Here are a few recommended readings before getting started.
Try your knowledge on these topics.
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?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} $
This formula is called the Distance Formula.
$a=x_{2}−x_{1}$, $b=y_{2}−y_{1}$
$d=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} $
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.
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.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.
Substitute $(-3,0)$ & $(4,-1)$
$d=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} $ | |||
---|---|---|---|
Segment | Points | Substitute | Simplify |
$AC$ | $A(-3,0)$ $&$ $C(4,-1)$ | $AC=$ $(4−(-3))_{2}+(-1−0)_{2} $ | $AC=$ $52 $ |
$AB$ | $A(-3,0)$ $&$ $B(-2,2)$ | $AB=$ $(-2−(-3))_{2}+(2−0)_{2} $ | $AB=$ $5 $ |
$BC$ | $B(-2,2)$ $&$ $C(4,-1)$ | $BC=$ $(4−(-2))_{2}+(-1−2)_{2} $ | $BC=$ $35 $ |
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.
Start by finding the radius of the circle.
Substitute $(0,0)$ & $(0,2)$
Substitute $(0,0)$ & $(1,3 )$
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.
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.
$M(2x_{1}+x_{2} ,2y_{1}+y_{2} )$
The formula above is called the Midpoint Formula.
$LHS+x_{m}=RHS+x_{m}$
$LHS+x_{1}=RHS+x_{1}$
Commutative Property of Addition
$LHS/2=RHS/2$
$M(2x_{1}+x_{2} ,2y_{1}+y_{2} )$
Use the Midpoint Formula to calculate the coordinates of the midpoint $M$ between the points plotted on the coordinate plane.
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.See solution.
Use the Midpoint Formula to show that the diagonals intersect at their midpoint.
Substitute $(-2,2)$ & $(3,-3)$
Add terms
$a+(-b)=a−b$
Put minus sign in front of fraction
$ba =a÷b$
$M(2x_{1}+x_{2} ,2y_{1}+y_{2} )$ | |||
---|---|---|---|
Diagonal | Endpoints | Substitute | Simplify |
$BD$ | $B(-2,2)$ $&$ $D(3,-3)$ | $M(2-2+3 ,22+(-3) )$ | $M(0.5,-0.5)$ $✓$ |
$AC$ | $A(-3,0)$ $&$ $D(4,-1)$ | $M(2-3+4 ,20+(-1) )$ | $M(0.5,-0.5)$ $✓$ |
The midpoint of both diagonals is the same as their point of intersection. Therefore, the diagonals bisect each other.
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 — located at any position on the coordinate plane.
In the applet below, the polygon can be translated by dragging it. Even more, it can be rotated around the center of rotation. Move the center of rotation around any of the polygon's vertices. Try to place a vertex of the polygon at the origin and a consecutive vertex on the positive $x-axis!$
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!See 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. |
$M(2x_{1}+x_{2} ,2y_{1}+y_{2} )$ | |||
---|---|---|---|
Side | Endpoints | Substitute | Simplify |
$AB$ | $A(0,0)$ and $B(2,2.5)$ | $M_{1}(20+2 ,20+2.5 )$ | $M_{1}(1,1.25)$ |
$BC$ | $B(2,2.5)$ and $C(6,0)$ | $M_{2}(22+6 ,22.5+0 )$ | $M_{2}(4,1.25)$ |
It can be seen above that both $M_{1}$ and $M_{2}$ have the same $y-$coordinate $1.25.$ This means that $M_{1}M_{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_{1}M_{2} ∥AC $ Finally, it needs to be proven that $M_{1}M_{2}$ is half $AC.$ To do this, the Distance Formula will be used.
$d=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} $ | |||
---|---|---|---|
Segment | Endpoints | Substitute | Simplify |
$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} $ | $M_{1}M_{2}=3$ |
$BC$ | $A(0,0)$ and $C(6,0)$ | $AC=(6−0)_{2}+(0−0)_{2} $ | $AC=6$ |
Since $3$ is half $6,$ the length of $M_{1}M_{2} $ is half the length of $AC.$ $M_{1}M_{2}=21 AC $ 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!
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?
See solution.
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.
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 |
$AD$ | $A(0,0)$ $&$ $D(3,0)$ | $AD=$ $(3−0)_{2}+(0−0)_{2} $ | $AD=$ $3$ |
$BC$ | $B(2,2)$ $&$ $C(5,2)$ | $BC=$ $(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.