Question

Point Q'Q ′ Q, prime is the image of Q(0,6)Q(0,6)Q, left parenthesis, 0, comma, 6, right parenthesis under the translation (x,y)\to(x+7,y-5)(x,y)→(x+7,y−5)left parenthesis, x, comma, y, right parenthesis, \to, left parenthesis, x, plus, 7, comma, y, minus, 5, right parenthesis. What are the coordinates of Q'Q

Answer 1

Q is located at (0,6)

The translation rule is $(x,y) \to (x+7,y-5)$ which says to add 7 to the x coordinate and subtract 5 from the y coordinate. Doing that to (0,6) moves it to (7,1) which is where point Q' is located.

In other words, if you shift point Q(0,6) seven units to the right and five units down, then it arrives at Q ' (7,1)

Answer: (7, 1)

Answer 2

The coordinates of Q' are $Q'(x,y) = (7, 1)$.

Note - There are typing mistakes in the question. Correct statement is presented below:

Point Q' is the image of $Q(0,6)$ under the translation $(x,y) \to (x+7, y-5)$. What are the coordinates of Q'.

In this question we should use the point and the operation described in statement. Vectorially speaking, a translation is defined by the following operation:

$P'(x,y) = P(x,y) + T(x,y)$ (1)

Where:

• $P(x,y)$ - Original point.
• $P'(x,y)$ - Translated point.
• $T(x,y)$ - Translation vector.

If we know that $Q(x,y) = (0,6)$, then the coordinates of $Q'(x,y)$ are:

$Q'(x,y) = Q(x,y) + (7,-5)$

$Q'(x,y) = (0,6) + (7,-5)$

$Q'(x,y) = (7, 1)$

The coordinates of Q' are $Q'(x,y) = (7, 1)$.

We kindly invite to see this problem on rigid transformations: https://brainly.com/question/18613109