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3 years ago

7

Which reflection rule, if any, can be used to prove that rectangle A(-8, -3), B(-2,-3), C (-2,-6), D(-8,-6) and rectangle A'(8.-

3), B (2, -3).
C'(2-6), D'(8,6) are congruent?

2 answers:

Pavlova-9 [17]3 years ago

7 0

Answer:

Step-by-step explanation:

For usatestprep grade 8 the answer is a not d

Zina [86]3 years ago

5 0

Answer:

d

Step-by-step explanation:

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1. S(–4, –4), P(4, –2), A(6, 6) and Z(–2, 4) a) Apply the distance formula for each side to determine whether SPAZ is equilatera

Aleksandr [31]

Answer:

a) SPAZ is equilateral.

b) Diagonals SA and PZ are perpendicular to each other.

c) Diagonals SA and PZ bisect each other.

Step-by-step explanation:

At first we form the triangle with the help of a graphing tool and whose result is attached below. It seems to be a paralellogram.

a) If figure is equilateral, then SP = PA = AZ = ZS:

$SP = \sqrt{[4-(-4)]^{2}+[(-2)-(-4)]^{2}}$

$SP \approx 8.246$

$PA = \sqrt{(6-4)^{2}+[6-(-2)]^{2}}$

$PA \approx 8.246$

$AZ =\sqrt{(-2-6)^{2}+(4-6)^{2}}$

$AZ \approx 8.246$

$ZS = \sqrt{[-4-(-2)]^{2}+(-4-4)^{2}}$

$ZS \approx 8.246$

Therefore, SPAZ is equilateral.

b) We use the slope formula to determine the inclination of diagonals SA and PZ:

$m_{SA} = \frac{6-(-4)}{6-(-4)}$

$m_{SA} = 1$

$m_{PZ} = \frac{4-(-2)}{-2-4}$

$m_{PZ} = -1$

Since $m_{SA}\cdot m_{PZ} = -1$ , diagonals SA and PZ are perpendicular to each other.

c) The diagonals bisect each other if and only if both have the same midpoint. Now we proceed to determine the midpoints of each diagonal:

$M_{SA} = \frac{1}{2}\cdot S(x,y) + \frac{1}{2}\cdot A(x,y)$

$M_{SA} = \frac{1}{2}\cdot (-4,-4)+\frac{1}{2}\cdot (6,6)$

$M_{SA} = (-2,-2)+(3,3)$

$M_{SA} = (1,1)$

$M_{PZ} = \frac{1}{2}\cdot P(x,y) + \frac{1}{2}\cdot Z(x,y)$

$M_{PZ} = \frac{1}{2}\cdot (4,-2)+\frac{1}{2}\cdot (-2,4)$

$M_{PZ} = (2,-1)+(-1,2)$

$M_{PZ} = (1,1)$

Then, the diagonals SA and PZ bisect each other.

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2 years ago

Figure ABCD is a parallelogram with point C (−4, 1). Figure ABCD is rotated 90° clockwise to form figure A′B′C′D′. What coordina

Svet_ta [14]

There are certain rules to follow when rotating a point 90 deg clockwise. Since it is given that ABCD is a parallelogram and the coordinates of point C are given, we just have to follow this simple relation:

R(90 deg) : (X,Y) ---> (-Y,X)

Using the given coordinates:
R(90 deg<span>) : (-4,1) ---> (-1,-4)
</span>
Therefore, Point C' will be located at (-1,-4)

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