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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.
- Coordinate plane
- Basic definitions
- Transformations
- Congruence
- Theorems about parallelograms
- Pythagorean Theorem
- Circles
Try your knowledge on these topics.
a State the coordinates of the points plotted on the plane.
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b Match each concept with its corresponding diagram.
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c True or false?
A parallelogram has exactly two pairs of parallel sides.
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In a parallelogram, adjacent angles are congruent.
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The four sides of a parallelogram are congruent.
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In a parallelogram, opposite sides and opposite angles are congruent.
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The diagonals of a parallelogram bisect each other.
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Adjacent angles in a parallelogram are supplementary.
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d Use the Pythagorean Theorem to find the missing side length.
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Challenge
Investigating the Properties of a Parallelogram
The figure below looks like 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 = sqrt((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
c=sqrt((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= sqrt((x_2-x_1)^2+(y_2-y_1)^2) c= d ⇓ d= sqrt((x_2-x_1)^2+(y_2-y_1)^2)
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 ABCD shown below is a rectangle.
Ramsha says that ABCD is a rectangle. Davontay says it is not. Use the Distance Formula to decide who is correct.
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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 ABCD is a rectangle. First, it will be determined whether ∠ B is a right angle. To do so, start by drawing the diagonal AC and considering △ABC.
By the Converse of the Pythagorean Theorem, if AC squared is equal to the sum of the squares of AB and BC, then △ ABC is a right triangle. In this case, then ∠ B is a right angle. To find AC, the coordinates of A(- 3,0) and C(4,,- 1) will be substituted into the Distance Formula.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
SubstitutePoints
Substitute ( - 3,0) & ( 4,- 1)
d=sqrt(( 4-( - 3))^2+( - 1- 0)^2)
d=5sqrt(2)
The distance d between A and C, and therefore the length of the diagonal AC, is 5sqrt(2). The same procedure can be used to find AB and BC.
| d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) | |||
|---|---|---|---|
| Segment | Points | Substitute | Simplify |
| AC | A( - 3, 0) & C( 4, - 1) | AC= sqrt(( 4-( - 3))^2+( - 1- 0)^2) | AC= 5sqrt(2) |
| AB | A( - 3, 0) & B( - 2, 2) | AB= sqrt(( - 2-( - 3))^2+( 2- 0)^2) | AB= sqrt(5) |
| BC | B( - 2, 2) & C( 4, - 1) | BC= sqrt(( 4-( - 2))^2+( - 1- 2)^2) | BC= 3sqrt(5) |
Finally, as previously mentioned, it needs to be seen whether AC squared is equal to the sum of the squares of AB and BC.
AC^2? =AB^2+BC^2
SubstituteValues
Substitute values
( 5sqrt(2))^2? =( sqrt(5))^2+( 3sqrt(5))^2
50=50 ✓
Since a true statement was obtained, it can be said that AC^2=AB^2+BC^2. By the Converse of the Pythagorean Theorem, △ABC 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 ABCD 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).
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Hint
Start by finding the radius of the circle.
Solution
The radius of a circle is constant. Therefore, if P(1,sqrt(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 = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
SubstitutePoints
Substitute ( 0,0) & ( 0,2)
d=sqrt(( 0- 0)^2+( 2- 0)^2)
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,sqrt(3)) will be calculated.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
SubstitutePoints
Substitute ( 0,0) & ( 1,sqrt(3))
d=sqrt(( 1- 0)^2+( sqrt(3)- 0)^2)
d=2
The distance between P(1,sqrt(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.
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(x_1 + x_2/2,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 Δ x2 and Δ y2. 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 Δ x2. Therefore, by the Transitive Property of Equality, they are equal.
x_m-x_1= Δ x2 x_2-x_m= Δ x2 ⇓ 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
2x_m - x_1 = x_2
AddEqn
LHS+x_1=RHS+x_1
2x_m = x_2+x_1
CommutativePropAdd
Commutative Property of Addition
2x_m = x_1+x_2
DivEqn
.LHS /2.=.RHS /2.
x_m = x_1+x_2/2
The x-coordinate of M is x_m = x_1+x_22. In the same way, it can be shown that the y-coordinate of M is y_m= y_1+y_22. With this information, the coordinates of M can be expressed in terms of the coordinates of A and B.
M(x_1 + x_2/2,y_1 + y_2/2 )
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 BD and AC. Then, identify the coordinates of their point of intersection.
It is seen above that the point of intersection of BD and AC is (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 BD, the coordinates of the endpoints B(- 2,2) and D(3,- 3) can be substituted into the Midpoint Formula.
M(x_1+x_2/2,y_1+y_2/2)
SubstitutePoints
Substitute ( - 2,2) & ( 3,- 3)
M(- 2+ 3/2,2+( - 3)/2)
▼
Evaluate
AddTerms
Add terms
M(1/2,2+(- 3)/2)
AddNeg
a+(- b)=a-b
M(1/2,- 1/2)
MoveNegNumToFrac
Put minus sign in front of fraction
M(1/2,- 1/2)
FracToDiv
a/b=a÷ b
M(0.5,- 0.5) ✓
The midpoint of the diagonal BD is the point of intersection of the diagonals. By following the same procedure, it can be shown that the midpoint of AC is also (0.5,- 0,5).
| M(x_1+x_2/2,y_1+y_2/2) | |||
|---|---|---|---|
| Diagonal | Endpoints | Substitute | Simplify |
| BD | B( - 2, 2) & D( 3, - 3) | M(- 2+ 3/2,2+( - 3)/2) | M(0.5,- 0.5) ✓ |
| AC | A( - 3, 0) & D( 4, - 1) | M(- 3+ 4/2,0+( - 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.
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.
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, △ ABC 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 AB and BC, respectively.
The coordinates of the midpoints of AB and BC can be found by using the Midpoint Formula.
| M(x_1+x_2/2,y_1+y_2/2) | |||
|---|---|---|---|
| Side | Endpoints | Substitute | Simplify |
| AB | A( 0, 0) and B( 2, 2.5) | M_1(0+ 2/2,0+ 2.5/2) | M_1(1, 1.25) |
| BC | B( 2, 2.5) and C( 6, 0) | M_2(2+ 6/2,2.5+ 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 M_1M_2 is a horizontal segment. Since AC is on the x-axis, it can be said that it is also a horizontal segment. Therefore, the midsegment is parallel to AC. M_1M_2 ∥ AC Finally, it needs to be proven that M_1M_2 is half AC. To do this, the Distance Formula will be used.
| d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) | |||
|---|---|---|---|
| Segment | Endpoints | Substitute | Simplify |
| M_1M_2 | M_1( 1, 1.25) and M_2( 4, 1.25) | M_1M_2=sqrt(( 4- 1)^2+( 1.25- 1.25)^2) | M_1M_2= 3 |
| BC | A( 0, 0) and C( 6, 0) | AC=sqrt(( 6- 0)^2+( 0- 0)^2) | AC=6 |
Since 3 is half 6, the length of M_1M_2 is half the length of AC. M_1M_2=1/2AC By following the same procedure, it can be proven that the other two midsegments of △ ABC 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 ABCD has one pair of congruent parallel sides to prove that ABCD is a parallelogram. Note that the y-coordinate of B and C is 2. Therefore, BC is parallel to the x-axis. Also, A and D lie on the x-axis, so BC and AD are parallel.
To prove that the sides have the same length, the Distance Formula will be used.
| d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) | |||
|---|---|---|---|
| Segment | Points | Substitute | Simplify |
| AD | A( 0, 0) & D( 3, 0) | AD= sqrt(( 3- 0)^2+( 0- 0)^2) | AD= 3 |
| BC | B( 2, 2) & C( 5, 2) | BC= sqrt(( 5- 2)^2+( 2- 2)^2) | BC= 3 |
The length of the opposite sides AD and BC is 3. This means that AD and BC are congruent.
The quadrilateral has been proven to have a pair of congruent parallel sides. Therefore, it is a parallelogram.
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