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{{ printedBook.courseTrack.name }} {{ printedBook.name }} Using Coordinates in Proofs

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.

Try your knowledge on these topics.

a Find the standard equation of the circle. Write the answer in the form (xh)2+(yk)2=r2, where (h,k) are the coordinates of the center, and r is the radius. b Consider lines m and r on the coordinate plane. Select all the statements that are true.

c Consider lines a, b and c on the coordinate plane. Select all the statements that are true.

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? 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 Q(0,2) 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 (h,k) and radius r is (xh)2+(yk)2=r2. The distance d 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 P 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.

1. Find the equation of the circle.
2. 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 (h,k) and radius r can be expressed.
Since the center is O(0,0), it is known that h=0 and k=0. Furthermore, since Q(2,0) 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.
Simplify right-hand side
r=2
The radius of the circle is r=2. With this information, the standard equation of the circle can be written.
The circle with center at the origin and radius 2 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 x=1 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.
(x0)2+(y0)2=22
Evaluate
A true statement was obtained. Therefore, Ramsha can conclude point is on the circle that passes through Q(0,2) 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.

1. Find the radius of the circle.
2. 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 O(0,0) is the center of the circle and Q(2,0) is a point on the circle, the distance between these points is the radius r of the circle.
Simplify right-hand side
r=2
Dylan found that the radius of the circle is 2.

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

If point is on the circle, then its distance from O(0,0) must also be 2. Once again, the Distance Formula can be used.
Simplify right-hand side
The distance between O(0,0) and is equal to 2, which is the same as the radius of the circle. Therefore, Dylan can conclude that the point is on the circle that passes through Q(0,2) and whose center is the origin. 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 1 and Road 2. 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 3. 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 1 and Road 2.

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 B(4,0), M(-4,4), and P(-2,-2), respectively. To find the coordinates of the midpoints of Road 1 and Road 2, the Midpoint Formula can be used.
Midpoint Formula:
The midpoint of Road 1 is M1(0,2) and the midpoint of Road 2 is M2(-3,1). Now that the coordinates of the midpoints of Road 1 and Road 2 are known, the length of the new road can be calculated. Also, the length of Road 3 will be calculated to compare their lengths. These calculations can be done by using the Distance Formula.
Distance Formula:
Points Substitute Simplify
M1(0,2) and
B(4,0) and
It can be seen in the above table that the distance between points M1 and M2 is half the distance between points B and P.
Therefore, the new road would be half the length of Road 3. Finally, the last step is to check whether this new road would be parallel to road 3. To do so, the slope between points M1 and M2 will be compared to the slope between points B and P. For this, the Slope Formula can be used.
Slope Formula:
Points Substitute Simplify
M1(0,2) and
B(4,0) and
It was found that the slope between M1 and M2 is the same as the slope between B and P.
By the Slopes of Parallel Lines Theorem, the new road would be parallel to Road 3. In conclusion, if the new road connects the midpoints of Road 1 and Road 2, then it would be parallel to Road 3 and half its length. Therefore, it is possible to construct.

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. If DE is a midsegment of ABC, then the following statement holds true.

DEBC and

Proof

Using Coordinates

This theorem can be proven by placing the triangle on a coordinate plane. For simplicity, vertex A will be placed at the origin and vertex B on the x-axis. Since A lies on the origin, its coordinates are (0,0). Point B is on the x-axis, meaning its y-coordinate is 0. The remaining coordinates are unknown and can be named a, b, and c.
If DE is the midsegment from AC to BC, then by the definition of a midpoint, D and E are the midpoints of AC and BC, respectively. To prove this theorem, it must be proven that DE is parallel to AB and that DE is half of AB.

AB∥DE

If the slopes of these two segments are equal, then they are parallel. The y-coordinate of both A(0,0) and B(a,0) is 0. Therefore, AB is a horizontal segment. Next, the coordinates of D and E will be found using the Midpoint Formula.

Segment Endpoints Substitute Simplify
AC A(0,0) and
BC B(a,0) and
The y-coordinate of both and is Therefore, DE is also a horizontal segment. Since all horizontal segments are parallel, it can be said that AB and DE are parallel.

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

Segment Endpoints Length Simplify
AB A(0,0) and B(a,0) AB=a0 AB=a
DE and
Since is half of a, it can be stated that the midsegment DE is half the length of AB.
Therefore, a midsegment of a triangle is parallel to the third side of the triangle and half its length.

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!

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 A(3,5), and respectively. These coordinates can be substituted into the Distance Formula to calculate the side lengths of the quadrilateral. The length of SP will be calculated first.
Evaluate right-hand side
By following the same procedure, all side lengths can be found.
Side Endpoints Substitute Simplify
SP and SP=
PA and A(3,5) PA=
AT A(3,5) and AT=
TS and TS=
By the Transitive Property of Equality, SP and AT are equal, and PA and TS 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 I 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 I is the midpoint of the diagonals, the Midpoint Formula can be used. The midpoint of PT will be calculated first.
Evaluate
The midpoint of the diagonal PT is By following the same procedure, the midpoint of the diagonal SA can be calculated.
Diagonal Endpoints Substitute Simplify
PT and
SA and A(3,5)
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.

Properties of Parallelograms

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

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 P will be placed at the origin and vertex S on the x-axis. Since P lies on the origin, its coordinates are (0,0). Point S is on the x-axis, meaning its y-coordinate is 0. Let a be the x-coordinate of S. Furthermore, let b and c be the coordinates of Q.
Note that both P and S lie on the x-axis. Therefore, SP is a horizontal segment. Since opposite sides of a parallelogram are parallel, QR is also a horizontal segment. This means that Q and R have the same y-coordinate. Let x be the x-coordinate of R. Next, the x-coordinate of R will be determined. Since PQ and SR are parallel, they have the same slope. The slope of PQ can be found using the Slope Formula.
The slope of PQ is By following the same procedure, the slope of SR can be expressed in terms of x.
Side Endpoints Substitute Simplify
PQ P(0,0) and Q(b,c)
SR S(a,0) and R(x,c)
As it has been previously stated, since PQ and SR are parallel, their slopes are equal.
The above equation can be solved for x.
Solve for x
c(xa)=c(b)
xa=b
x=b+a
x=a+b
The x-coordinate of R is a+b. Finally, by using the Distance Formula, the length of each side of the parallelogram can be calculated. The length of PQ will be calculated first.
By following the same procedure, all the side lengths can be calculated.
Side Endpoints Substitute Simplify
PQ P(0,0) and Q(b,c) PQ=
QR Q(b,c) and R(a+b,c) QR= QR=a
SR S(a,0) and R(a+b,c) SR=
PS P(0,0) and S(a,0) PS= PS=a
By the Transitive Property of Equality, it can be said that PQ=SR and that QR=PS.
By definition of congruent segments, it can be stated that the opposite sides of a parallelogram are congruent.

Parallelogram Diagonals Theorem

In a parallelogram, the diagonals bisect each other. If PQRS 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 P will be placed at the origin and vertex S will be placed on the x-axis. Since P lies on the origin, its coordinates are (0,0). Point S is on the x-axis, meaning its y-coordinate is 0. Let a be the x-coordinate of S. Furthermore, let b and c be the coordinates of Q.
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 R are (a+b,c). 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 QS will be calculated first.
Simplify
By following the same procedure, the midpoint of PR can also be found.
Diagonal Endpoints Substitute Simplify
QS Q(b,c) and S(a,0)
PR P(0,0) and R(a+b,c)

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 PM=RM and QM=SM. Finally, by the definition of congruent segments it can be said that PM and RM are congruent, and that QM and SM are also congruent.

Therefore, the diagonals of a parallelogram bisect each other.

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. Solution

For simplicity, the vertices will be labeled A, B, C, and D. Next, the slopes of the sides will be found by using the Slope Formula. The slope of AB will be calculated first.
Evaluate right-hand side
By following the same procedure, the slopes of the remaining sides can be found.
Side Endpoints Substitute Simplify
AB and
BC and
DC 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
and
and
and

Since the product of their slopes is -1, the quadrilateral's consecutive sides are perpendicular. Therefore, they form right angles. Consequently, by its definition, the quadrilateral is a rectangle.