Geo Chapter 8 Coordinate Proofs Part 2 Notes
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•CHAPTER 8 PART 2
•Coordinate Proofs
•8H Parallelogram
•8I Rectangle
•8J Rhombus
•8K Square
•8L Trapezoid
•8M Regents Practice
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8H Coordinate Geometry - Parallelogram
How to prove: - Parallelogram - Show the diagonals bisect each other (midpoint twice)
Examples:
1) Prove that quadrilateral ABCD is a parallelogram. A(−4,3) B (2,7) C (5,1) D (−1, −3) Parallelogram - Show the diagonals bisect each other (midpoint twice)
To prove that quadrilateral ABCD is a parallelogram, we need to show that its opposite sides are parallel. One way to demonstrate this is by showing that the midpoints of its diagonals are the same.
Let's find the midpoints of the diagonals AC and BD :
1. Midpoint of diagonal AC :
2. Midpoint of diagonal BD :
As we can see, both midpoints are (\frac{1}{2}, 2).
Since the midpoints of the diagonals are the same, ABCD is a parallelogram. This is because in a parallelogram, the diagonals bisect each other.
2) Prove that quadrilateral EFGH is a parallelogram. E(−9, 4) F (−2, 7) G (4, −2) H (−3, −5) Parallelogram - Show the diagonals bisect each other (midpoint twice)
To prove that quadrilateral EFGH is a parallelogram, we need to show that its opposite sides are parallel. One way to demonstrate this is by showing that the midpoints of its diagonals are the same.
Let's find the midpoints of the diagonals EG and FH:
1. Midpoint of diagonal EG:
2. Midpoint of diagonal FH:
As we can see, both midpoints are (-2.5,1).
Since the midpoints of the diagonals are the same, EFGH is a parallelogram. This is because in a parallelogram, the diagonals bisect each other.
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3) Determine if quadrilateral JKLM with coordinates J(−2, 3), K (1, 5), L (2, 9) and M (−1,9) is a parallelogram. Parallelogram - Show the diagonals bisect each other (midpoint twice)
To determine if quadrilateral JKLM is a parallelogram, we need to show that its opposite sides are parallel. One way to demonstrate this is by showing that the midpoints of its diagonals are the same.
Let's find the midpoints of the diagonals JL and KM :
1. Midpoint of diagonal JL :
2. Midpoint of diagonal KM :
As we can see, both midpoints are (0, 6) & ( 0,7)
Since the midpoints of the diagonals are NOT same, JKLM is NOT a parallelogram.
4) Prove that quadrilateral PQRS is a parallelogram. P(4,4) Q (6, 1) R (0, −3) S (−2, 0) Parallelogram - Show the diagonals bisect each other (midpoint twice)
To prove that quadrilateral PQRS is a parallelogram, we need to show that its opposite sides are parallel. One way to demonstrate this is by showing that the midpoints of its diagonals are the same.
Let's find the midpoints of the diagonals PR and QS :
1. Midpoint of diagonal PR :
2. Midpoint of diagonal QS :
As we can see, both midpoints are (2, \frac{1}{2}).
Since the midpoints of the diagonals are the same, PQRS is a parallelogram. This is because in a parallelogram, the diagonals bisect each other.
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5) Determine if quadrilateral ABCD with A ( 1 , 4 ) , B ( 7 , 0 ) , C ( 5 , − 3 ) and D ( − 1 , 1 ) is a parallelogram. Parallelogram - Show the diagonals bisect each other (midpoint twice)
To determine if quadrilateral ABCD is a parallelogram, we need to show that its opposite sides are parallel. One way to demonstrate this is by showing that the midpoints of its diagonals are the same.
Let's find the midpoints of the diagonals AC and BD:
1. Midpoint of diagonal AC:
2. Midpoint of diagonal BD:
As we can see, both midpoints are (3, \frac{1}{2}).
Since the midpoints of the diagonals are the same, ABCD is a parallelogram. This is because in a parallelogram, the diagonals bisect each other.
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8H Coordinate Geometry Parallelogram HW
1) Determine if quadrilateral JACK is a parallelogram. J (1, 10) A(−4, 0) C(7, 2) K(12, 9) Parallelogram - Show the diagonals bisect each other (midpoint twice)
To determine if quadrilateral JACK is a parallelogram, we need to show that its opposite sides are parallel. One way to demonstrate this is by showing that the midpoints of its diagonals are the same.
Let's find the midpoints of the diagonals JK and AC :
1. Midpoint of diagonal JK :
2. Midpoint of diagonal AC :
As we can see, the midpoints are not the same. Therefore, the diagonals of JACK do not bisect each other at the same point.
Since the midpoints of the diagonals are not the same, we cannot conclude that JACK is a parallelogram.
2) Determine if quadrilateral ABCD with A(1, 2), B (2, 5), C (5, 7) and D (4,4) is a parallelogram Parallelogram - Show the diagonals bisect each other (midpoint twice)
To determine if quadrilateral ABCD is a parallelogram, we need to show that its opposite sides are parallel. One way to demonstrate this is by showing that the midpoints of its diagonals are the same.
Let's find the midpoints of the diagonals AC and BD:
1. Midpoint of diagonal AC:
2. Midpoint of diagonal BD:
As we can see, both midpoints are (3, \frac{9}{2}).
Since the midpoints of the diagonals are the same, ABCD is a parallelogram. This is because in a parallelogram, the diagonals bisect each other.
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3) Prove that quadrilateral ALEX is a parallelogram. A (−2, 2) L(1, 4) E(2, 8) X(−1, 6) Parallelogram - Show the diagonals bisect each other (midpoint twice)
To prove that quadrilateral ALEX is a parallelogram, we need to show that its opposite sides are parallel. One way to demonstrate this is by showing that the midpoints of its diagonals are the same.
Let's find the midpoints of the diagonals AC and BX:
1. Midpoint of diagonal AC:
2. Midpoint of diagonal BX:
As we can see, both midpoints are (0, 5).
Since the midpoints of the diagonals are the same, ALEX is a parallelogram. This is because in a parallelogram, the diagonals bisect each other.
Review Questions:
4) In the accompanying diagram of parallelogram ABCD, m∠B = 5x, and m ∠ C = 2x + 12. Find the number of degrees in ∠D.
To find the measure of angle D in parallelogram ABCD , we need to use the properties of parallelograms.
In a parallelogram, consecutive angles are supplementary. This means that the sum of adjacent angles is 180° .
Given that m∠B = 5x, and m ∠ C = 2x + 12. , we have:
Substituting the given expressions for m\angle B and m\angle C , we get:
Simplify and solve for x :
Now that we have found the value of x , we can find the measure of angle D by substituting x = 24^\circ into the expression for m\angle B :
Therefore, the measure of angle D is 120^\circ .
5) In the diagram below of ∆ABC, BD is drawn to side AC. If M ∠ A = 35, M ∠ ABD = 25, and M ∠ C = 60, which type of triangle is ∆BCD?
1) equilateral (2) scalene (3) obtuse (4) right
Based on the given information, we can determine the type of triangle \triangle BCD by analyzing its angles:
1. \angle A = 35^\circ
2. \angle ABD = 25^\circ
3. \angle C = 60^\circ
To find \angle BCD , we'll use the fact that the sum of the angles in a triangle is 180^\circ :
Now, let's analyze the triangle \triangle BCD based on its angles:
- \angle BCD = 60^\circ
- \angle C = 60^\circ
Since two angles are equal, \triangle BCD is an isosceles triangle.
Therefore, the correct answer is (2) scalene.
6) What is the length of the line segment whose endpoints are (1,-4) and (9,2)?
To find the length of the line segment with endpoints (1, -4) and (9, 2) , we can use the distance formula:
If the endpoints of a line segment in two-dimensional Cartesian coordinates are (x_1, y_1) and (x_2, y_2) , then the distance d between these two points is given by:
Applying this formula to the given endpoints:
Let (x_1, y_1) = (1, -4) and (x_2, y_2) = (9, 2) .
So, the length of the line segment is 10 units.
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8I Coordinate Geometry Rectangle
How to prove: - Rectangle - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are congruent (distance twice)
Examples: 1) Prove that quadrilateral ABCD is a rectangle. A(4, 4) B (6, 1) C (0, −3) D (−2, 0) Rectangle - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are congruent (distance twice)
To prove that quadrilateral ABCD is a rectangle, we need to show that it satisfies the properties of a rectangle, which include having congruent diagonals that bisect each other.
Let's start by calculating the midpoints of the diagonals. The midpoints of the diagonals will coincide if the diagonals bisect each other.
The diagonals of a quadrilateral are formed by connecting opposite vertices. For quadrilateral ABCD, the diagonals are AC and BD.
1. Midpoint of diagonal AC:
So, the midpoint of diagonal AC is (2, \frac{1}{2}) .
2. Midpoint of diagonal BD:
So, the midpoint of diagonal BD is also (2, \frac{1}{2}) .
Since the midpoints of both diagonals are the same, we have shown that the diagonals bisect each other.
Next, we'll calculate the lengths of the diagonals to confirm that they are congruent.
1. Length of diagonal AC:
2. Length of diagonal BD:
Since AC = BD and the diagonals bisect each other, we have proven that quadrilateral ABCD is a rectangle.
2) Determine if quadrilateral PQRS with P(0, 2), Q(4, 8), R(7, 6) and S( 3, 0) is a rectangle Rectangle - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are congruent (distance twice)
To determine if quadrilateral PQRS is a rectangle, we need to check if it satisfies the properties of a rectangle, including having congruent diagonals that bisect each other.
1. Midpoint of diagonal PR:
So, the midpoint of diagonal PR is (3.5, 4) .
2. Midpoint of diagonal QS:
So, the midpoint of diagonal QS is also (3.5, 4) .
Since the midpoints of both diagonals are the same, we have shown that the diagonals bisect each other.
Next, we'll calculate the lengths of the diagonals to confirm that they are congruent.
1. Length of diagonal PR:
2. Length of diagonal QS:
Since PR = QS and the diagonals bisect each other, we have proven that quadrilateral PQRS is a rectangle.
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3) Determine if parallelogram WXYZ is a rectangle. W(1, 2) X(−4, 8) Y(−8,6) Z(−3, 0) Rectangle - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are congruent (distance twice)
To determine if parallelogram WXYZ is a rectangle, we need to check if it satisfies the properties of a rectangle, including having congruent diagonals that bisect each other.
1. Midpoint of diagonal WY:
So, the midpoint of diagonal WY is (-3.5, 4) .
2. Midpoint of diagonal XZ:
So, the midpoint of diagonal XZ is also (-3.5, 4) .
Since the midpoints of both diagonals are the same, we have shown that the diagonals bisect each other.
Next, we'll calculate the lengths of the diagonals to confirm that they are congruent.
1. Length of diagonal WY:
2. Length of diagonal XZ:
Since WY = XZ and the diagonals bisect each other, we have proven that parallelogram WXYZ is a rectangle.
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8I Coordinate Geometry – Rectangle HW
1) Determine if quadrilateral EFGH with E(1, 4), F(7,0), G(5, −3) and H( −1, 1) is a rectangle.
Rectangle - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are congruent (distance twice)
To determine if quadrilateral EFGH is a rectangle, we need to check if it satisfies the properties of a rectangle, including having congruent diagonals that bisect each other.
1. Midpoint of diagonal EG:
So, the midpoint of diagonal EG is (3, 0.5) .
2. Midpoint of diagonal FH:
So, the midpoint of diagonal FH is also (3, 0.5) .
Since the midpoints of both diagonals are the same, we have shown that the diagonals bisect each other.
Next, we'll calculate the lengths of the diagonals to confirm that they are congruent.
1. Length of diagonal EG:
2. Length of diagonal FH:
Since EG = FH and the diagonals bisect each other, we have proven that quadrilateral EFGH is a rectangle.
2) Prove that ABCD is a parallelogram. Is ABCD a rectangle? A (−5,6) B(6, 6) C(8, −3) D(−3, −3) Rectangle - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are congruent (distance twice)
To prove that quadrilateral ABCD is a parallelogram, we need to show that both pairs of opposite sides are parallel.
Given the coordinates:
A (−5,6)
B(6, 6)
C(8, −3)
D(−3, −3)
Let's find the slopes of each side to determine if they are parallel.
1. Slope of side AB:
2. Slope of side BC:
3. Slope of side CD:
4. Slope of side DA:
Now, let's check if opposite sides have the same slopes:
Since both pairs of opposite sides have the same slopes, we have shown that ABCD is a parallelogram.
Now, let's check if ABCD is a rectangle by verifying that its diagonals bisect each other and are congruent.
1. Midpoint of diagonal AC:
2. Midpoint of diagonal BD:
Both diagonals have the same midpoint \left(\frac{3}{2}, \frac{3}{2}\right) , so they bisect each other.
Next, let's calculate the lengths of the diagonals to verify if they are congruent.
1. Length of diagonal AC:
2. Length of diagonal BD:
Since AC \neq BD , the diagonals of ABCD are not congruent. Therefore, ABCD is not a rectangle.
In conclusion, we have proven that ABCD is a parallelogram, but it is not a rectangle.
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3) Prove that parallelogram WXYZ is a rectangle. W (−5, −1) X(−4, 3) Y(8, 0) Z(7, −4) Rectangle - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are congruent (distance twice)
To prove that parallelogram WXYZ is a rectangle, we need to show that it satisfies the properties of a rectangle:
1. Diagonals bisect each other.
2. Diagonals are congruent.
Given the coordinates:
W (−5, −1)
X(−4, 3)
Y(8, 0)
Z(7, −4)
Let's start by finding the midpoint of each diagonal to verify if they bisect each other.
Midpoint of diagonal WY:
Midpoint of diagonal XZ:
Both diagonals have the same midpoint \left(\frac{3}{2}, -\frac{1}{2}\right) , so they bisect each other.
Next, let's calculate the lengths of the diagonals to verify if they are congruent.
1. Length of diagonal WY:
2. Length of diagonal XZ:
Since WY = XZ , the diagonals of WXYZ are congruent.
Therefore, WXYZ satisfies both properties of a rectangle, and we have proven that it is a rectangle.
Review Questions:
4) As shown in the diagram of rectangle ABCD below, diagonals AC and BD intersect at E. If AE = X + 2 and BD = 4X − 16, then the length of AC is (1) 6 (2) 10 (3) 12 (4) 24
Given that AE = X + 2 and BD = 4X − 16 in rectangle ABCD, and the diagonals AC and BD intersect at E, we can use the properties of rectangles to find the length of AC.
In a rectangle, the diagonals are congruent and bisect each other. Therefore, AE = EC and BD = AC.
We are given that AE = X + 2 and BD = 4X − 16.
Since AE = EC, we have EC = X + 2.
Since BD = AC, we have AC = 4X − 16.
Since the diagonals of a rectangle are congruent, we can set AE equal to BD and solve for X:
X + 2 = 4X − 16
Subtracting X from both sides, we get:
2 = 3X - 16
Adding 16 to both sides, we get:
18 = 3X
Dividing both sides by 3, we find:
X = 6
Now that we have found the value of X, we can find the length of AC:
AC = 4X - 16
AC = 4(6) - 16
AC = 24 - 16
AC = 8
Therefore, the length of AC is 8. So, the correct answer is (1) 6.
5) Through a given point, P, on a plane, how many lines can be drawn that are perpendicular to that plane? (1) 1 (2) 2 (3) more than 2 (4) none
Through a given point, P, on a plane, infinitely many lines can be drawn that are perpendicular to that plane. So, the correct answer is (3) more than 2.
6) Given ∆ABC, construct the perpendicular bisector of side AB. [Leave all construction marks]
To construct the perpendicular bisector of side AB in triangle ABC, follow these steps:
1. Use a compass to mark arcs on both sides of segment AB.
2. Set the compass width to be greater than half the length of segment AB.
3. Place the compass point on point A and draw an arc that intersects the line containing AB. Label this point of intersection as X.
4. Repeat step 3, placing the compass point on point B, and label the point of intersection with the line containing AB as Y.
5. Draw a straight line connecting points X and Y. This line is the perpendicular bisector of side AB.
The construction marks should include the arcs drawn from points A and B to intersect the line containing AB, and the line drawn connecting points X and Y, representing the perpendicular bisector.
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8J Coordinate Geometry- Rhombus
How to prove: - Rhombus - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are perpendicular (slope twice)
Examples: 1) Prove that quadrilateral ABCD is a rhombus. A(−3, 2) B (−2, 6) C(2, 7) D (1, 3)
Rhombus - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are perpendicular (slope twice)
To prove that quadrilateral ABCD is a rhombus, we need to demonstrate that it satisfies the properties of a rhombus:
1. Diagonals bisect each other (midpoint twice).
2. Diagonals are perpendicular (slope twice).
Let's examine each property:
1. Diagonals bisect each other:
We need to find the midpoints of the diagonals AC and BD and check if they coincide.
For diagonal AC: Midpoint of AC = ( (-3 + 2) / 2, (2 + 7) / 2 ) = (-0.5, 4.5)
For diagonal BD: Midpoint of BD = ( (-2 + 1) / 2, (6 + 3) / 2 ) = (-1.5, 4.5)
The midpoints of both diagonals are the same, so the diagonals bisect each other.
2. Diagonals are perpendicular:
We need to find the slopes of the diagonals AC and BD and check if they are negative reciprocals of each other.
For diagonal AC: Slope of AC = (7 - 2) / (2 - (-3)) = 5 / 5 = 1
For diagonal BD: Slope of BD = (3 - 6) / (1 - (-2)) = -3 / 3 = -1
Since the slope of AC is 1 and the slope of BD is -1, they are perpendicular.
Since quadrilateral ABCD satisfies both properties of a rhombus, we can conclude that it is indeed a rhombus.
2) Determine if quadrilateral PQRS with P(3, 2), Q(7, 0), R (11, 2) and S(7, 4) is a rhombus
- Rhombus - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are perpendicular (slope twice)
To determine if quadrilateral PQRS is a rhombus, we need to verify the following properties:
1. Diagonals bisect each other (midpoint twice).
2. Diagonals are perpendicular (slope twice).
Let's examine each property:
1. Diagonals bisect each other:
We need to find the midpoints of the diagonals PR and QS and check if they coincide.
For diagonal PR: Midpoint of PR = ( (3 + 11) / 2, (2 + 2) / 2 ) = (7, 2)
For diagonal QS: Midpoint of QS = ( (7 + 7) / 2, (0 + 4) / 2 ) = (7, 2)
The midpoints of both diagonals are the same, so the diagonals bisect each other.
2. Diagonals are perpendicular:
We need to find the slopes of the diagonals PR and QS and check if they are negative reciprocals of each other.
For diagonal PR: Slope of PR = (2 - 2) / (11 - 3) = 0 / 8 = 0
For diagonal QS: Slope of QS = (4 - 0) / (7 - 7) = 4 / 0 (undefined)
Since the slope of QS is undefined, it cannot be compared to the slope of PR to determine perpendicularity.
Since we couldn't verify the second property, we cannot conclude whether quadrilateral PQRS is a rhombus based on the given information. We would need additional information to determine the perpendicularity of the diagonals.
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3) Prove that parallelogram DEFG with D (2, 2), E(5, −2) F (9,1) and G(6,5) is a rhombus.
- Rhombus - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are perpendicular (slope twice)
To prove that parallelogram DEFG is a rhombus, we need to show that it satisfies the following properties:
1. Diagonals bisect each other.
2. Diagonals are perpendicular.
Let's start by examining each property:
1. Diagonals bisect each other:
We need to find the midpoints of the diagonals DE and FG and check if they coincide.
For diagonal DE: Midpoint of DE = ( (2 + 9) / 2, (2 + 1) / 2 ) = (11/2, 3/2)
For diagonal FG: Midpoint of FG = ( (5 + 6) / 2, (-2 + 5) / 2 ) = (11/2, 3/2)
The midpoints of both diagonals are the same, so the diagonals bisect each other.
2. Diagonals are perpendicular:
We need to find the slopes of the diagonals DE and FG and check if they are negative reciprocals of each other.
For diagonal DE: Slope of DE = (2 - (-2)) / (2 - 5) = 4 / (-3) = -4/3
For diagonal FG: Slope of FG = (5 - 1) / (6 - 9) = 4 / (-3) = -4/3
The slopes of the diagonals DE and FG are equal, which means they are parallel. Since the product of the slopes of parallel lines is -1, the diagonals are perpendicular.
Since both properties are satisfied, we can conclude that parallelogram DEFG is indeed a rhombus.
4) Determine if parallelogram JACK with J(3, 1), A(3, −3) , C (−2, −3) and K (−2, 1) is a rhombus.
- Rhombus - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are perpendicular (slope twice)
To determine if parallelogram JACK is a rhombus, we need to check if it satisfies the properties of a rhombus:
1. Diagonals bisect each other.
2. Diagonals are perpendicular.
Let's start by finding the midpoint of the diagonals JA and KC:
1. Diagonals bisect each other:
Midpoint of JA = ((3 - 2) / 2, (1 - 3) / 2) = (1/2, -1)
Midpoint of KC = ((-2 + (-2)) / 2, (-3 + 1) / 2) = (-2, -1)
Since the midpoints are different, the diagonals JA and KC do not bisect each other.
2. Diagonals are perpendicular:
Let's calculate the slopes of the diagonals JA and KC:
Slope of JA = (-3 - 1) / (3 - 3) = -4 / 0 (undefined)
Slope of KC = (1 + 1) / (-2 + 2) = 2 / 0 (undefined)
Since both slopes are undefined, we cannot determine if the diagonals are perpendicular.
Based on the calculations, we cannot conclude that parallelogram JACK is a rhombus because it does not meet the criteria for the diagonals to bisect each other or be perpendicular.
prove that quadrilateral MATH with M (1, 1), A(−2, 5) T(3, 5) and H(6, 1) is a rhombus
Prove that quadrilateral MATH with M (1, 1), A(−2, 5) T(3, 5) and H(6, 1) is a rhombus
•slope m = "y2 -y1" /"x2 -x1"
•Mid point Formula
M = ("x2 + x1" /"2" , "y2 +y 1" /"2" )
•M(1,1), T(3, 5)
•
M = ((1+3)/"2" , "1+5" /"2" )
M = (4/"2" , "6" /"2" )
M = (2, 3)
• Slope of MT = (5-1)/(3-1)= 4/2 =2
•A(−2, 5) H(6, 1)
M = ("-2 +6" /"2" , "1+5" /"2" )
M = (4/"2" , "6" /"2" )
M = (2, 3)
• Slope of AH = (1-5)/(6+2) = -" -4" /"8" = (-1)/"2"
•
•The diagonals bisect each other, Hence MATH is a Parallelogram.
•Slope of MT . Slope of AH = -1,So the diagonals are perpendicular. Hence MATH is a Rhombus.
••D = √(〖"(x2" -"x1" )〗^2+〖"(y2" -"y1" )〗^2 ) slope m = "y2 -y1" /"x2 -x1"
•M(1,1), A(-2,5),
•MA=√(〖"(-2" -"1" )〗^2+〖"(5" -"1" )〗^2 ) =√(〖"(-3" )〗^2+〖"(4" )〗^2 ) = √(9+16) = √25 =5
• Slope of MA = (5-1)/(-2-1)= (-4)/3
•A(−2, 5) T(3, 5)
•AT=√(〖"(" 3+2)〗^2+〖"(5" -"5" )〗^2 ) =√("5" ^2 ) = √(9+16) = √25 =5
• Slope of AT = (5-5)/(3+2) = 0
•T(3, 5) and H(6, 1)
•TH=√(〖"(" 6-3)〗^2+〖"(1" -5)〗^2 ) =√(〖"(3" )〗^2+〖"(-4" )〗^2 ) = √(9+16) = √25 =5
• Slope of AT = (1-5)/(6-3)= (-4)/3
•H(6, 1) and M (1, 1)
•HM=√(〖"(1" -"6" )〗^2+〖"(" 1-"1" )〗^2 ) =√("5" ^2 ) = √(9+16) = √25 =5
•Slope of HM = (1-1)/(1-6) = 0
•MA = AT = TH = HM =5 units
•Slope of MA = Slope of TH = (-4)/3
•Slope of AT = Slope of HM = 0
•All 4 sides have equal length, opposite sides parallel
adjacent sides are not perpendicular(slopes are NOT negative reciprocals)
Rhombus but not a Square.
Determine if parallelogram HOPE with H(1, 2), O(2, 5) P(5, 7) and E (4, 4) is a rhombus
•slope m = "y2 -y1" /"x2 -x1"
•Mid point Formula
M = ("x2 + x1" /"2" , "y2 +y 1" /"2" )
•H(1, 2), P(5, 7)
•
M = ((1+5)/"2" , ("7+" 2)/"2" )
M = (6/"2" , "9" /"2" )
M = (3, 4.5)
• Slope of HP = (7-2)/(5-1)= 5/4
•O(2, 5) E (4, 4)
M = (("2 +" 4)/"2" , ("5+" 4)/"2" )
M = (6/"2" , "9" /"2" )
M = (3, 4.5)
• Slope of OE = (4-5)/(4-2) = -(" -" 1)/2
•
•The diagonals bisect each other, Hence HOPE is a Parallelogram.
•Slope of MT . Slope of AH IS NOT EQUAL TO -1,So the diagonals are NOT perpendicular. Hence HOPE is NOT a Rhombus.
ANOTHER METHOD
•
Determine if parallelogram MIKE with M (5, 2) I(1, 9) K(−3, 2) and E (1, −5) is a rhombus
•slope m = "y2 -y1" /"x2 -x1"
•Mid point Formula
M = ("x2 + x1" /"2" , "y2 +y 1" /"2" )
•M (5, 2), K(−3, 2)
•
M = ((-3+5)/"2" , (2+2)/"2" )
M = (2/"2" , "4" /"2" )
M = (1,2)
• Slope of MK = (2-2)/(-3-5)=0
•I(1, 9) & E (1, −5)
M = (("1 +" 1)/"2" , ("-5+" 9)/"2" )
M = (2/"2" , "4" /"2" )
M = (1, 2)
• Slope of IE = (-5-9)/(1-1) = 0
•
•The diagonals bisect each other, Hence MIKE is a Parallelogram.
•Slope of MK . Slope of IE = -1,So the diagonals are perpendicular. Hence MIKE is a Rhombus.
•ANOTHER METHOD
Review Questions:
4) The vertices of triangle ABC are A(-8,1), B(-6,2), and C(-1,2).
a) State the coordinates of A’B’C’, the image of ABC, after a reflection in the line y = x.
b) State the coordinates of A”B”C”, the image of A’B’C’, after T-2,3.
c) State the coordinate of A’’’B’’’C’’’, the image of A”B”C”, after D2.
•a) To find the coordinates of the image of triangle ABC after a reflection in the line y = x, we switch the x-coordinates with the y-coordinates of each vertex.
•For vertex A(-8, 1): The image of A, A', will have coordinates (1, -8).
•For vertex B(-6, 2): The image of B, B', will have coordinates (2, -6).
•For vertex C(-1, 2):The image of C, C', will have coordinates (2, -1).
•So, the coordinates of A'B'C' are (1, -8), (2, -6), and (2, -1).
•b) To find the coordinates of the image of A'B'C' after a translation by T(-2, 3), we add -2 to the x-coordinates and 3 to the y-coordinates.
•For A'(1, -8): A" will have coordinates (1 - 2, -8 + 3) = (-1, -5).
•For B'(2, -6): B" will have coordinates (2 - 2, -6 + 3) = (0, -3).
•For C'(2, -1): C" will have coordinates (2 - 2, -1 + 3) = (0, 2).
•So, the coordinates of A"B"C" are (-1, -5), (0, -3), and (0, 2).
•c) To find the coordinates of the image of A"B"C" after a dilation by a scale factor of 2 with respect to the point D(2), we multiply the distances from D by 2.
•For A"(-1, -5):A"' will have coordinates (2 + 2(-1 - 2), 3 + 2(-5 - 3)) = (2 - 6, 3 - 16) = (-4, -13).
•For B"(0, -3):B"' will have coordinates (2 + 2(0 - 2), 3 + 2(-3 - 3)) = (2 + 4, 3 - 12) = (6, -9).
•For C"(0, 2):C"' will have coordinates (2 + 2(0 - 2), 3 + 2(2 - 3)) = (2 + 4, 3 - 2) = (6, 1).
•So, the coordinates of A"'B"'C"' are (-4, -13), (6, -9), and (6, 1).
5) What is the solution of the system of equations Y = X² + 2X − 3 and Y = 2X + 1?
(1) (0,-3)
(2) (-1,-4)
(3) (-3,0) and (1,0)
(4) (-2,-3) and (2,5)
•Since both equations represent Y as a function of X, we can set them equal to each other:
•X^2 + 2X - 3 = 2X + 1
•X^2 + 2X - 3 - 2X - 1 = 0
•X^2 - 4 = 0
• (X - 2)(X + 2) = 0
•X - 2 = 0 X = 2
•X + 2 = 0 X = -2
•Now, we can plug these values of X into either equation to find the corresponding values of Y. Let's use X = 2 for simplicity:
•Y = 2(2) + 1 = 5
•So, one solution is (2, 5).
•Now, let's use X = -2:
•Y = 2(-2) + 1 = -3
•So, another solution is (-2, -3).
•Therefore, the solution of the system of equations is (4) (-2, -3) and (2, 5).
6) As shown in the diagram below, when hexagon ABCDEF is reflected over line m, the image is hexagon A’B’C’D’E’F’. Under this transformation, which property is not preserved? (1) area (2) distance (3) orientation (4) angle measure
When a figure is reflected over a line, several properties are preserved, but one that is not preserved is orientation.
Orientation refers to the arrangement of the vertices or the direction in which they are ordered. In a reflection, the arrangement of the vertices may change because they are flipped across the reflecting line. However, the distances between the vertices and the angles formed by the sides are preserved.
Therefore, the property that is not preserved under a reflection is (3) orientation.
7) What is the slope of a line parallel to 6Y − 12X = 24?
To find the slope of a line parallel to the given line 6Y - 12X = 24, we first need to rewrite the equation in slope-intercept form (Y = MX + B), where M is the slope.
Starting with 6Y - 12X = 24, let's isolate Y:
6Y = 12X + 24
Y = 2X + 4
Now that the equation is in slope-intercept form (Y = 2X + 4), we can see that the slope of the line is M = 2.
Parallel lines have the same slope, so any line parallel to 6Y - 12X = 24 will also have a slope of 2.
Therefore, the slope of a line parallel to 6Y - 12X = 24 is 2.
Page 15 of 30
8K Coordinate Geometry - Square
How to prove: - Square - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are congruent (distance twice) (Rectangle) - Show that the diagonals are perpendicular (slope twice)
Examples: 1) Prove that quadrilateral ABCD is a square. A(−1,1 ) B (3, 3) C (5, −1) D (1, −3)
To prove that quadrilateral ABCD is a square, we need to show that it satisfies the properties of a square:
1. All sides are congruent. 2. Opposite sides are parallel. 3. All angles are right angles.
1. Length of side AB:
2. Length of side BC:
3. Length of side CD:
4. Length of side DA:
Now, we see that all sides AB, BC, CD, and DA are of equal length, which satisfies property 1 of a square.
Next, let's check if opposite sides are parallel. By calculating the slopes of AB and CD, and BC and DA:
1. Slope of AB:
2. Slope of CD:
3. Slope of BC:
4. Slope of DA:
The slopes of opposite sides AB and CD are equal (\frac{1}{2}), and the slopes of opposite sides BC and DA are equal (-2). Therefore, opposite sides are parallel, which satisfies property 2 of a square.
Lastly, we need to check if all angles are right angles. We can check this by calculating the slopes of the adjacent sides and verifying that they are negative reciprocals of each other:
1. Slope of AB:
2. Slope of BC:
3. Slope of CD:
4. Slope of DA:
The slopes of AB and BC are not negative reciprocals of each other, but the slopes of BC and CD, as well as the slopes of CD and DA, are negative reciprocals of each other. This implies that BC and CD, as well as CD and DA, form right angles at the vertices C and D, respectively.
Therefore, quadrilateral ABCD satisfies all the properties of a square, and hence, quadrilateral ABCD is a square.
ANOTHER METHOD
To prove that quadrilateral ABCD is a square, we need to show that it satisfies the properties of a square:
1. The diagonals bisect each other. 2. The diagonals are congruent. 3. The diagonals are perpendicular.
Let's start by calculating the midpoints of the diagonals AC and BD:
Midpoint of diagonal AC:
Midpoint of diagonal BD:
Since both midpoints are (2, 0), the diagonals bisect each other.
Next, let's calculate the lengths of the diagonals AC and BD:
Length of diagonal AC:
Length of diagonal BD:
Since AC = BD = 2\sqrt{10} , the diagonals are congruent.
Now, let's calculate the slopes of the diagonals AC and BD:
Slope of diagonal AC:
Slope of diagonal BD:
Since the product of the slopes of the diagonals is (-\frac{1}{3}) \times 3 = -1 , the diagonals are perpendicular.
Therefore, since quadrilateral ABCD satisfies all the properties of a square, we can conclude that it is indeed a square.
Page 16 of 30
2) Prove that quadrilateral TJLK is a square. T(−1, 0 ) J(3, 3) L (6, −1) K (2, −4)
Square - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are congruent (distance twice) (Rectangle) - Show that the diagonals are perpendicular (slope twice)
To prove that quadrilateral TJLK is a square, we need to demonstrate that it satisfies the properties of a square, namely:
1. The diagonals bisect each other.2. The diagonals are congruent.3. The diagonals are perpendicular.
Let's start by calculating the coordinates of the midpoints of the diagonals and the distances between the opposite midpoints:
Midpoint of diagonal TK:
Midpoint of diagonal JL:
Now, let's calculate the distances between the opposite midpoints:
Distance between M_1 and M_2:
Now, let's calculate the slopes of the diagonals:
Slope of diagonal TK (T to K):
Slope of diagonal JL (J to L):
Since both diagonals have the same slope, they are parallel.
Now, let's calculate the distances between the endpoints of the diagonals:
Distance between T and K:
Distance between J and L:
Since both distances are equal, the diagonals are congruent.
Finally, let's calculate the slopes of the diagonals:
Slope of diagonal TK:
Slope of diagonal JL:
The product of the slopes of the diagonals is (-4/3) \times (-4/3) = 16/9, which is not equal to -1. Therefore, the diagonals are not perpendicular.
Since TJLK satisfies only two out of the three properties of a square (the diagonals bisect each other and are congruent), we cannot conclude that TJLK is a square. It may be a rhombus or a rectangle, but it is not necessarily a square.
Page 17 of 30
3) Determine if rhombus KATE with K(−2, −1) A(−1, 3) T(3, 2) and E(2, −2) is a square.
Square - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are congruent (distance twice) (Rectangle) - Show that the diagonals are perpendicular (slope twice)
To determine if rhombus KATE is a square, we need to verify if it satisfies the properties of a square:
1. The diagonals bisect each other. 2. The diagonals are congruent.3. The diagonals are perpendicular.
Let's start by calculating the midpoints of the diagonals KT and AE:
Midpoint of diagonal KT:
Midpoint of diagonal AE:
Since both midpoints are \left(\frac{1}{2}, \frac{1}{2}\right), the diagonals bisect each other.
Next, let's calculate the lengths of the diagonals KT and AE:
Length of diagonal KT:
Length of diagonal AE:
Since KT = AE = \sqrt{34} , the diagonals are congruent.
Now, let's calculate the slopes of the diagonals KT and AE:
Slope of diagonal KT:
Slope of diagonal AE:
Since the product of the slopes of the diagonals is \left(\frac{3}{5}\right) \times \left(-\frac{5}{3}\right) = -1 , the diagonals are perpendicular.
Therefore, since rhombus KATE satisfies all the properties of a square, we can conclude that it is indeed a square.
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4) Prove that parallelogram MATH is a square. M(5, 0) A(2, 4) T(−2, 1) H(1, −3)
Square - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are congruent (distance twice) (Rectangle) - Show that the diagonals are perpendicular (slope twice)
To prove that parallelogram MATH is a square, we need to show that it satisfies the properties of a square:
1. Opposite sides are parallel.2. Opposite sides are congruent.3. Diagonals are congruent.4. Diagonals bisect each other.5. All angles are right angles.
Let's start by calculating the slopes of the sides MA, AT, TH, and HM to verify that opposite sides are parallel.
Slope of side MA:
Slope of side AT:
Slope of side TH:
Slope of side HM:
Since the slopes of MA and TH are equal (-\frac{4}{3}), and the slopes of AT and HM are equal (\frac{3}{4}), opposite sides are parallel.
Next, let's calculate the lengths of the sides to verify that opposite sides are congruent.
Length of side MA:
Length of side AT:
Length of side TH:
Length of side HM:
Since MA = AT = TH = HM = 5 , opposite sides are congruent.
Now, let's calculate the lengths of the diagonals to verify that they are congruent.
Length of diagonal MT:
Length of diagonal AH:
Since MT = AH = 5 , the diagonals are congruent.
Finally, let's calculate the midpoint of the diagonals to verify that they bisect each other.
Midpoint of diagonal MT:
Midpoint of diagonal AH:
Since the midpoints of the diagonals are \left(\frac{3}{2}, \frac{1}{2}\right) and (3, -\frac{3}{2}), they are not the same point, indicating that the diagonals bisect each other.
Therefore, since parallelogram MATH satisfies all the properties of a square, we can conclude that it is indeed a square.
Page 19 of 30
8K Coordinate Geometry – Square HW
1) Determine if rectangle BETH with B(−10, 0 ), E(10, 0), T (10, 7) and H (−10, 7) is a square.
Square - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are congruent (distance twice) (Rectangle) - Show that the diagonals are perpendicular (slope twice)
To determine if rectangle BETH is a square, we need to verify if it satisfies the properties of a square:
1. The diagonals bisect each other.2. The diagonals are congruent.3. The diagonals are perpendicular.
Let's start by calculating the midpoints of the diagonals BT and EH:
Midpoint of diagonal BT:
Midpoint of diagonal EH:
Since both midpoints are (10, \frac{7}{2}) and (-10, \frac{7}{2}), the diagonals bisect each other.
Next, let's calculate the lengths of the diagonals BT and EH:
Length of diagonal BT:
Length of diagonal EH:
Since BT = 7 and EH = \sqrt{449} , the diagonals are not congruent.
Now, let's calculate the slopes of the diagonals BT and EH:
Slope of diagonal BT:
Slope of diagonal EH:
Since one of the slopes is undefined and the other is a fraction, the diagonals are not perpendicular.
Therefore, since rectangle BETH does not satisfy all the properties of a square, we can conclude that it is not a square.
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2) Prove that quadrilateral DAVE is a square. D(−2, −1) A(2,2) V(5, −2) E(1, −5)
Square - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are congruent (distance twice) (Rectangle) - Show that the diagonals are perpendicular (slope twice)
To prove that quadrilateral DAVE is a square, we need to demonstrate that it satisfies the properties of a square:
1. The diagonals bisect each other.2. The diagonals are congruent.3. The diagonals are perpendicular.
Let's start by calculating the midpoints of the diagonals DV and AE:
Midpoint of diagonal DV:
Midpoint of diagonal AE:
Since both midpoints are \left(\frac{3}{2}, \frac{-3}{2}\right), the diagonals bisect each other.
Next, let's calculate the lengths of the diagonals DV and AE:
Length of diagonal DV:
Length of diagonal AE:
Since DV = AE = \sqrt{50} , the diagonals are congruent.
Now, let's calculate the slopes of the diagonals DV and AE:
Slope of diagonal DV:
Slope of diagonal AE:
Since the product of the slopes of the diagonals is \left(\frac{-1}{7}\right) \times 7 = -1 , the diagonals are perpendicular.
Therefore, since quadrilateral DAVE satisfies all the properties of a square, we can conclude that it is indeed a square.
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3) Determine if rectangle ABCD with A (−4, −1) B(−2, −5), C(4, −2) and Da(2, 2) is a square
Square - Show the diagonals bisect each other (midpoint twice) (Parallelogram) - Show that the diagonals are congruent (distance twice) (Rectangle) - Show that the diagonals are perpendicular (slope twice).
To determine if rectangle ABCD is a square, we need to check if it satisfies the properties of a square:
1. The diagonals bisect each other.2. The diagonals are congruent.3. The diagonals are perpendicular.
Let's start by calculating the midpoints of the diagonals AC and BD:
Midpoint of diagonal AC:
Midpoint of diagonal BD:
Since both midpoints are (0, -\frac{3}{2}), the diagonals bisect each other.
Next, let's calculate the lengths of the diagonals AC and BD:
Length of diagonal AC:
Length of diagonal BD:
Since AC = \sqrt{65} and BD = 7 , the diagonals are not congruent.
Now, let's calculate the slopes of the diagonals AC and BD:
Slope of diagonal AC:
Slope of diagonal BD:
Since one of the slopes is undefined and the other is a fraction, the diagonals are not perpendicular.
Therefore, since rectangle ABCD does not satisfy all the properties of a square, we can conclude that it is not a square.
Review Questions:
4) The lines represented by the equations 4X + 6Y = 6 and Y = 2/3 X − 1 are (1) parallel (2) the same line (3) perpendicular (4) intersecting but not perpendicular
To determine the relationship between the two lines represented by the equations 4X + 6Y = 6 and Y = \frac{2}{3}X - 1, let's analyze their slopes.
The first equation 4X + 6Y = 6 can be rearranged to slope-intercept form, Y = MX + B, where M is the slope:
So, the slope of the first line is -\frac{2}{3}.
The second equation Y = \frac{2}{3}X - 1 is already in slope-intercept form. Its slope is \frac{2}{3}.
Since the slopes of the two lines are negative reciprocals of each other (-\frac{2}{3} and \frac{2}{3}), the lines are perpendicular to each other.
Therefore, the correct answer is:(3) perpendicular
5) Determine and state the measure, in degrees, of an interior angle of a regular decagon
To determine the measure of an interior angle of a regular decagon, we can use the formula:
Where n is the number of sides of the polygon.
For a regular decagon (a polygon with 10 sides), we substitute n = 10 into the formula:
So, an interior angle of a regular decagon measures 144^\circ.
6) The sides of a right triangle are 3, 4, and 5. Could these sides represent the sides of the right triangle? Explain.
Yes, the sides of lengths 3, 4, and 5 could represent the sides of a right triangle. This is because these lengths satisfy the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.
In this case, if we square the lengths of the sides:
- 3^2 = 9
- 4^2 = 16
- 5^2 = 25
And then check if the sum of the squares of the two shorter sides equals the square of the longest side:
Since 9 + 16 = 25, the Pythagorean theorem is satisfied, and the lengths 3, 4, and 5 could represent the sides of a right triangle. Therefore, a triangle with side lengths 3, 4, and 5 forms a right triangle.
7) Write an equation of a line that is perpendicular to 2X + 4Y = −10 and passes through the point (-6,10)
To find the equation of a line that is perpendicular to 2X + 4Y = -10 and passes through the point (-6, 10), we first need to rewrite the given equation in slope-intercept form (Y = MX + B), where M is the slope.
Starting with 2X + 4Y = -10, let's isolate Y:
Now, the slope of the given line is -\frac{1}{2}.
Since we want to find a line perpendicular to this one, we need to find the negative reciprocal of -\frac{1}{2} to get the slope of the perpendicular line.
The negative reciprocal of -\frac{1}{2} is \frac{2}{1} = 2.
Now, we have the slope (M = 2) of the perpendicular line and a point (-6, 10) through which it passes.
Using the point-slope form of a linear equation, which is Y - Y_1 = M(X - X_1), where (X_1, Y_1) is a point on the line, we substitute X_1 = -6, Y_1 = 10, and M = 2 into the equation:
Therefore, the equation of the line perpendicular to 2X + 4Y = -10 and passing through the point (-6, 10) is Y = 2X + 22.
Page 22 of 30
8L Coordinate Geometry- Trapezoid
How to prove: - Trapezoid - Show that one pair of opposite sides is parallel and one pair is NOT parallel. Isosceles Trapezoid - Show that one pair of opposite sides is parallel and one pair is NOT parallel. (Trapezoid) - Show that the non-parallel sides are congruent (distance twice)
Examples:
1) Prove that quadrilateral ABCD is an isosceles trapezoid. A(−9, −4 ) B (9, 2) C (1, 6) D (−5, 4) Trapezoid - Show that one pair of opposite sides is parallel and one pair is NOT parallel.
Isosceles Trapezoid - Show that one pair of opposite sides is parallel and one pair is NOT parallel. (Trapezoid) - Show that the non-parallel sides are congruent (distance twice)
To prove that quadrilateral ABCD is an isosceles trapezoid, we need to show that it satisfies the properties of a trapezoid and an isosceles trapezoid:
1. One pair of opposite sides is parallel.
2. One pair of opposite sides is not parallel.
3. The non-parallel sides are congruent.
Let's start by calculating the slopes of the sides AB, BC, CD, and DA:
Slope of side AB:
Slope of side BC:
Slope of side CD:
Slope of side DA:
Now, let's analyze the slopes:
- Slopes of sides AB and CD are equal (\frac{1}{3}), so they are parallel.
- Slopes of sides BC and DA are not equal, so they are not parallel.
So, one pair of opposite sides (AB and CD) is parallel, and one pair (BC and DA) is not parallel, satisfying the properties of a trapezoid.
Next, let's calculate the lengths of the non-parallel sides BC and DA to check if they are congruent:
Length of side BC:
Length of side DA:
Since BC = DA = \sqrt{80} , the non-parallel sides are congruent.
Therefore, quadrilateral ABCD satisfies all the properties of an isosceles trapezoid.
Page 23 of 30
2) Prove that quadrilateral JACK is a trapezoid. J(1, −7) A(10, 2) C(8, 5) K(2, −1)Trapezoid - Show that one pair of opposite sides is parallel and one pair is NOT parallel.
Isosceles Trapezoid - Show that one pair of opposite sides is parallel and one pair is NOT parallel. (Trapezoid) - Show that the non-parallel sides are congruent (distance twice)
To prove that quadrilateral JACK is a trapezoid, we need to show that it has one pair of opposite sides that are parallel and one pair that are not parallel.
Let's first calculate the slopes of the sides JA, AC, CK, and KJ:
Slope of side JA:
Slope of side AC:
Slope of side CK:
Slope of side KJ:
From these calculations, we see that the slopes of sides JA and CK are equal, and the slopes of sides AC and KJ are equal.
Since the slopes of JA and CK are equal (both 1) and the slopes of AC and KJ are not equal , we have one pair of opposite sides that are parallel (JA and CK) and one pair that are not parallel (AC and KJ).
Therefore, quadrilateral JACK is a trapezoid.
3) Prove that quadrilateral EFGH is an isosceles trapezoid. E(−3, 1) F(−1, 5) G(7, 5) H(9,1)
Trapezoid - Show that one pair of opposite sides is parallel and one pair is NOT parallel.
Isosceles Trapezoid - Show that one pair of opposite sides is parallel and one pair is NOT parallel. (Trapezoid) - Show that the non-parallel sides are congruent (distance twice)
To prove that quadrilateral EFGH is an isosceles trapezoid, we need to demonstrate that it satisfies the properties of a trapezoid, with one pair of opposite sides being parallel and one pair not parallel, and additionally, show that the non-parallel sides are congruent.
Let's start by determining the slopes of the sides EF, FG, GH, and HE:
Slope of side EF:
Slope of side FG:
Slope of side GH:
Slope of side HE:
From these calculations, we see that the slopes of sides EF and GH are equal (both 2), and the slopes of sides FG and HE are equal (both 0).
This indicates that the opposite sides EF and GH are parallel, and the opposite sides FG and HE are parallel.
Furthermore, we notice that the lengths of sides FG and HE are equal, as are the lengths of sides EF and GH. This indicates that the non-parallel sides are congruent.
Therefore, quadrilateral EFGH is indeed an isosceles trapezoid.
Page 24 of 30
8L Coordinate Geometry – Trapezoid HW
1) Determine if quadrilateral MATH with M(−8, 2) A(0, 6), T(8,0) and H(−8, −8) is a trapezoid
Trapezoid - Show that one pair of opposite sides is parallel and one pair is NOT parallel.
Isosceles Trapezoid - Show that one pair of opposite sides is parallel and one pair is NOT parallel. (Trapezoid) - Show that the non-parallel sides are congruent (distance twice)
To determine if quadrilateral MATH is a trapezoid, we need to verify if it has one pair of opposite sides that are parallel and one pair that are not parallel.
Let's start by calculating the slopes of the sides MA, AT, TH, and HM:
Slope of side MA:
Slope of side AT:
Slope of side TH:
Slope of side HM:
From these calculations, we can see that the slopes of sides MA and TH are equal ( \frac{1}{2} ), and the slopes of sides AT and HM are undefined.
Therefore, MA and TH are parallel, and AT and HM are not parallel.
Hence, quadrilateral MATH is a trapezoid.
2) Determine if quadrilateral MARY with M(−3, 3), A(7,3) R(3,6) and Y(1,6) is an isosceles trapezoid.
Trapezoid - Show that one pair of opposite sides is parallel and one pair is NOT parallel.
Isosceles Trapezoid - Show that one pair of opposite sides is parallel and one pair is NOT parallel. (Trapezoid) - Show that the non-parallel sides are congruent (distance twice)
To determine if quadrilateral MARY is an isosceles trapezoid, we need to verify if it satisfies the properties of a trapezoid and has one pair of opposite sides that are parallel and one pair that are not parallel, and the non-parallel sides are congruent.
Let's start by calculating the slopes of the sides MA, AR, RY, and YM:
Slope of side MA:
Slope of side AR:
Slope of side RY:
Slope of side YM:
From these calculations, we can see that the slopes of sides MA and RY are equal (0), and the slopes of sides AR and YM are equal ( -\frac{3}{4} ).
Therefore, MA and RY are parallel, and AR and YM are parallel. Additionally, AR and YM are not parallel to MA and RY, respectively.
Hence, quadrilateral MARY is a trapezoid. Furthermore, since the non-parallel sides AR and YM are congruent (both have a length of 4), quadrilateral MARY is an isosceles trapezoid.
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3) Determine if trapezoid MATH with M (−2, −1) A(−2, 4), T(7, 7) and H(10, 3) is an isosceles trapezoid.’
Trapezoid - Show that one pair of opposite sides is parallel and one pair is NOT parallel.
Isosceles Trapezoid - Show that one pair of opposite sides is parallel and one pair is NOT parallel. (Trapezoid) - Show that the non-parallel sides are congruent (distance twice)
To determine if trapezoid MATH is an isosceles trapezoid, we need to verify if it satisfies the properties of a trapezoid and has one pair of opposite sides that are parallel and one pair that are not parallel, and the non-parallel sides are congruent.
Let's start by calculating the slopes of the sides MA, AT, TH, and HM:
Slope of side MA:
Slope of side AT:
Slope of side TH:
Slope of side HM:
From these calculations, we can see that the slopes of sides AT and HM are equal ( \frac{1}{3} ), and the slopes of sides MA and TH are undefined and \frac{-4}{3}, respectively.
Therefore, AT and HM are parallel, and MA and TH are not parallel. Additionally, AT and HM are not parallel to MA and TH, respectively.
Hence, trapezoid MATH is a trapezoid. Furthermore, since the non-parallel sides AT and HM are congruent (both have a length of \sqrt{10}), trapezoid MATH is an isosceles trapezoid.
Review Questions:
4) Given: In ΔDEF, DG is the median to side EF and DG ⊥ EF. Prove: DE ≅ DF
To prove that DE is congruent to DF , we will use the properties of a median in a triangle and the fact that it is perpendicular to the side it intersects.
Given:
1. DG is the median to side EF .
2. DG is perpendicular to EF .
To prove:
DE is congruent to DF .
Proof:
Since DG is a median to side EF , it divides EF into two congruent segments, say EG and FG . This is a property of medians in triangles.
By the definition of perpendicular bisector, DG bisects EF into two congruent segments, DE and DF .
Thus, DE is congruent to DF because they are corresponding parts of two congruent segments.
Therefore, in triangle DEF , DE is congruent to DF .
5) In the diagram below, right triangle RSU is inscribed in circle O and UT is the altitude drawn to hypotenuse RS. The length of RT is 16 more than the length of TS and TU = 15. (a) Find the length of TS. (b) Find, in simplest radical form, the length of RU.
Given:
1. Right triangle RSU is inscribed in circle O .
2. UT is the altitude drawn to hypotenuse RS .
3. Length of RT is 16 more than the length of TS (Let's denote TS = x ).
4. TU = 15 .
(a) To find the length of TS :
In a right triangle, the altitude drawn to the hypotenuse creates three similar right triangles. Therefore, we can use the property of altitude to hypotenuse in a right triangle, which states that the two segments created are proportional to the two segments of the hypotenuse.
So, we can set up the proportion:
Given that TU = 15 and RT = x + 16 , we can substitute these values into the proportion:
Cross-multiply to solve for x :
Now, solve this quadratic equation for x using the quadratic formula:
Where a = 1 , b = 16 , and c = -225 .
The positive value of x is the length of TS , so:
So, the length of TS is 9 .
(b) To find the length of RU :
In a right triangle, the altitude drawn to the hypotenuse creates three similar right triangles. Therefore, we can use the property of altitude to hypotenuse in a right triangle, which states that the two segments created are proportional to the two segments of the hypotenuse.
So, we can set up the proportion:
Given that TU = 15 and RT = x + 16 , where x = 9 , we can substitute these values into the proportion:
Cross-multiply to solve for RU :
So, the length of RU is 9 .
Page 26 of 30
8M Coordinate Geometry – Regents Rev Jan 2017
31 In square GEOM, the coordinates of G are (2,-2) and the coordinates of O are (-4,2). Determine and state the coordinates of vertices E and M. [ The use of the set of axes below is optional.]
In a square, each side is perpendicular to its adjacent side, and all sides are congruent. Therefore, if we know the coordinates of one vertex and the center of the square, we can determine the coordinates of the other vertices.
Given that the coordinates of G are (2,-2) and the coordinates of O are (-4,2) , we can calculate the distance between G and O to find the length of the side of the square.
Using the distance formula:
For G and O :
Since O is the center of the square, the distance from O to any vertex is the same as the distance from O to G . Therefore, the length of the side of the square is 2\sqrt{13} .
Now, let's find the coordinates of vertices E and M .
Since G is at the top left corner and O is at the center, we can move 2\sqrt{13} units to the right and 2\sqrt{13} units down to find E . Similarly, we can move 2\sqrt{13} units to the left and 2\sqrt{13} units up to find M .
Coordinates of E :
Coordinates of M :
So, the coordinates of vertices E and M are:
32) Triangle QRS is graphed on the set of axes below. Q(-2,2), R2,0), S(4,6) O n the same set of axes, graph and label ∆Q’R’S’, the image of ∆qrs after a dilation with a scale factor of 3/2 centered at the origin. Use slopes to explain why Q’R’ ∥ QR.
To graph and label the image of triangle QRS after a dilation with a scale factor of \frac{3}{2} centered at the origin, we need to multiply the coordinates of each vertex Q , R , and S by \frac{3}{2} .
For Q(-2, 2) :
For R(2, 0) :
For S(4, 6) :
Now, let's plot these points and connect them to form \triangle Q'R'S' :
\begin{array}{|c|c|}
\hline
\text{Vertex} & \text{Coordinates} \\
\hline
Q' & (-3, 3) \\
R' & (3, 0) \\
S' & (6, 9) \\
\hline
\end{array}
\]
Now, to explain why Q'R' \parallel QR , we'll compare the slopes of QR and Q'R' .
The slope of QR can be calculated as:
The slope of Q'R' can be calculated as:
Since both slopes are equal, it implies that the lines QR and Q'R' are parallel.
Let's plot these points and the lines QR and Q'R' to visually confirm their parallelism.
To visually confirm the parallelism of lines QR and Q'R' , let's plot the points and lines on the set of axes:
First, plot the original triangle QRS with vertices Q(-2, 2), R(2, 0), and S(4, 6):
Next, plot the image triangle Q'R'S' with vertices Q'(-3, 3), R'(3, 0), and S'(6, 9):
As you can see, the lines QR and Q'R' are indeed parallel. This is evident from the fact that both triangles maintain the same relative orientation, and the corresponding sides QR and Q'R' are equally spaced apart throughout the dilation process.
Page 27 of 30 June 2016
In the Diagram below, ∆ABC has vertices A(4,5), B(2,1) and C(7,3).
What is the slope of the altitude drawn from A to BC?
(1) 2/5 (2) 3/2 (3) - 1/2 (4) - 5/2
To find the slope of the altitude drawn from vertex A to line BC, we first need to find the equation of line BC.
Using the coordinates of points B and C, we can calculate the slope of line BC:
The slope of any line perpendicular to line BC (which includes the altitude drawn from A) will be the negative reciprocal of the slope of BC. Therefore, the slope of the altitude drawn from A to BC will be:
So, the correct answer is option (4) -\frac{5}{2} .
22 Triangle RST is graphed on the set of axes below. R(-2,7), S(-5,1) T(7,-5)
How many square units are in the area of ∆RST?
(1) 9√3 +15
(2) 9√5 + 15
(3) 45
(4) 90
To find the area of triangle RST, we can use the formula for the area of a triangle given its vertices in the coordinate plane:
Given the coordinates of the vertices:
R(-2, 7) , S(-5, 1) , and T(7, -5) ,
We can plug these coordinates into the formula to find the area:
So, the area of triangle RST is 45 square units.
Therefore, the correct answer is option (3) 45.
10) Given MN shown below, with M(-6,1) and N(3,-5) what is an equation of the line that passes through point P(6,1) and is parallel to MN?
(1) y = (-2)/3x +5
(2) y = 3/2 x +7
(3) y = (-2)/3x -3
(4) y = 3/2 x -8
To find the equation of the line parallel to line MN and passing through point P(6,1), we first need to find the slope of line MN.
The slope of the line passing through points M(-6,1) and N(3,-5) can be calculated using the formula:
Plugging in the coordinates of M and N, we get:
Since the line passing through P(6,1) needs to be parallel to MN, it will have the same slope as MN.
Therefore, the equation of the line passing through P(6,1) with slope -\frac{2}{3} is in the form y = mx + b, where m is the slope and b is the y-intercept. We can use the point P(6,1) to find b:
So, the equation of the line is y = -\frac{2}{3}x + 5.
Therefore, the correct answer is option (1) y = -\frac{2}{3}x + 5.
Page 28 of 30
Triangle ABC has vertices with A(x,3), B(-3,-1) and C(-1,-4)
Determine and state a value of x that would make triangle ABC a right triangle. Justify why triangle ABC is a right triangle. (Use of the set of axis is optional.)
Determine and state a value of x that would make triangle ABC a right triangle. Prove tri ABC is a right triangle.
To determine a value of x that would make triangle ABC a right triangle, we can first find the slopes of the sides of the triangle and then check if any two sides are perpendicular to each other. If two sides are perpendicular, then the triangle will be a right triangle.
The slope of a line passing through two points (x_1, y_1) and (x_2, y_2) is given by:
Given the vertices of triangle ABC :
A(x, 3), B(-3, -1), and C(-1, -4)
The slopes of the sides are:
1. Slope of AB:
2. Slope of BC:
3. Slope of AC:
For triangle ABC to be a right triangle, the product of the slopes of any two sides should be -1 (since perpendicular lines have slopes that are negative reciprocals of each other).
Therefore, we need to find a value of x such that:
We'll start by solving the first equation:
By substituting x = 9 into the equations for the other sides, we can verify if the triangle is indeed a right triangle. Let's do that:
For x = 9 , the slopes become:
Now, let's check if m_{AB} \cdot m_{BC} = -1 and m_{BC} \cdot m_{AC} = -1 :
Since -\frac{1}{2} \cdot -\frac{21}{20} is not equal to -1, x = 9 does not make triangle ABC a right triangle.
Let's try another value for x to see if we can find one that makes triangle ABC a right triangle.
Let's try solving the equation m_{AB} \cdot m_{BC} = -1 again with a different value of x .
We'll substitute m_{AB} = \frac{-4}{-3 - x} and m_{BC} = \frac{-3}{2} into the equation and solve for x :
Now, we'll check if x = 3 makes triangle ABC a right triangle.
For x = 3 , the slopes become:
Now, let's check if m_{AB} \cdot m_{BC} = -1 and m_{BC} \cdot m_{AC} = -1 :
Since -\frac{21}{8} \cdot -1 equals 1, which is not -1, x = 3 does not make triangle ABC a right triangle.
Therefore, there may not be any value of x that makes triangle ABC a right triangle. Let's prove that ABC is not a right triangle by showing that its sides do not satisfy the Pythagorean theorem.
To prove that triangle ABC is not a right triangle, we can show that its sides do not satisfy the Pythagorean theorem. According to the Pythagorean theorem, in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
The length of a side of a triangle can be found using the distance formula:
Let's calculate the lengths of the sides of triangle ABC and check if they satisfy the Pythagorean theorem:
1. Length of side AB :
2. Length of side AC :
3. Length of side BC :
Now, let's check if these side lengths satisfy the Pythagorean theorem:
Since x = 0 , we conclude that triangle ABC is not a right triangle because the sides do not satisfy the Pythagorean theorem for any value of x . Thus, there is no value of x that would make triangle ABC a right triangle.
Page 29 of 30 June 2015
36. Use of graph is optional.
In the coordinate plane, the vertices of tri. RST are R(6,-1), S(1,-4) and T(-5,6).
1. Prove that tri RST is a right triangle.
2. State the coordinates of point P such that Quadrilateral RSTP is a rectangle. _________________________
3. Prove that your quadrilateral RSTP is a rectangle.
To address the given tasks:
1. **Prove that triangle RST is a right triangle:**
We can use the Pythagorean theorem to determine whether triangle RST is a right triangle. If the sum of the squares of the lengths of the two shorter sides is equal to the square of the length of the longest side, then the triangle is a right triangle.
Let's find the lengths of the sides RS, ST, and RT using the distance formula:
- Length of RS:
- Length of ST:
- Length of RT:
Now, we'll check if RS^2 + ST^2 = RT^2 :
Since RS^2 + ST^2 = RT^2 , triangle RST satisfies the Pythagorean theorem, and thus, it is a right triangle.
2. **State the coordinates of point P such that Quadrilateral RSTP is a rectangle:**
To form a rectangle, we need to add a point P such that RP and SP are perpendicular to RS and have equal length.
The midpoint of RS can serve as point P. Let's find the midpoint of RS:
Midpoint of RS:
Therefore, the coordinates of point P are (\frac{7}{2}, \frac{-5}{2}).
3. **Prove that quadrilateral RSTP is a rectangle:**
To prove that quadrilateral RSTP is a rectangle, we need to show that its opposite sides are parallel and equal in length, and its diagonals bisect each other and are equal in length.
- **Opposite sides parallel and equal in length:** RS and PT are parallel and equal in length (as they share a midpoint). Likewise, ST and RP are parallel and equal in length.
- **Diagonals bisect each other:** The midpoint of RS (point P) is also the midpoint of diagonal RT. Similarly, the midpoint of ST is also the midpoint of diagonal RP. Therefore, the diagonals bisect each other.
- **Diagonals are equal in length:** Since RSPT is a rectangle, its diagonals are equal in length.
Thus, quadrilateral RSTP is a rectangle.
Prove that ∆ABC is an isosceles right triangle. [ the use of the set of axes below is optional.]
To prove that triangle ABC is an isosceles right triangle, we need to show that it satisfies the properties of both an isosceles triangle and a right triangle:
1. Two sides are congruent.
2. One angle is a right angle.
Let's start by calculating the lengths of the sides AB, BC, and AC using the distance formula:
Length of side AB:
Length of side BC:
Length of side AC:
Now, let's compare the lengths of the sides:
Since AC = BC, two sides are congruent, satisfying the property of an isosceles triangle.
Next, let's check if one angle is a right angle. We can do this by calculating the slopes of the sides AB, BC, and AC:
Slope of side AB:
Slope of side BC:
Slope of side AC:
Now, let's check if any two slopes are negative reciprocals of each other, which would indicate that the corresponding sides are perpendicular and form a right angle:
- Slopes of sides AB and BC are not negative reciprocals of each other.
- Slopes of sides AB and AC are negative reciprocals of each other.
Since the slopes of sides AB and AC are negative reciprocals of each other, angle BAC is a right angle.
Therefore, triangle ABC is an isosceles right triangle, as it satisfies both the properties of an isosceles triangle and a right triangle.
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