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Quadrilateral ABCD is translated to get quadrilateral A’B’C’D’. Vertex A is at (-5, 2), and vertex A’ is at (2, -2). Quadrilateral ABCD is translated blank. If vertex B is at (-6, 5), then vertex B’ is at blank

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Ashraf82

Answer:

vertex B' is at (1, 1)

Step-by-step explanation:

Let us revise the rules of translation of a point

  • If the point (x, y) translated horizontally to the right by h units then its image is (x + h, y) ⇒ T (x, y) → (x + h, y)
  • If the point (x, y) translated horizontally to the left by h units then its image is (x - h, y) ⇒ T (x, y) → (x - h, y)
  • If the point (x, y) translated vertically up by k units then its image is (x, y + k)→ (x + h, y) ⇒ T (x, y) → (x, y + k)
  • If the point (x, y) translated vertically down by k units then its image is (x, y - k) ⇒ T (x, y) → (x, y - k)

Let us use the rules to solve our question

∵ Vertex A = (-5, 2)

∵ Vertex A' = (2, -2)

→ The coordinates of the two points changed which means the

  quadrilateral is translated horizontally and vertically

∵ The x-coordinate of A = -5

∵ The x-coordinate of A' = 2

→ Which means it moves to the right, so use the first rule above

∴ the rule of translation is T (x, y) → (x + h, y)

∴ x → x + h

∵ x = -5 and x + h = 2

∴ -5 + h = 2

→ Add 5 to both sides

∴ -5 + 5 + h = 2 + 5

∴ h = 7

∴ The quadrilateral is translated 7 units right

∵ The y-coordinate of A = 2

∵ The y-coordinate of A' = -2

→ Which means it moves down, so use the 4th rule above

∴ the rule of translation is T (x, y) → (x, y - k)

∴ y → y - k

∵ y = 2 and y - k = -2

∴ 2 - k = -2

→ Add K to both sides

∴ 2 - k + k = -2 + k

∴ 2 = -2 + k

→ Add 2 to both sides

∴ 2 + 2 = -2 + 2 + k

∴ 4 = k

∴ The quadrilateral is translated 4 units down

→ Let us find B'

∵ T (x, y) → (x + h, y - k) is the rule of translation

∵ B = (-6, 5)

∵ h = 7 and k = 4

∴ B' = (-6 + 7, 5 - 4)

∴ B' = (1, 1)

∴ vertex B' is at (1, 1)

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