# A triangle has corners at (1, 7 ), ( 8, 3 ), and ( 4 , 8 ). If the triangle is dilated by  2 x around (2, 5), what will the new coordinates of its corners be?

Feb 17, 2018

$\left(0 , 9\right) , \left(14 , 1\right) \text{ and } \left(6 , 11\right)$

#### Explanation:

$\text{label the vertices of the triangle}$

$A \left(1 , 7\right) , B \left(8 , 3\right) \text{ and } C \left(4 , 8\right)$

$\text{and A', B', C' the images of A, B and C}$

$\text{let the centre of dilatation be } D \left(2 , 5\right)$

$\Rightarrow \vec{D A '} = \textcolor{red}{2} \vec{D A}$

$\Rightarrow \underline{a} ' - \underline{d} = 2 \left(\underline{a} - \underline{d}\right) = 2 \underline{a} - 2 \underline{d}$

$\Rightarrow \underline{a} ' = 2 \underline{a} - \underline{d}$

$\textcolor{w h i t e}{\Rightarrow \underline{a} '} = 2 \left(\begin{matrix}1 \\ 7\end{matrix}\right) - \left(\begin{matrix}2 \\ 5\end{matrix}\right) = \left(\begin{matrix}0 \\ 9\end{matrix}\right)$

$\text{the coordinates of A' are the components of } \underline{a} '$

$\Rightarrow A ' = \left(0 , 9\right)$

$\text{similarly }$

$\vec{D B '} = \textcolor{red}{2} \vec{D B}$

$\Rightarrow \underline{b} ' = 2 \underline{b} - \underline{d}$

$\textcolor{w h i t e}{\Rightarrow \underline{b} '} = 2 \left(\begin{matrix}8 \\ 3\end{matrix}\right) - \left(\begin{matrix}2 \\ 5\end{matrix}\right) = \left(\begin{matrix}14 \\ 1\end{matrix}\right)$

$\Rightarrow B ' = \left(14 , 1\right)$

$\text{and } \vec{D C '} = \textcolor{red}{2} \vec{D C}$

$\Rightarrow \underline{c} ' = 2 \underline{c} - \underline{d}$

$\textcolor{w h i t e}{\Rightarrow \underline{c} '} = 2 \left(\begin{matrix}4 \\ 8\end{matrix}\right) - \left(\begin{matrix}2 \\ 5\end{matrix}\right) = \left(\begin{matrix}6 \\ 11\end{matrix}\right)$

$\Rightarrow C ' = \left(6 , 11\right)$

$\textcolor{b l u e}{\text{Result after dilatation}}$

$\left(1 , 7\right) \to \left(0 , 9\right) , \left(8 , 3\right) \to \left(14 , 1\right) , \left(4 , 8\right) \to \left(6 , 11\right)$