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
Exercise 3.1
The incenter of a triangle is the point of intersection of the triangle's angle bisectors. Find the coordinates of the incenter of the equilateral triangle ABC without using the Incenter Theorem. Answer in exact form.
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An equilateral triangle has three congruent sides. Let's find the side length for our triangle by substituting the coordinates for A and B into the Distance Formula.
Let's write this into the diagram. We will also extend the angle bisectors all the way to the opposite side.
Inside △ ABC we have six small triangles. These are right triangles and they are all congruent.
We have been given the coordinates of C and we see that the angle bisector of C is a vertical line. Therefore, the vertex C and the incenter will have the same x-coordinate. Furthermore, we notice that the y-coordinate of the incenter equals the length of one leg of the small triangles. Let's call this length h.
Next we need to find the height of the triangle, which equals the length of the angle bisector of C. Notice that the angle bisector intersects the opposite side at the point (4,0).
Notice that the height of △ ABC also can be written as the sum of the lengths of h and the hypotenuse of the small triangles. Furthermore, we notice that the longer leg of the small triangles is half the side of △ ABC. Therefore, it has a length of 62= 3 units.
Let's label the hypotenuse x. Using the Pythagorean Theorem we can write an equation for the sides of the right triangles. 3^2+ h^2= x^2 Recall that the height of △ ABC is the sum of h and the hypotenuse x. Using this and that we know that the height is 3sqrt(3), we can write a second equation. x+ h= 3sqrt(3) Together these equations form a system of equations. 3^2+ h^2= x^2 & (I) x+ h= 3sqrt(3) & (II) Let's use Equation (II) to rewrite Equation (I).
To find the y-coordinate of the incenter we only need to find the value of h. Therefore, we will now continue by solving Equation (I).
The coordinates of the incenter are (4,sqrt(3))
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Determine the coordinates of Q, in terms of b.
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We will find the coordinates of Q one at the time.
Finding the x-coordinate
Let's examine the given triangle.
In an equilateral triangle all sides are congruent. By substituting the coordinates of P and R into the Distance Formula we can write an expression for the side lengths.
Notice that R is on the positive side of the x-axis. Therefore, we know that b must be a positive number. With that in mind, we can further simplify the expression.
Since △ PQR is equilateral, to find the x-coordinate of vertex Q, we can draw the altitude to the triangle's base PR. Since PQ ≅ QR, by the Perpendicular Bisector Theorem, we know that the altitude cuts PR in two congruent segments.
As we can see, the x-coordinate of Q is x_Q= 2b.
Finding the y-coordinate
The y-coordinate, y_Q, is the same as the triangle's height h. Let's use the Pythagorean Theorem to find it.
The coordinates of Q are (2b,2bsqrt(3)).
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Using Coordinates in Proofs
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