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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.
In a parallelogram, adjacent angles are congruent.
The four sides of a parallelogram are congruent. {"type":"choice","form":{"alts":["True","False"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}
In a parallelogram, opposite sides and opposite angles are congruent. {"type":"choice","form":{"alts":["True","False"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
The diagonals of a parallelogram bisect each other. {"type":"choice","form":{"alts":["True","False"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
Adjacent angles in a parallelogram are supplementary. {"type":"choice","form":{"alts":["True","False"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}

A parallelogram has exactly two pairs of parallel sides.

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d Use the Pythagorean Theorem to find the missing side length.

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The figure below *looks* like 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} $

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=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} $

${c=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} c=d ⇓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.{"type":"choice","form":{"alts":["Ramsha","Davontay"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}

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.

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 $△ABC.$
The distance $d$ between $A$ and $C,$ and therefore the length of the diagonal $AC,$ is $52 .$ The same procedure can be used to find $AB$ and $BC.$

Finally, as previously mentioned, it needs to be seen whether $AC$ squared is equal to the sum of the squares of $AB$ and $BC.$
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 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=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} $

SubstitutePoints

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

$d=(4−(-3))_{2}+(-1−0)_{2} $

$d=52 $

$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.

Consider the circle centered at the origin and containing the point $Q(0,2).$
### Hint

### Solution

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.

{"type":"choice","form":{"alts":["The point lies on the circle","The point does not lie on the circle"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}

Start by finding the radius of the circle.

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.
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.
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.

$d=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} $

SubstitutePoints

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

$d=(0−0)_{2}+(2−0)_{2} $

$d=2$

$d=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} $

SubstitutePoints

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

$d=(1−0)_{2}+(3 −0)_{2} $

$d=2$

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.

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 $AB$ since the distance from $A$ to $M$ is the same as the distance from $M$ to $B.$

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.

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.
The $x-$coordinate of $M$ is $x_{m}=2x_{1}+x_{2} .$ In the same way, it can be shown that the $y-$coordinate of $M$ is $y_{m}=2y_{1}+y_{2} .$ With this information, the coordinates of $M$ can be expressed in terms of the coordinates of $A$ and $B.$

Consider the horizontal distance $Δx$ and the vertical distance $Δy$ between $A$ and $B$. Since $M$ is the midpoint, $M$ splits each distance, $Δx$ and $Δy$, in half. Therefore, the horizontal and vertical distances from each endpoint to the midpoint are $2Δx $ and $2Δy .$ 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}andx_{2}−x_{m} $

The graph above shows that these distances are both equal to $2Δx .$ Therefore, by the Transitive Property of Equality, they are equal.
$⎩⎪⎪⎨⎪⎪⎧ x_{m}−x_{1}=2Δx 21 x_{2}−x_{m}=2Δx 21 ⇓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}=2x_{1}+x_{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.

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.

$M(2x_{1}+x_{2} ,$