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An illustration shows a coordinate plane. Both x- and y-axes range from negative 10 to 10 in one unit increments. A quadrilateral with vertices P Q R S is drawn on it. Vertex P is at coordinates left parenthesis 4 comma 6 right parenthesis, vertex Q is at coordinates left parenthesis 7 comma 2 right parenthesis, vertex R is at coordinates left parenthesis 5 comma negative 2 right parenthesis, and vertex S is at coordinates left parenthesis 2 comma 2 right parenthesis.

Translate figure PQRS
6 units down and 2 units to the right. What are the coordinates of P′
?

(1 point)
Responses

(2, 12)
left parenthesis 2 comma 12 right parenthesis

(10, 4)
left parenthesis 10 comma 4 right parenthesis

(6, 0)
left parenthesis 6 comma 0 right parenthesis

(2, 0)
left parenthesis 2 comma 0 right parenthesis
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1 answer

To find the coordinates of point P' after translating point P(4, 6) six units down and two units to the right, we can apply the translation to the coordinates of point P.

  1. Translation Down: To move down by 6 units, we subtract 6 from the y-coordinate:

    • y-coordinate = 6 - 6 = 0

  2. Translation Right: To move right by 2 units, we add 2 to the x-coordinate:

    • x-coordinate = 4 + 2 = 6

Thus, the new coordinates of P' after the translation are (6, 0).

The correct answer is: (6, 0) or left parenthesis 6 comma 0 right parenthesis.