9. Using Coordinates in Proofs
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Chapter 5
9.
Using Coordinates in Proofs
Understanding how to use coordinates in proofs opens up a new dimension in geometry. The lesson delves into the application of geometric theorems, the distance formula, and the circle equation to solve various problems. For example, you can use coordinates to prove theorems about lines, angles, and shapes. The distance formula helps in calculating the length between two points, which is crucial in real-world applications like navigation and construction. The circle equation is particularly useful in astronomy and engineering where precise calculations are needed. Overall, mastering these concepts equips you with the tools to tackle a wide range of challenges in both academic and professional settings.
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Lesson Settings & Tools
| | 8 Theory slides |
| | 13 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Using Coordinates in Proofs
Slide of 8
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.
{"type":"multichoice","form":{"alts":["a \u22a5 b","a \u22a5 c","b \u22a5 c"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":[0]}
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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?
Example
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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 P(1,sqrt(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.
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Hint
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 = sqrt((x_2-x_1)^2 + (y_2-y_1)^2).
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.
- 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.
Finding the Equation of the Circle
The standard equation of a circle with center at ( h, k) and radius r can be expressed. (x- h)^2+(y- k)^2= r^2 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.d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
SubstituteValues
Substitute values
r=sqrt(( 0- 2)^2+( 0- 0)^2)
r=2
Verifying If the Coordinates of the Point Satisfy the Equation
Finally, to determine whether point P( 1,sqrt(3)) is on the circle, Ramsha will substitute x= 1 and y=sqrt(3) 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. A true statement was obtained. Therefore, Ramsha can conclude point P(1,sqrt(3)) 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.
- 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.
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.d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
SubstituteValues
Substitute values
r=sqrt(( 0- 2)^2+( 0- 0)^2)
r=2
Calculating the Distance From the Point to the Center of the Circle
If point P( 1,sqrt(3)) is on the circle, then its distance from O( 0, 0) must also be 2. Once again, the Distance Formula can be used.d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
SubstituteValues
Substitute values
2? =sqrt(( 0- 1)^2+( 0-sqrt(3))^2)
2=2 ✓
Example
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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.
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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.
The midpoint of Road 1 is M_1(0,2) and the midpoint of Road 2 is 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.
| Midpoint Formula: (x_1+x_2/2,y_1+y_2/2) | |||
|---|---|---|---|
| Road | Points | Substitute | Simplify |
| Road 1 | B( 4, 0) and M( - 4, 4) | M_1(4+( - 4)/2,0+ 4/2) | M_1(0,2) |
| Road 2 | M( - 4, 4) and P( - 2, - 2) | M_2( - 4+( - 2)/2,4+( - 2)/2 ) | M_2(- 3,1) |
| Distance Formula: d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) | |||
|---|---|---|---|
| Points | Substitute | Simplify | |
| M_1( 0, 2) and M_2( - 3, 1) | d_(M_1M_2)=sqrt(( - 3- 0)^2+( 1- 2)^2) | d_(M_1M_2)= sqrt(10) | |
| B( 4, 0) and P( - 2, - 2) | d_(BP)=sqrt(( - 2- 4)^2+( - 2- 0)^2) | d_(BP)=2sqrt(10) | |
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. sqrt(10)=1/2(2sqrt(10)) ⇓ d_(M_1M_2)=1/2d_(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 = y_2-y_1/x_2-x_1 | |||
|---|---|---|---|
| Points | Substitute | Simplify | |
| M_1( 0, 2) and M_2( - 3, 1) | m_(M_1M_2)=1- 2/- 3- 0 | m_(M_1M_2)= 1/3 | |
| B( 4, 0) and P( - 2, - 2) | m_(BP)=- 2- 0/- 2- 4 | m_(BP)=1/3 | |
It was found that the slope between M_1 and M_2 is the same as the slope between B and P. 1/3=1/3 ⇓ m_(M_1M_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.
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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.
DE ∥ BC and DE=1/2BC
Proof
Using CoordinatesThis theorem can be proven by placing the triangle on a coordinate plane. For simplicity, vertex B will be placed at the origin and vertex C on the x-axis.
Since B lies on the origin, its coordinates are (0,0). Point C is on the x-axis, meaning its y-coordinate is 0. The remaining coordinates are unknown and can be named a, b, and c. B(0,0) C(a,0) A(b,c) If DE is the midsegment from BA to CA, then by the definition of a midpoint, D and E are the midpoints of BA and CA, respectively.
To prove this theorem, it must be proven that DE is parallel to BC and that DE is half of BC.
BC ∥ DE
If the slopes of these two segments are equal, then they are parallel. The y-coordinate of both B(0, 0) and C(a, 0) is 0. Therefore, BC is a horizontal segment. Next, the coordinates of D and E will be found using the Midpoint Formula.
| M(x_1+x_2/2,y_1+y_2/2) | |||
|---|---|---|---|
| Segment | Endpoints | Substitute | Simplify |
| BA | B( 0, 0) and A( b,c) | D(0+ b/2,0+c/2) | D(b/2,c/2) |
| CA | C( a, 0) and A( b,c) | E(a+ b/2,0+c/2) | E(a+b/2,c/2) |
The y-coordinate of both D( b2,c2) and E( a+b2,c2) is c2. Therefore, DE is also a horizontal segment. Since all horizontal segments are parallel, it can be said that BC and DE are parallel. BC ∥ DE ✓
DE=1/2BC
Since both BC and DE are horizontal, their lengths are given by the difference of the x-coordinates of their endpoints.
| Segment | Endpoints | Length | Simplify |
|---|---|---|---|
| BC | B(0,0) and C(a,0) | BC=a-0 | BC=a |
| DE | D(b/2,c/2) and E(a+b/2,c/2) | DE=a+b/2-b/2 | DE=1/2a |
Since 12a is half of a, it can be stated that the midsegment DE is half the length of BC. DE=1/2BC ✓ Therefore, a midsegment of two sides of a triangle is parallel to the third side of the triangle and half its length.
Example
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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.
Answer
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.
By following the same procedure, all side lengths can be found.
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.
The midpoint of the diagonal PT is M_1( 12,1). 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 ( 12,1). Therefore, they intersect at their midpoint.
It can be stated that the diagonals bisect each other.
PI≅ IT and SI≅ IA ✓
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.d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
SubstituteValues
Substitute values
SP=sqrt(( -5-( - 2))^2+( 4-( -3))^2)
SP=sqrt(58)
| d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) | |||
|---|---|---|---|
| Side | Endpoints | Substitute | Simplify |
| SP | S( -2, -3) and P( - 5, 4) | SP= sqrt(( -5-( - 2))^2+( 4-( -3))^2) | SP=sqrt(58) |
| PA | P( - 5, 4) and A( 3, 5) | PA= sqrt(( 3-( -5))^2+( 5- 4)^2) | PA=sqrt(69) |
| AT | A( 3, 5) and T( 6, - 2) | AT= sqrt(( 6- 3)^2+( - 2- 5)^2) | AT=sqrt(58) |
| TS | T( 6, - 2) and S( -2, -3) | TS= sqrt(( -2- 6)^2+( -3-( -2))^2) | TS=sqrt(69) |
By the Transitive Property of Equality, SP and AT are equal, and PA and TS are also equal. cc SP=sqrt(58) AT=sqrt(58) & PA=sqrt(69) TS=sqrt(69) ⇓ & ⇓ SP=AT & 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≅ AT and PA≅ TS ✓ 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.M_1(x_1+x_2/2,y_1+y_2/2)
SubstitutePoints
Substitute ( -5,4) & ( 6,-2)
M_1(-5+ 6/2,4+( -2)/2)
▼
Evaluate
M_1(1/2,1)
| M(x_1+x_2/2,y_1+y_2/2) | |||
|---|---|---|---|
| Diagonal | Endpoints | Substitute | Simplify |
| PT | P( - 5, 4) and T( 6, - 2) | M_1(-5+ 6/2,4+( -2)/2) | M_1(1/2,1) |
| SA | S( -2, -3) and A( 3, 5) | M_2(-2+ 3/2,-3+ 5/2) | M_2(1/2,1) |
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Properties of Parallelograms
The results of the previous example can be generalized to any parallelogram.
Rule
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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.
PQ≅SR and QR≅PS
Proof
Using CoordinatesThis 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.
| m = y_2-y_1/x_2-x_1 | |||
|---|---|---|---|
| Side | Endpoints | Substitute | Simplify |
| PQ | P( 0, 0) and Q( b, c) | m_(PQ)=c- 0/b- 0 | m_(PQ)=c/b |
| SR | S( a, 0) and R( x, c) | m_(SR)=c- 0/x- a | m_(SR)=c/x-a |
c/b=c/x-a
▼
Solve for x
MultEqn
LHS * b=RHS* b
c=c/x-a(b)
MoveRightFacToNum
a/c* b = a* b/c
c=c(b)/x-a
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 = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
SubstituteValues
Substitute values
PQ=sqrt(( b- 0)^2+( c- 0)^2)
SubTerms
Subtract terms
PQ=sqrt(b^2+c^2)
| d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) | |||
|---|---|---|---|
| Side | Endpoints | Substitute | Simplify |
| PQ | P( 0, 0) and Q( b, c) | PQ= sqrt(( b- 0)^2+( c- 0)^2) | PQ=sqrt(b^2+c^2) |
| QR | Q( b, c) and R( a+b, c) | QR= sqrt(( a+b- b)^2+( c- c)^2) | QR=a |
| SR | S( a, 0) and R( a+b, c) | SR= sqrt(( a+b- a)^2+( c- 0)^2) | SR=sqrt(b^2+c^2) |
| PS | P( 0, 0) and S( a, 0) | PS= sqrt(( 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. cc PQ=sqrt(b^2+c^2) SR=sqrt(b^2+c^2) & QR=a PS=a ⇓ & ⇓ PQ=SR & QR=PS By definition of congruent segments, it can be stated that the opposite sides of a parallelogram are congruent.
PQ≅SR and QR≅PS
Rule
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Parallelogram Diagonals Theorem
In a parallelogram, the diagonals bisect each other.
If PQRS is a parallelogram, then the following statement holds true.
PM≅RM and QM≅SM
Proof
Using CoordinatesThis 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).
M_(QS)(x_1+x_2/2,y_1+y_2/2)
SubstitutePoints
Substitute ( b, c) & ( a, 0)
M_(QS)(b+ a/2,c+ 0/2)
▼
Simplify
IdPropAdd
Identity Property of Addition
M_(QS)(b+a/2,c/2)
CommutativePropAdd
Commutative Property of Addition
M_(QS)(a+b/2,c/2)
| M(x_1+x_2/2,y_1+y_2/2) | |||
|---|---|---|---|
| Diagonal | Endpoints | Substitute | Simplify |
| QS | Q( b, c) and S( a, 0) | M_(QS)(b+ a/2,c+ 0/2) | M_(QS)(a+b/2,c/2) |
| PR | P( 0, 0) and R( a+b, c) | M_(PR)(0+ a+b/2,0+ c/2) | M_(PR)(a+b/2,c/2) |
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≅RM and QM≅SM
Therefore, the diagonals of a parallelogram bisect each other.
Closure
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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.
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Hint
By the Slopes of Perpendicular Lines Theorem, if the slopes of two lines or segments are opposite reciprocals, then lines are perpendicular.
Solution
For simplicity, the vertices will be labeled A, B, C, and D.
m = y_2-y_1/x_2-x_1
SubstituteValues
Substitute values
m_(AB)=6- 2/-3-( -6)
▼
Evaluate right-hand side
m_(AB)=4/3
| m = y_2-y_1/x_2-x_1 | |||
|---|---|---|---|
| Side | Endpoints | Substitute | Simplify |
| AB | A( -6, 2) and B( -3, 6) | m_(AB)=6- 2/-3-( -6) | m_(AB)=4/3 |
| BC | B( -3, 6) and C( 9, -3) | m_(BC)=-3- 6/9-( -3) | m_(BC)=- 3/4 |
| DC | D( 6, -7) and C( 9, -3) | m_(DC)=-3-( -7)/9- 6 | m_(DC)=4/3 |
| AD | A( -6, 2) and D( 6, -7) | m_(AD)=-7- 2/6-( -6) | m_(AD)=- 3/4 |
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 | 4/3( - 3/4)=- 1 ✓ |
| BC and DC | - 3/4(4/3)=- 1 ✓ |
| DC and AD | 4/3(- 3/4)=- 1 ✓ |
| AD and AB | - 3/4( 4/3)=- 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.
Using Coordinates in Proofs
Exercises
What is the radius of the student's circle? Give an exact answer.
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The radius of a circle is the distance between the center and any point on the circle. Since we know the center of Zosia's circle and one point contained in it we can find the radius by substituting the given coordinates into the Distance Formula and simplify.
We have been given the center of Tearrik's circle and a point on its circumference. The radius is the distance between these points. Let's substitute the given coordinates into the Distance Formula and simplify.
For Tiffaniqua's circle we know the endpoints of a diameter. We will first calculate the length of the diameter. Then, to find the radius, we will divide the result by 2.
The radius is half the length of the diameter. r=d/2=13/2
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Find the point M where the diagonals intersect.
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Since the opposite sides in the quadrilateral are congruent we know that it is a parallelogram. Let's recall the Parallelogram Diagonals Theorem.
Parallelogram Diagonals Theorem |- In a parallelogram, the diagonals bisect each other.
This means that the point of intersection M of the diagonals also is the midpoint of the diagonals. Let's illustrate this in the diagram. Let's also identify the coordinates of the parallelogram's vertices.
We can find the midpoint by substituting the coordinates of two opposite vertices into the Midpoint Formula.
We have found the coordinates of the point of intersection of the diagonals. As an extra step for more clarity, we will now show that the midpoint of the other diagonal is the same point.
The midpoint has the coordinates (- 12,4).
Since we have a parallelogram we know that the diagonals intersection point M is also the midpoint of each diagonal.
Let's substitute the coordinates of the points A and C into the Midpoint Formula to find the coordinates of M.
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Find the coordinates of the circumcenter of a triangle with vertices in P(1,4), Q(- 3,4), and R(-3,- 2).
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Let's start by graphing the triangle using the given coordinates.
To find the circumcenter, we need to draw the perpendicular bisectors of at least two sides of the triangle. The point of intersection of the perpendicular bisectors is the circumcenter.
Finding Perpendicular Bisectors
A perpendicular bisector is a line that is perpendicular to a side at its midpoint. Let's find their midpoints. To do so, we can use the Midpoint Formula.
| Side | Points | M(x_1+x_2/2,y_1+y_2/2) | Evaluate |
|---|---|---|---|
| PQ | ( 1,4), ( - 3,4) | M(1+( -3)/2,4+ 4/2) | M(- 1,4) |
| QR | ( - 3,4), ( - 3,- 2) | M(-3+( -3)/2,4+( -2)/2) | M(-3,1) |
We know that horizontal and vertical lines are perpendicular. Since AB is horizontal, any perpendicular line will be vertical. Similarly, since BC is vertical, any perpendicular line will be horizontal. Let's add the midpoints and the perpendicular bisectors to our graph.
Finding the Circumcenter
The triangle's circumcenter is the point at which the perpendicular bisectors intersect.
We can see that the circumcenter is located at (- 1,1).
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For the triangle ABC, the side BC is 8 units long and DE has a length of 4 units.
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Let's recall the Triangle Midsegment Theorem.
Triangle Midsegment Theorem |- The line segment that connects the midpoints of two sides of a triangle is parallel to the third side of the triangle and half its length.
Notice that BC is parallel to and half as long as DE. DE=1/2BC and DE ∥ BC Using this we can tell that DE is a midsegment to the triangle and that D is the midpoint on AB. Let's identify the endpoints of this segment.
Let's use the Midpoint Formula to find the coordinates of D.
From Part A we know that DE is a midsegment to the triangle. Therefore, we know that E is the midpoint on AC. By substituting the coordinates of A and C into the Midpoint Formula we can find the coordinates of E.
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A quadrilateral ABCD has vertices in (- 2,- 1), (4,2), (3,4), and (- 3,1). Determine if it is a parallelogram, a rhombus, a rectangle, or a square.
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Let's plot the given points on a coordinate plane and graph the quadrilateral.
Next, we will review the definitions of the types of quadrilaterals.
| Quadrilateral | Definition |
|---|---|
| Parallelogram | Quadrilateral with two pairs of parallel sides |
| Rhombus | Parallelogram with four congruent sides |
| Rectangle | Parallelogram with four right angles |
| Square | Parallelogram with four congruent sides and four right angles |
To determine what kind of quadrilateral it is we have we need to find the slope and the length of each of its sides.
Finding Slopes
Now, let's find the slope of each side using the Slope Formula.
| Side | Points | Slope Formula | Simplified |
|---|---|---|---|
| AB | ( - 2,- 1), ( 4,2) | 2-( - 1)/4-( - 2) | 1/2 |
| BC | ( 4,2), ( 3, 4) | 4- 2/3- 4 | - 2 |
| CD | ( 3, 4), ( - 3, 1) | 1- 4/- 3- 3 | 1/2 |
| DA | ( - 3, 1), ( - 2,- 1) | - 1- 1/- 2-( - 3) | - 2 |
We can tell that adjacent sides are perpendicular, as their slopes are opposite reciprocals. - 2(1/2) = - 1 Therefore, our parallelogram can be either a rectangle or a square.
Finding Lengths
Now we will check the lengths of its sides using the Distance Formula.
| Side | Points | Distance Formula | Simplified |
|---|---|---|---|
| AB | ( - 2,- 1), ( 4,2) | sqrt(( 4-( - 2))^2+( 2-( - 1))^2) | 3sqrt(5) |
| BC | ( 4,2), ( 3, 4) | sqrt(( 3- 4)^2+( 4- 2)^2) | sqrt(5) |
| CD | ( 3, 4), ( - 3, 1) | sqrt(( - 3- 3)^2+( 1- 4)^2) | 3sqrt(5) |
| DA | ( - 3, 1), ( - 2,- 1) | sqrt(( - 2-( - 3))^2+( - 1- 1)^2) | sqrt(5) |
Our parallelogram has two pairs of congruent sides. Therefore, this parallelogram is a rectangle.
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The radius of a regular polygon is the distance from the center of the polygon to any vertex. Find the radius of the given polygon.
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Let's start by finding vertices that pass through lattice points.
Let's draw segments between opposite vertices.
These segments intersect in the center of the polygon. The radius is half the length of either of these segments.
Let's use the Distance Formula to find the distance from the vertices with coordinates ( 2,6) and ( 6,2).
The radius is half this length. r=d/2 ⇒ r= 4sqrt(2)/2=2sqrt(2)
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Consider the parallelogram.
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To find x we will first find the length of the segment. Then we will form an equation we can solve for x.
Finding the Length
Let's identify the endpoints of the segment.
By substituting these coordinates into the Distance Formula we can find the length of the parallelogram's side.
Solving the Equation
Now that we know the length of the segment we can write an equation for x. sqrt(6x-8)= sqrt(10) Let's solve this radical equation.
This process of solving radical equations can produce extraneous solutions, or solutions that do not actually satisfy the equation. Therefore, we need to verify our solution in the original equation.
As in part A, we first need to find the length of the parallelogram's side. Then we will write an equation.
Finding the Length
To find the length of the segment we need to know the coordinates of its endpoints.
Let's substitute these coordinates into the Distance Formula.
Solving the Equation
Let's equate the length of the segment with the given expression. sqrt(2y-1)= sqrt(17) To find y we will now solve this equation.
Let's check our result for extraneous solutions.
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Using Coordinates in Proofs
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