# A triangle has corners at (3, 8 ), ( 2, 1 ), and ( 9 , 6 ). If the triangle is dilated by  3 x around (1, 2), what will the new coordinates of its corners be?

Oct 28, 2017

$\left(7 , 20\right) , \left(4 , - 1\right) \text{ and } \left(25 , 14\right)$

#### Explanation:

$\text{let the 3 corners be}$

$A = \left(3 , 8\right) , B = \left(2 , 1\right) \text{ and } C = \left(9 , 6\right)$

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

$\text{let "D=(1,2)" be the centre of dilatation}$

•color(white)(x)A=(3,8)

$\vec{D A '} = 3 \vec{D A}$

$\vec{D A} = \underline{a} - \underline{d} = \left(\begin{matrix}3 \\ 8\end{matrix}\right) - \left(\begin{matrix}1 \\ 2\end{matrix}\right) = \left(\begin{matrix}2 \\ 6\end{matrix}\right)$

$\Rightarrow \vec{D A '} = 3 \left(\begin{matrix}2 \\ 6\end{matrix}\right) = \left(\begin{matrix}6 \\ 18\end{matrix}\right)$

$\Rightarrow A ' = \left(1 + 6 , 2 + 18\right) = \left(7 , 20\right)$

•color(white)(x)B=(2,1)

$\vec{D B '} = 3 \vec{D B}$

$\vec{D B} = \underline{b} - \underline{d} = \left(\begin{matrix}2 \\ 1\end{matrix}\right) - \left(\begin{matrix}1 \\ 2\end{matrix}\right) = \left(\begin{matrix}1 \\ - 1\end{matrix}\right)$

$\Rightarrow \vec{D B '} = 3 \left(\begin{matrix}1 \\ - 1\end{matrix}\right) = \left(\begin{matrix}3 \\ - 3\end{matrix}\right)$

$\Rightarrow B ' = \left(1 + 3 , 2 - 3\right) = \left(4 , - 1\right)$

•color(white)(x)C=(9,6)

$\vec{D C '} = 3 \vec{D C}$

$\vec{D C} = \underline{c} - \underline{d} = \left(\begin{matrix}9 \\ 6\end{matrix}\right) - \left(\begin{matrix}1 \\ 2\end{matrix}\right) = \left(\begin{matrix}8 \\ 4\end{matrix}\right)$

$\Rightarrow \vec{D C '} = 3 \left(\begin{matrix}8 \\ 4\end{matrix}\right) = \left(\begin{matrix}24 \\ 12\end{matrix}\right)$

$\Rightarrow C ' = \left(1 + 24 , 2 + 12\right) = \left(25 , 14\right)$