Trapezoid WKLX has vertices W(2, −3), K(4, −3), L(5, −2), and X(1, −2). Trapezoid WKLX​ is translated 4 units right producing trapezoid W′K′L′X′.

Question

Which coordinates describe the vertices of the image?

Responses

W′(6, −3), K′(8, −3), L′(9, −2), X′(5, −2)
W prime open parentheses 6 comma negative 3 close parentheses , K prime open parentheses 8 comma negative 3 close parentheses , L prime open parentheses 9 comma negative 2 close parentheses , X prime open parentheses 5 comma negative 2 close parentheses

W′(2, −7), K′(4, −7), L′(5, −6), X′(1, −6)
W prime open parentheses 2 comma negative 7 close parentheses , K prime open parentheses 4 comma negative 7 close parentheses , L prime open parentheses 5 comma negative 6 close parentheses , X prime open parentheses 1 comma negative 6 close parentheses

W′(−2, −3), K′(0, −3), L′(1, −2), X′(−3, −2)
W prime open parentheses negative 2 comma negative 3 close parentheses , K prime open parentheses 0 comma negative 3 close parentheses , L prime open parentheses 1 comma negative 2 close parentheses , X prime open parentheses negative 3 comma negative 2 close parentheses

W′(2, 1), K′(4, 1), L′(5, 2), X′(1, 2)

Answer 1

To find the coordinates of the vertices of trapezoid WKLX after translating it 4 units to the right, we need to add 4 to the x-coordinate of each vertex.

The original coordinates of the vertices are:

• \( W(2, -3) \)
• \( K(4, -3) \)
• \( L(5, -2) \)
• \( X(1, -2) \)

Now we perform the translation:

• \( W′(2 + 4, -3) = W′(6, -3) \)
• \( K′(4 + 4, -3) = K′(8, -3) \)
• \( L′(5 + 4, -2) = L′(9, -2) \)
• \( X′(1 + 4, -2) = X′(5, -2) \)

Thus, the coordinates of the translated vertices are:

• \( W′(6, -3) \)
• \( K′(8, -3) \)
• \( L′(9, -2) \)
• \( X′(5, -2) \)

The correct response is:

W′(6, −3), K′(8, −3), L′(9, −2), X′(5, −2)