Question

I need help with this one .Y’all got some answers?

Answer 1

`\begin{array}c\cline{1-4} x & y & \textsf{new $x$} & \textsf{new $y$} \\\cline{1-4} 1 & 1 & -1.5 & 0\\\cline{1-4} 3 & 1 & 0.5 & 0\\\cline{1-4} 3 & 2 & 0.5 & 1\\\cline{1-4} 2 & 2 & -0.5 & 1\\\cline{1-4} 2 & 4 & -0.5 & 3\\\cline{1-4} 1 & 4 & -1.5 & 3\\\cline{1-4}\end{array}`

Explanation:

Coordinates of the vertices of the original figure:

• (1, 1)
• (3, 1)
• (3, 2)
• (2, 2)
• (2, 4)
• (1, 4)

Required translation: 2.5 units left and 1 unit down.

Therefore, the mapping rule is: (x, y) → (x - 2.5, y - 1)

Apply the mapping rule to find the coordinates of the new vertices:

• (1, 1) → (1-2.5, 1-1) = (-1.5, 0)
• (3, 1) → (3-2.5, 1-1) = (0.5, 0)
• (3, 2) → (3-2.5, 2-1) = (0.5, 1)
• (2, 2) → (2-2.5, 2-1) = (-0.5, 1)
• (2, 4) → (2-2.5, 4-1) = (-0.5, 3)
• (1, 4) → (1-2.5, 4-1) = (-1.5, 3)

Therefore:

`\begin{array}r\cline{1-4} x & y & \textsf{new $x$} & \textsf{new $y$} \\\cline{1-4} 1 & 1 & -1.5 & 0\\\cline{1-4} 3 & 1 & 0.5 & 0\\\cline{1-4} 3 & 2 & 0.5 & 1\\\cline{1-4} 2 & 2 & -0.5 & 1\\\cline{1-4} 2 & 4 & -0.5 & 3\\\cline{1-4} 1 & 4 & -1.5 & 3\\\cline{1-4}\end{array}`