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{{ printedBook.courseTrack.name }} {{ printedBook.name }} 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 $(x−h)_{2}+(y−k)_{2}=r_{2},$ where $(h,k)$ are the coordinates of the center, and $r$ is the radius.

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b Consider lines $ℓ,$ $m$ and $r$ on the coordinate plane.

Select all the statements that are true.

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

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

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 $P(1,3 )$ 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.

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The standard equation of the circle with center at $(h,k)$ and radius $r$ is $(x−h)_{2}+(y−k)_{2}=r_{2}.$ The distance $d$ between two points with coordinates $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is $d=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} .$

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.

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.

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

SubstituteValues

Substitute values

$r=(0−2)_{2}+(0−0)_{2} $

$r=2$

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.

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

SubstituteValues

Substitute values

$r=(0−2)_{2}+(0−0)_{2} $

$r=2$

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

SubstituteValues

Substitute values

$2=?(0−1)_{2}+(0−3 )_{2} $

$2=2✓$

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.

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After placing the diagram on a coordinate plane, use the Midpoint Formula to find the coordinates of the midpoints of Road $1$ and Road $2.$

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

The midpoint of Road $1$ is $M_{1}(0,2)$ and the midpoint of Road $2$ is $M_{2}(-3,1).$

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: $(2x_{1}+x_{2} ,2y_{1}+y_{2} )$ | |||
---|---|---|---|

Road | Points | Substitute | Simplify |

Road $1$ | $B(4,0)$ and $M(-4,4)$ | $M_{1}(24+(-4) ,20+4 )$ | $M_{1}(0,2)$ |

Road $2$ | $M(-4,4)$ and $P(-2,-2)$ | $M_{2}(2-4+(-2) ,24+(-2) )$ | $M_{2}(-3,1)$ |

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

Points | Substitute | Simplify | |

$M_{1}(0,2)$ and $M_{2}(-3,1)$ | $d_{M_{1}M_{2}}=(-3−0)_{2}+(1−2)_{2} $ | $d_{M_{1}M_{2}}=10 $ | |

$B(4,0)$ and $P(-2,-2)$ | $d_{BP}=(-2−4)_{2}+(-2−0)_{2} $ | $d_{BP}=210 $ |

It can be seen in the above table that the distance between points $M_{1}$ and $M_{2}$ is half the distance between points $B$ and $P.$ $10 =21 (210 )⇓d_{M_{1}M_{2}}=21 d_{BP} $ 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 $M_{1}$ and $M_{2}$ will be compared to the slope between points $B$ and $P.$ For this, the Slope Formula can be used.

Slope Formula: $m=x_{2}−x_{1}y_{2}−y_{1} $ | |||
---|---|---|---|

Points | Substitute | Simplify | |

$M_{1}(0,2)$ and $M_{2}(-3,1)$ | $m_{M_{1}M_{2}}=-3−01−2 $ | $m_{M_{1}M_{2}}=31 $ | |

$B(4,0)$ and $P(-2,-2)$ | $m_{BP}=-2−4-2−0 $ | $m_{BP}=31 $ |

It was found that the slope between $M_{1}$ and $M_{2}$ is the same as the slope between $B$ and $P.$ $31 =31 ⇓m_{M_{1}M_{2}}=m_{BP} $ 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.

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.
### Proof

Using Coordinates ### $AB∥DE$

### $DE=21 AB$

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

$DE∥BC$ and $DE=21 BC$

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 $c.$ $A(0,0)B(a,0)C(b,c) $ If $DE$ is the midsegment from $AC$ to $BC,$ then by 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 $AB.$

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.

$M(2x_{1}+x_{2} ,2y_{1}+y_{2} )$ | |||
---|---|---|---|

Segment | Endpoints | Substitute | Simplify |

$AC$ | $A(0,0)$ and $C(b,c)$ | $C(20+b ,20+c )$ | $C(2b ,2c )$ |

$BC$ | $B(a,0)$ and $C(b,c)$ | $E(2a+b ,20+c )$ | $E(2a+b ,2c )$ |

The $y-$coordinate of both $C(2b ,2c )$ and $E(2a+b ,2c )$ is $2c .$ Therefore, $DE$ is also a horizontal segment. Since all horizontal segments are parallel, it can be said that $AB$ and $DE$ are parallel. $AC∥DE✓ $

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 $B(a,0)$ | $AB=a−0$ | $AB=a$ |

$DE$ | $C(2b ,2c )$ and $E(2a+b ,2c )$ | $CE=2a+b −2b $ | $CE=21 a$ |

Since $21 a$ is half of $a,$ it can be stated that the midsegment $DE$ is half the length of $AB.$ $DE=21 AB✓ $ Therefore, a midsegment of a triangle is parallel to the third side of the triangle and half its length.

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
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.

Place the diagram on a coordinate plane.

First, the diagram will be placed on a coordinate plane.
### 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 $S(-2,-3),$ $P(-5,4),$ $A(3,5),$ and $T(6,-2),$ 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.
By following the same procedure, all side lengths can be found.

### 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 the diagonal $PT$ is $M_{1}(21 ,1).$ By following the same procedure, the midpoint of the diagonal $SA$ can be calculated.

It can be seen that the midpoint of each diagonal has coordinates $(21 ,1).$ Therefore, they intersect at their midpoint.

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.

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

SubstituteValues

Substitute values

$SP=(-5−(-2))_{2}+(4−(-3))_{2} $

$SP=58 $

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

Side | Endpoints | Substitute | Simplify |

$SP$ | $S(-2,-3)$ and $P(-5,4)$ | $SP=$ $(-5−(-2))_{2}+(4−(-3))_{2} $ | $SP=58 $ |

$PA$ | $P(-5,4)$ and $A(3,5)$ | $PA=$ $(3−(-5))_{2}+(5−4)_{2} $ | $PA=69 $ |

$AT$ | $A(3,5)$ and $T(6,-2)$ | $AT=$ $(6−3)_{2}+(-2−5)_{2} $ | $AT=58 $ |

$TS$ | $T(6,-2)$ and $S(-2,-3)$ | $TS=$ $(-2−6)_{2}+(-3−(-2))_{2} $ | $TS=69 $ |

By the Transitive Property of Equality, $SP$ and $AT$ are equal, and $PA$ and $TS$ are also equal. ${SP=58 AT=58 ⇓SP=AT {PA=69 TS=69 ⇓PA=TS $ 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. $SP≅ATandPA≅TS✓ $ This confirms that the quadrilateral is a parallelogram.

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.

$M_{1}(2x_{1}+x_{2} ,2y_{1}+y_{2} )$

SubstitutePoints

Substitute $(-5,4)$ & $(6,-2)$

$M_{1}(2-5+6 ,24+(-2) )$

Evaluate

AddNeg

$a+(-b)=a−b$

$M_{1}(2-5+6 ,24−2 )$

AddSubTerms

Add and subtract terms

$M_{1}(21 ,22 )$

QuotOne

$aa =1$

$M_{1}(21 ,1)$

$M(2x_{1}+x_{2} ,2y_{1}+y_{2} )$ | |||
---|---|---|---|

Diagonal | Endpoints | Substitute | Simplify |

$PT$ | $P(-5,4)$ and $T(6,-2)$ | $M_{1}(2-5+6 ,24+(-2) )$ | $M_{1}(21 ,1)$ |

$SA$ | $S(-2,-3)$ and $A(3,5)$ | $M_{2}(2-2+3 ,2-3+5 )$ | $M_{2}(21 ,1)$ |

It can be stated that the diagonals bisect each other.
$PI≅ITandSI≅IA✓ $

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

The opposite sides of a parallelogram are congruent.

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

$PQ ≅SRandQR ≅PS$

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 $Q.$ $P(0,0)Q(b,c)S(a,0) $ 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.$m=x_{2}−x_{1}y_{2}−y_{1} $

SubstituteValues

Substitute values

$m=b−0c−0 $

SubTerms

Subtract terms

$m=bc $

$m=x_{2}−x_{1}y_{2}−y_{1} $ | |||
---|---|---|---|

Side | Endpoints | Substitute | Simplify |

$PQ $ | $P(0,0)$ and $Q(b,c)$ | $m_{PQ}=b−0c−0 $ | $m_{PQ}=bc $ |

$SR$ | $S(a,0)$ and $R(x,c)$ | $m_{SR}=x−ac−0 $ | $m_{SR}=x−ac $ |

$bc =x−ac $

Solve for $x$

MultEqn

$LHS⋅b=RHS⋅b$

$c=x−ac (b)$

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$c=x−ac(b) $

MultEqn

$LHS⋅(x−a)=RHS⋅(x−a)$

$c(x−a)=c(b)$

DivEqn

$LHS/c=RHS/c$

$x−a=b$

AddEqn

$LHS+a=RHS+a$

$x=b+a$

CommutativePropAdd

Commutative Property of Addition

$x=a+b$

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

SubstituteValues

Substitute values

$PQ=(b−0)_{2}+(c−0)_{2} $

SubTerms

Subtract terms

$PQ=b_{2}+c_{2} $

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

Side | Endpoints | Substitute | Simplify |

$PQ $ | $P(0,0)$ and $Q(b,c)$ | $PQ=$ $(b−0)_{2}+(c−0)_{2} $ | $PQ=b_{2}+c_{2} $ |

$QR $ | $Q(b,c)$ and $R(a+b,c)$ | $QR=$ $(a+b−b)_{2}+(c−c)_{2} $ | $QR=a$ |

$SR$ | $S(a,0)$ and $R(a+b,c)$ | $SR=$ $(a+b−a)_{2}+(c−0)_{2} $ | $SR=b_{2}+c_{2} $ |

$PS$ | $P(0,0)$ and $S(a,0)$ | $PS=$ $(a−0)_{2}+(0−0)_{2} $ | $PS=a$ |

By the Transitive Property of Equality, it can be said that $PQ=SR$ and that $QR=PS.$ ${PQ=b_{2}+c_{2} SR=b_{2}+c_{2} ⇓PQ=SR {QR=aPS=a ⇓QR=PS $ By definition of congruent segments, it can be stated that the opposite sides of a parallelogram are congruent.

$PQ ≅SRandQR ≅PS$

In a parallelogram, the diagonals bisect each other.

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

$PM≅RMandQM ≅SM$

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 $Q.$ $P(0,0)Q(b,c)S(a,0) $ 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.$M_{QS}(2x_{1}+x_{2} ,2y_{1}+y_{2} )$

SubstitutePoints

Substitute $(b,c)$ & $(a,0)$

$M_{QS}(2b+a ,2c+0 )$

Simplify

IdPropAdd

Identity Property of Addition

$M_{QS}(2b+a ,2c )$

CommutativePropAdd

Commutative Property of Addition

$M_{QS}(2a+b ,2c )$

$M(2x_{1}+x_{2} ,2y_{1}+y_{2} )$ | |||
---|---|---|---|

Diagonal | Endpoints | Substitute | Simplify |

$QS $ | $Q(b,c)$ and $S(a,0)$ | $M_{QS}(2b+a ,2c+0 )$ | $M_{QS}(2a+b ,2c )$ |

$PR$ | $P(0,0)$ and $R(a+b,c)$ | $M_{PR}(20+a+b ,20+c )$ | $M_{PR}(2a+b ,2c )$ |

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.

$PM≅RMandQM ≅SM$

Therefore, the diagonals of a parallelogram bisect each other.

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.

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

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.$m=x_{2}−x_{1}y_{2}−y_{1} $

SubstituteValues

Substitute values

$m_{AB}=-3−(-6)6−2 $

Evaluate right-hand side

$m_{AB}=34 $

$m=x_{2}−x_{1}y_{2}−y_{1} $ | |||
---|---|---|---|

Side | Endpoints | Substitute | Simplify |

$AB$ | $A(-6,2)$ and $B(-3,6)$ | $m_{AB}=-3−(-6)6−2 $ | $m_{AB}=34 $ |

$BC$ | $B(-3,6)$ and $C(9,-3)$ | $m_{BC}=9−(-3)-3−6 $ | $m_{BC}=-43 $ |

$DC$ | $D(6,-7)$ and $C(9,-3)$ | $m_{DC}=9−6-3−(-7) $ | $m_{DC}=34 $ |

$AD$ | $A(-6,2)$ and $D(6,-7)$ | $m_{AD}=6−(-6)-7−2 $ | $m_{AD}=-43 $ |

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 |
---|---|

$AB$ and $BC$ | $34 (-43 )=-1✓$ |

$BC$ and $DC$ | $-43 (34 )=-1✓$ |

$DC$ and $AD$ | $34 (-43 )=-1✓$ |

$AD$ and $AB$ | $-43 (34 )=-1✓$ |

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.

{{ 'mldesktop-placeholder-grade' | message }} {{ article.displayTitle }}!

{{ exercise.headTitle }}

{{ 'ml-heading-exercise' | message }} {{ focusmode.exercise.exerciseName }}