To solve the problem of describing translations using column vectors, we need to determine how each shape has been moved from its original position (Shape A) to its new position. The movement is described by a column vector in the form:
\[
\begin{pmatrix}
x \\ y
\end{pmatrix}
\]
where:
- \( x \) represents the horizontal movement (right if positive, left if negative),
- \( y \) represents the vertical movement (up if positive, down if negative).
Let's analyze each part step by step:
---
1)
-
Original Shape A: Bottom-left corner at \((0, 0)\).
-
New Position: Bottom-left corner at \((3, 0)\).
-
Movement: Right by 3 units, no vertical movement.
-
Column Vector:
\[
\begin{pmatrix}
3 \\ 0
\end{pmatrix}
\]
---
2)
-
Original Shape A: Bottom-left corner at \((0, 0)\).
-
New Position: Bottom-left corner at \((0, 3)\).
-
Movement: Up by 3 units, no horizontal movement.
-
Column Vector:
\[
\begin{pmatrix}
0 \\ 3
\end{pmatrix}
\]
---
3)
-
Original Shape A: Bottom-left corner at \((0, 0)\).
-
New Position: Bottom-left corner at \((3, 3)\).
-
Movement: Right by 3 units and up by 3 units.
-
Column Vector:
\[
\begin{pmatrix}
3 \\ 3
\end{pmatrix}
\]
---
4)
-
Original Shape A: Bottom-left corner at \((0, 0)\).
-
New Position: Bottom-left corner at \((-2, -1)\).
-
Movement: Left by 2 units and down by 1 unit.
-
Column Vector:
\[
\begin{pmatrix}
-2 \\ -1
\end{pmatrix}
\]
---
5)
-
Original Shape A: Bottom-left corner at \((0, 0)\).
-
New Position: Bottom-left corner at \((4, 2)\).
-
Movement: Right by 4 units and up by 2 units.
-
Column Vector:
\[
\begin{pmatrix}
4 \\ 2
\end{pmatrix}
\]
---
6)
-
Original Shape A: Bottom-left corner at \((0, 0)\).
-
New Position: Bottom-left corner at \((2, -2)\).
-
Movement: Right by 2 units and down by 2 units.
-
Column Vector:
\[
\begin{pmatrix}
2 \\ -2
\end{pmatrix}
\]
---
7)
-
Original Shape A: Bottom-left corner at \((0, 0)\).
-
New Position: Bottom-left corner at \((2, -1)\).
-
Movement: Right by 2 units and down by 1 unit.
-
Column Vector:
\[
\begin{pmatrix}
2 \\ -1
\end{pmatrix}
\]
---
8)
-
Original Shape A: Bottom-left corner at \((0, 0)\).
-
New Position: Bottom-left corner at \((3, 2)\).
-
Movement: Right by 3 units and up by 2 units.
-
Column Vector:
\[
\begin{pmatrix}
3 \\ 2
\end{pmatrix}
\]
---
9)
-
Original Shape A: Bottom-left corner at \((0, 0)\).
-
New Position: Bottom-left corner at \((3, -2)\).
-
Movement: Right by 3 units and down by 2 units.
-
Column Vector:
\[
\begin{pmatrix}
3 \\ -2
\end{pmatrix}
\]
---
Final Answers
\[
\boxed{
\begin{array}{ccc}
1) & \begin{pmatrix} 3 \\ 0 \end{pmatrix} & 2) \begin{pmatrix} 0 \\ 3 \end{pmatrix} & 3) \begin{pmatrix} 3 \\ 3 \end{pmatrix} \\
4) & \begin{pmatrix} -2 \\ -1 \end{pmatrix} & 5) \begin{pmatrix} 4 \\ 2 \end{pmatrix} & 6) \begin{pmatrix} 2 \\ -2 \end{pmatrix} \\
7) & \begin{pmatrix} 2 \\ -1 \end{pmatrix} & 8) \begin{pmatrix} 3 \\ 2 \end{pmatrix} & 9) \begin{pmatrix} 3 \\ -2 \end{pmatrix}
\end{array}
}
\]
Parent Tip: Review the logic above to help your child master the concept of transformational geometry worksheet.
