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egoroff_w [7]
3 years ago
14

Drag the tiles to the correct boxes to complete the pairs. Find the distance between each pair of points. 5 units 4 units 2 unit

s 3 units 6 units A (5, 4) and B( 5, -2) arrowRight E(-2, -1) and F(-2, -5) arrowRight C(-4, 1) and D(1, 1) arrowRight G(3, -5) and H(6, -5) arrowRight
Mathematics
1 answer:
Olin [163]3 years ago
8 0

Answer:

a) Distance between points A (5, 4) and B( 5, -2) is 6 units

b) Distance between points E (-2, -1) and F( -2, -5) is 4 units

c) Distance between points C (-4, 1) and D( 1, 1) is 5 units

d) Distance between points G(3, -5) and H(6, -5) is 3 units

Step-by-step explanation:

We need to find the distance between each pair

a) A (5, 4) and B( 5, -2)

the distance formula is:

d(A,B) = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Putting values: x₁ = 5, x₂=5 and y₁= 4 and y₂= -2

d(A,B) = \sqrt{(5-5)^2+(-2-4)^2}

d(A,B) = \sqrt{(0)^2+(-6)^2}

d(A,B) = \sqrt{36}

d(A,B) = 6

Distance between points A (5, 4) and B( 5, -2) is 6 units

b)  E (-2, -1) and F( -2, -5)

the distance formula is:

d(E,F) = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Putting values: x₁ = -2, x₂=-2 and y₁= -1 and y₂= -5

d(E,F) = \sqrt{(-2-(-2))^2+(-5-(-1))^2}

d(E,F) = \sqrt{(0)^2+(-4)^2}

d(E,F) = \sqrt{16}

d(E,F) = 4

Distance between points E (-2, -1) and F( -2, -5) is 4 units

c)  C (-4, 1) and D( 1, 1)

the distance formula is:

d(C,D) = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Putting values: x₁ = -4, x₂=1 and y₁= 1 and y₂= 1

d(C,D)= \sqrt{(1-(-4))^2+(1-(1))^2}

d(C,D) = \sqrt{(5)^2+(0)^2}

d(C,D) = \sqrt{25}

d(C,D) = 5

Distance between points C (-4, 1) and D( 1, 1) is 5 units

d) G(3, -5) and H(6, -5)

the distance formula is:

d(G,H) = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Putting values: x₁ = 3, x₂=6 and y₁= -5 and y₂= -5

d(G,H)= \sqrt{(6-(3))^2+(-5-(-5))^2}

d(G,H) = \sqrt{(3)^2+(0)^2}

d(G,H) = \sqrt{9}

d(G,H) = 3

Distance between points G(3, -5) and H(6, -5) is 3 units

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    |\

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h  |            \      10 ft

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The vertices of a parallelogram are A(x1, y1), B(x2, y2), C(x3, y3), and D(x4, y4). Which of the following must be true if paral
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The answer is really none of the above, because all of these have divisions by difference in x, and there are perfectly fine rectangles where those will mean dividing by zero.

To do the problem the way they want us to, we note segments with the same slope are parallel (or collinear); segments with slopes which multiply to -1 are perpendicular.

Let's figure out what each choice is trying to tell us by translating it into parallel and perpendicular. Remember A is point 1, ...

A. ((y_4-y_3)/(x_4-x_3)= (y_3-y_2)/(x_3-x_2)) and ((y_4-y_3)/(x_4-x_3)×(y_3-y_2)/(x_3-x_2))=-1

translated, that's CD║BC and CD⊥BC - contradictory; second half right

B. (y_4-y_3)/(x_4-x_3)= (y_2-y_1)/(x_2-x_1) and ((y_4-y_3)/(x_4-x_3)×(y_2-y_1)/(x_2-x_1))=-1

CD ║ AB and CD ⊥ AB - only the parallelogram half is right

C. (y_4-y_3)/(x_4-x_3)= (y_2-y_1)/(x_2-x_1) and (y_4-y_3)/(x_4-x_3)×(y_3-y_2)/(x_3-x_2)=-1

CD║AB and CD⊥BC - that's TRUE

D. (y_4-y_3)/(x_4-x_3)= (y_3-y_1)/(x_3-x_1) and (y_4-y_3)/(x_4-x_3)×(y_2-y_1)/(x_2-x_1)=-1

CD║AC and CD⊥AB - nope

Answer: C

The way I'd get the proper expression is with the dot product.

A parallelogram is a rectangle when adjacent sides are perpendicular. It only takes one right angle to make them all right.

Two sides are perpendicular when the dot product of their direction vectors is zero. All the answers have y_4 - y_3 so one side is CD. The other side must be BC or AD; AD would give y_4 - y_1 terms which don't appear among the answer, so let's go with BC.

The direction vectors are

D-C = (x_4 - x_3, y_4 - y_3)

C-B = (x_3 - x_2, y_3 - y_2)

For perpendicularity we need a zero dot product:

0 = (D-C)\cdot(C-B) = (x_4 - x_3)(x_3 - x_2) + (y_4 - y_3)(y_3 - y_2)

- (x_4 - x_3)(x_3 - x_2) = (y_4 - y_3)(y_3 - y_2)

That's the correct expression, given it's a parallelogram, to demonstrate it's a rectangle. The divisions make it iffy.

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