Question

Select the correct answer from each drop-down menu. Quadrilateral PQRS, with vertex P(-5, -3), undergoes a transformation to form quadrilateral P′Q′R′S′, with P′ at (5, 3). The type of transformation PQRS undergoes is a ( A. 180 rotation about the origins. B. Translation 2 units rights 7 units up. C. Reflection over the x - axis. D reflection over the y - axis. ) . If vertex Q is at (-4, -5), then vertex Q′ is at ( A. (-4,-5) B. ( -4,5) C. (4,5) D.(5, -4) .

Answer 1

Answer:

i) A. 180º rotation about the origin, ii) $Q' = (4, 5)$.

Step-by-step explanation:

i) In this case, we understand that vertex $P = (-5,-3)$ changed to $P' = (5,3)$ after doing an operation. At first we must calculate the distance of each point regarding origin by Pythagorean Theorem:

Point P:

$OP = \sqrt{(x_{P}-x_{O})^{2}+(y_{P}-y_{O})^{2}}$

If we know that $x_{P} = -5$, $y_{P} = -3$, $x_{O} = 0$ and $y_{O} = 0$, the distance $OP$ is:

$OP = \sqrt{(-5-0)^{2}+(-3-0)^{2}}$

$OP \approx 5.831$

Point P':

$OP' = \sqrt{(x_{P'}-x_{O})^{2}+(y_{P'}-y_{O})^{2}}$

If we know that $x_{P'} = 5$, $y_{P'} = 3$, $x_{O} = 0$ and $y_{O} = 0$, the distance $OP'$ is:

$OP' = \sqrt{(5-0)^{2}+(3-0)^{2}}$

$OP' \approx 5.831$

As $OP = OP'$, origin is the center of rotation.

Besides, $P'$ is a multiple of $P$, that is:

1) $(-5, -3)$ Given

2) $((-1)\cdot 5, (-1)\cdot 3)$ $(-a)\cdot b = -a\cdot b$

3) $(-1)\cdot (5, 3)$ Scalar multiplication of a vector/Result.

The value of the scalar proves that P experimented a 180º rotation about the origin. Hence, the correct answer is A.

ii) If $Q = (-4, -5)$ and the same operation described in item i) is used, then, the location of $Q'$ is:

$Q' = (-1)\cdot Q$

$Q' = (-1) \cdot (-4,-5)$

$Q' = ((-1)\cdot (-4), (-1)\cdot (-5))$

$Q' = (4, 5)$

Which corresponds to option C.