefore, Ramsha can conclude point

is on the circle that passes through

and whose center is the origin.

Dylan's Solution

As previously stated, Dylan has decided to use the Distance Formula. To do so, he will follow two steps.

Find the radius of the circle.
Verify whether the distance from the given point to the center of the circle is the same as the radius.

These steps will be done one at a time.

Finding the Radius of the Circle

Since

is the center of the circle and

is a point on the circle, the distance between these points is the radius

of the circle.

SubstituteValues

Substitute values

▼

Simplify right-hand side

SubTerms

Subtract terms

CalcPow

Calculate power

IdPropAdd

Identity Property of Addition

CalcRoot

Calculate root

Dylan found that the radius of the circle is

Calculating the Distance From the Point to the Center of the Circle

If point

is on the circle, then its distance from

must also be

Once again, the Distance Formula can be used.

SubstituteValues

Substitute values

▼

Simplify right-hand side

Subtract terms

Add terms

Calculate root

$2=2\ {\color{#009600}{\bm{\Large{\checkmark }}}}$

The distance between

and

is equal to

which is the same as the radius of the circle. Therefore, Dylan can conclude that the point

is on the circle that passes through

and whose center is the origin.

Example

Solving Real Life Problems Involving Triangles

Real life problems can also be solved by setting a coordinate plane and using geometric properties.

A small country has three main roads connecting the beach, the mountains, and a national park. To avoid traffic jams and car accidents during summer holidays, the government of the country is planning to build a fourth road to connect the midpoints of Road and Road

Due to financial and logistic issues, building the fourth road would only be possible if it is parallel to and half the length of Road

By placing the diagram on a coordinate plane, determine whether it is possible to build the new road.

Hint

After placing the diagram on a coordinate plane, use the Midpoint Formula to find the coordinates of the midpoints of Road and Road

Solution

First, the diagram will be placed on a coordinate plane.

It can be seen in the above diagram that the beach, the mountains, and the national park are located at

and

respectively. To find the coordinates of the midpoints of Road

and Road

the Midpoint Formula can be used.

Midpoint Formula:
Road	Points	Substitute	Simplify
Road	and
Road	and

The midpoint of Road

and the midpoint of Road

Now that the coordinates of the midpoints of Road

and Road

are known, the length of the new road can be calculated. Also, the length of Road

will be calculated to compare their lengths. These calculations can be done by using the Distance Formula.

Distance Formula:
Points	Substitute	Simplify
and
and

It can be seen in the above table that the distance between points

and

is half the distance between points

and

Therefore, the new road would be half the length of Road

Finally, the last step is to check whether this new road would be parallel to road

To do so, the slope between points

and

will be compared to the slope between points

and

For this, the Slope Formula can be used.

Slope Formula:
Points	Substitute	Simplify
and
and

It was found that the slope between

and

is the same as the slope between

and

By the Slopes of Parallel Lines Theorem, the new road would be parallel to Road

In conclusion, if the new road connects the midpoints of Road

and Road

then it would be parallel to Road

and half its length. Therefore, it is possible to construct.

Discussion

Triangle Midsegment Theorem

The result obtained in the previous example can be generalized to any triangle.

The line segment that connects the midpoints of two sides of a triangle — also known as a midsegment — is parallel to the third side of the triangle and half its length.

is a midsegment of

then the following statement holds true.

and

Proof

Using Coordinates

This theorem can be proven by placing the triangle on a coordinate plane. For simplicity, vertex will be placed at the origin and vertex on the axis.

Since

lies on the origin, its coordinates are

Point

is on the

axis, meaning its

coordinate is

The remaining coordinates are unknown and can be named

and

is the midsegment from

then by the definition of a midpoint,

and

are the midpoints of

and

respectively.

To prove this theorem, it must be proven that is parallel to and that is half of

If the slopes of these two segments are equal, then they are parallel. The coordinate of both and is Therefore, is a horizontal segment. Next, the coordinates of and will be found using the Midpoint Formula.


Segment	Endpoints	Substitute	Simplify
	and
	and

The

coordinate of both

and

Therefore,

is also a horizontal segment. Since all horizontal segments are parallel, it can be said that

and

are parallel.

Since both and are horizontal, their lengths are given by the difference of the coordinates of their endpoints.

Segment	Endpoints	Length	Simplify
	and
	and

Since

is half of

it can be stated that the midsegment

is half the length of

Therefore, a midsegment of two sides of a triangle is parallel to the third side of the triangle and half its length.

Example

Solving Real Life Problems Involving Rectangles

Not only can coordinates be used to prove properties of triangles, but also they can be used to prove properties of quadrilaterals.

Davontay lives in a neighborhood in Boston where the stadium, the amusement park, the airport, and the train station are the vertices of a quadrilateral that seems to be a parallelogram.

Davontay knows that to determine whether a quadrilateral is a parallelogram, it is enough with to prove that opposite sides are congruent. Additionally, he wants to determine whether the diagonals bisect each other. Help Davontay to prove or deny these statements!

Answer

See solution.

Hint

Place the diagram on a coordinate plane.

Solution

First, the diagram will be placed on a coordinate plane.

Davontay wants to see whether the opposite sides are congruent, and whether the diagonals bisect each other. These two things will be done one at a time.

Opposite Sides are Congruent

In the previous diagram, it can be seen that the stadium, the amusement park, the airport, and the train station are at

and

respectively. These coordinates can be substituted into the Distance Formula to calculate the side lengths of the quadrilateral. The length of

will be calculated first.

SubstituteValues

Substitute values

▼

Evaluate right-hand side

SubNeg

CalcPow

Calculate power

AddTerms

Add terms

By following the same procedure, all side lengths can be found.


Side	Endpoints	Substitute	Simplify
	and
	and
	and
	and

By the Transitive Property of Equality,

and

are equal, and

and

are also equal.

Finally, if two segments have the same length, then they are congruent. Therefore, it can be said that the opposite sides of the quadrilateral are congruent.

This confirms that the quadrilateral is a parallelogram.

Diagonals Bisect Each Other

The diagonals of the parallelogram are the line segments that connect opposite vertices. In the diagram below, the diagonals will be drawn and their point of intersection

will be plotted.

The midpoint of a line segment is the point that divides the segment into two congruent segments. Therefore, if the diagonals intersect at their midpoint, then they bisect each other. To determine whether

is the midpoint of the diagonals, the Midpoint Formula can be used. The midpoint of

will be calculated first.

SubstitutePoints

Substitute &

▼

Evaluate

AddNeg

AddSubTerms

Add and subtract terms

QuotOne

The midpoint of the diagonal

By following the same procedure, the midpoint of the diagonal

can be calculated.


Diagonal	Endpoints	Substitute	Simplify
	and
	and

It can be seen that the midpoint of each diagonal has coordinates

Therefore, they intersect at their midpoint.

It can be stated that the diagonals bisect each other.

Discussion

Properties of Parallelograms

The results of the previous example can be generalized to any parallelogram.

Rule

Parallelogram Opposite Sides Theorem

The opposite sides of a parallelogram are congruent.

In respects to the characteristics of the diagram, the following statement holds true.

Proof

Using Coordinates

This theorem can be proven by placing the parallelogram on a coordinate plane. For simplicity, vertex will be placed at the origin and vertex on the axis.

Since

lies on the origin, its coordinates are

Point

is on the

axis, meaning its

coordinate is

Let

be the

coordinate of

Furthermore, let

and

be the coordinates of

Note that both

and

lie on the

axis. Therefore,

is a horizontal segment. Since opposite sides of a parallelogram are parallel,

is also a horizontal segment. This means that

and

have the same

coordinate. Let

be the

coordinate of

Next, the

coordinate of

will be determined. Since

and

are parallel, they have the same slope. The slope of

can be found using the Slope Formula.

SubstituteValues

Substitute values

SubTerms

Subtract terms

The slope of

By following the same procedure, the slope of

can be expressed in terms of


Side	Endpoints	Substitute	Simplify
	and
	and

As it has been previously stated, since

and

are parallel, their slopes are equal.

The above equation can be solved for

▼

Solve for

Commutative Property of Addition

The

coordinate of

Finally, by using the Distance Formula, the length of each side of the parallelogram can be calculated. The length of

will be calculated first.

SubstituteValues

Substitute values

SubTerms

Subtract terms

By following the same procedure, all the side lengths can be calculated.


Side	Endpoints	Substitute	Simplify
	and
	and
	and
	and

By the Transitive Property of Equality, it can be said that

and that

By definition of congruent segments, it can be stated that the opposite sides of a parallelogram are congruent.

Rule

Parallelogram Diagonals Theorem

In a parallelogram, the diagonals bisect each other.

If is a parallelogram, then the following statement holds true.

Proof

Using Coordinates

This theorem can be proven by placing the parallelogram on a coordinate plane. For simplicity, vertex will be placed at the origin and vertex will be placed on the axis.

Since

lies on the origin, its coordinates are

Point

is on the

axis, meaning its

coordinate is

Let

be the

coordinate of

Furthermore, let

and

be the coordinates of

By the definition of a parallelogram and by the Parallelogram Opposite Sides Theorem, opposite sides of a parallelogram are parallel and congruent. With this information, it can be said that the coordinates of

are

Knowing the coordinates of the vertices, the coordinates of the midpoint of the diagonals can be found. Use the Midpoint Formula to do so. The midpoint of

will be calculated first.

SubstitutePoints

Substitute &

▼

Simplify

IdPropAdd

Identity Property of Addition

CommutativePropAdd

Commutative Property of Addition

By following the same procedure, the midpoint of

can also be found.


Diagonal	Endpoints	Substitute	Simplify
	and
	and

The midpoints of the diagonals are the same. Therefore, the diagonals intersect at their midpoints. By the definition of a midpoint, it can be stated that and Finally, by the definition of congruent segments it can be said that and are congruent, and that and are also congruent.

Therefore, the diagonals of a parallelogram bisect each other.

Closure

Classifying a Parallelogram in a Coordinate Plane

The challenge presented at the beginning of this lesson can be solved by using coordinates.

Using neither the Pythagorean Theorem nor the Distance Formula, it is desired to determine whether the quadrilateral below is a rectangle.

Hint

By the Slopes of Perpendicular Lines Theorem, if the slopes of two lines or segments are opposite reciprocals, then lines are perpendicular.

Solution

For simplicity, the vertices will be labeled and

Next, the slopes of the sides will be found by using the Slope Formula. The slope of

will be calculated first.

SubstituteValues

Substitute values

▼

Evaluate right-hand side

SubNeg

AddSubTerms

Add and subtract terms

By following the same procedure, the slopes of the remaining sides can be found.


Side	Endpoints	Substitute	Simplify
	and
	and
	and
	and

By the Slopes of Perpendicular Lines Theorem, if the slopes of two line segments are opposite reciprocals, then the segments are perpendicular. The slopes of consecutive sides will be multiplied to see if they are opposite reciprocals. Recall that, in a polygon, two sides are consecutive if they share a common vertex.

Sides	Product of Slopes
and	${\color{#0000FF}{\dfrac{4}{3}}}\left({\color{#009600}{\text{-} \dfrac{3}{4}}}\right)=\text{-} 1\ {\color{#009600}{\bm{\Large{\checkmark }}}}$
and	${\color{#009600}{\text{-} \dfrac{3}{4}}}\left(\textcolor{deepskyblue}{\dfrac{4}{3}}\right)=\text{-} 1\ {\color{#009600}{\bm{\Large{\checkmark }}}}$
and	$\textcolor{deepskyblue}{\dfrac{4}{3}}\left(\textcolor{olive}{\text{-} \dfrac{3}{4}}\right)=\text{-} 1\ {\color{#009600}{\bm{\Large{\checkmark }}}}$
and	$\textcolor{olive}{\text{-} \dfrac{3}{4}}\left({\color{#0000FF}{\dfrac{4}{3}}}\right)=\text{-} 1\ {\color{#009600}{\bm{\Large{\checkmark }}}}$

Since the product of their slopes is the quadrilateral's consecutive sides are perpendicular. Therefore, they form right angles. Consequently, by its definition, the quadrilateral is a rectangle.

	{{ '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

Hint

Solution

Ramsha's Solution

Finding the Equation of the Circle

Verifying If the Coordinates of the Point Satisfy the Equation

Dylan's Solution

Finding the Radius of the Circle

Calculating the Distance From the Point to the Center of the Circle

Hint

Solution

Proof

Answer

Hint

Solution

Opposite Sides are Congruent

Diagonals Bisect Each Other

Proof

Proof

Hint

Solution