# A triangle has corners at (2, 1 ), ( 4, 3 ), and ( 1 , 5 ). If the triangle is dilated by  2 x around (2, 5), what will the new coordinates of its corners be?

Jun 5, 2017

$\left(2 , - 3\right) , \left(6 , 1\right) , \left(0 , 5\right)$

#### Explanation:

$\text{let the vertices be } A \left(2 , 1\right) , B \left(4 , 3\right) , C \left(1 , 5\right)$

$\text{and " A',B',C' "be the images of A,B,C respectively under}$
$\text{the dilatation}$
$\text{let the centre of dilatation be } D \left(2 , 5\right)$

$\vec{D A} = \underline{a} - \underline{d} = \left(\begin{matrix}2 \\ 1\end{matrix}\right) - \left(\begin{matrix}2 \\ 5\end{matrix}\right) = \left(\begin{matrix}0 \\ - 4\end{matrix}\right)$

$\Rightarrow \vec{D A '} = \textcolor{red}{2} \left(\begin{matrix}0 \\ - 4\end{matrix}\right) = \left(\begin{matrix}0 \\ - 8\end{matrix}\right)$

$\Rightarrow A ' = \left(2 + 0 , 5 - 8\right) = \left(2 , - 3\right) \leftarrow \textcolor{red}{\text{ image of A}}$

$\vec{D B} = \underline{b} - \underline{d} = \left(\begin{matrix}4 \\ 3\end{matrix}\right) - \left(\begin{matrix}2 \\ 5\end{matrix}\right) = \left(\begin{matrix}2 \\ - 2\end{matrix}\right)$

$\Rightarrow \vec{D B '} = \textcolor{red}{2} \left(\begin{matrix}2 \\ - 2\end{matrix}\right) = \left(\begin{matrix}4 \\ - 4\end{matrix}\right)$

$\Rightarrow B ' = \left(2 + 4 , 5 - 4\right) = \left(6 , 1\right) \leftarrow \textcolor{red}{\text{ image of B}}$

$\vec{D C} = \underline{c} - \underline{d} = \left(\begin{matrix}1 \\ 5\end{matrix}\right) - \left(\begin{matrix}2 \\ 5\end{matrix}\right) = \left(\begin{matrix}- 1 \\ 0\end{matrix}\right)$

$\Rightarrow \vec{D C '} = \textcolor{red}{2} \left(\begin{matrix}- 1 \\ 0\end{matrix}\right) = \left(\begin{matrix}- 2 \\ 0\end{matrix}\right)$

$\Rightarrow C ' = \left(2 - 2 , 5 - 0\right) = \left(0 , 5\right) \leftarrow \textcolor{red}{\text{ image of C}}$