Question

? What translation vector represents the transformation from A (2,5) to A'(-5,3) ? 2​

Answer 1

`\begin{bmatrix}-7 \\-2\end{bmatrix}` (might also be denoted as
`\langle-7,\, -2\rangle` in some textbooks.)

Explanation:

Let
`a` denote the position vector of point
`A`
`(2,\, 5)`. Thus:

`a = \begin{bmatrix}2 \\ 5\end{bmatrix}`.

Similarly, let
`a^(\prime)` denote the position vector of point
`A^(\prime)`. Thus:

`\displaystyle a^(\prime) = \begin{bmatrix}-5 \\ 3\end{bmatrix}`.

Let
`x` be the translation vector from point
`A` to point
`A^(\prime)`. Adding this translation vector
`x\!` to the position vector of the original point should yield the positional vector of the new point. In other words:

`a + x = a^(\prime)`.

`\begin{aligned}x &= a^(\prime) - a \\ &= \begin{bmatrix}-5 \\ 3\end{bmatrix} - \begin{bmatrix}2 \\ 5\end{bmatrix}\\ &= \begin{bmatrix}-5 - 2 \\ 3 - 5\end{bmatrix} \\ &=\begin{bmatrix}-7 \\ -2\end{bmatrix}\end{aligned}`.

Thus, the required translation vector would be
`\begin{bmatrix}-7 \\ -2\end{bmatrix}` (or equivalently,
`\langle -7,\, -2 \rangle`.