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

Jun 9, 2017

The new coordinates are $\left(- 5 , 5\right)$, $\left(- 1 , 9\right)$ and $\left(- 5 , 17\right)$

#### Explanation:

Let the corners of the triangle be

$A = \left(1 , 2\right)$

$B = \left(2 , 3\right)$

$C = \left(1 , 5\right)$

And the point $D = \left(3 , 1\right)$

Let the corners of the triangle be

$A ' = \left(x , y\right)$, $B ' = \left(x ' , y '\right)$ and $C ' = \left(x ' ' , y ' '\right)$ after dilatation.

$\vec{D A '} = 4 \vec{D A}$

$\left(\begin{matrix}x - 3 \\ y - 1\end{matrix}\right) = 4 \left(\begin{matrix}1 - 3 \\ 2 - 1\end{matrix}\right) = \left(\begin{matrix}- 8 \\ 4\end{matrix}\right)$

$x - 3 = - 8$, $\implies$, $x = - 5$

$y - 1 = 4$, $\implies$, $y = 5$

So,

$A ' = \left(- 5 , 5\right)$

$\vec{D B '} = 4 \vec{D B}$

$\left(\begin{matrix}x ' - 3 \\ y ' - 1\end{matrix}\right) = 4 \left(\begin{matrix}2 - 3 \\ 3 - 1\end{matrix}\right) = \left(\begin{matrix}- 4 \\ 8\end{matrix}\right)$

$x ' - 3 = - 4$, $\implies$, $x ' = - 1$

$y ' - 1 = 8$, $\implies$, $y ' = 9$

$B ' = \left(- 1 , 9\right)$

$\vec{D C '} = 4 \vec{D C}$

$\left(\begin{matrix}x ' ' - 3 \\ y ' ' - 1\end{matrix}\right) = 4 \left(\begin{matrix}1 - 3 \\ 5 - 1\end{matrix}\right) = \left(\begin{matrix}- 8 \\ 16\end{matrix}\right)$

$x ' ' - 3 = - 8$, $\implies$, $x ' ' = - 5$

$y ' ' - 1 = 16$, $\implies$, $y ' ' = 17$

$C ' = \left(- 5 , 17\right)$

Jun 9, 2017

$\left(- 5 , 5\right) , \left(- 1 , 9\right) , \left(- 5 , 17\right)$

#### Explanation:

$\text{let the vertices of the triangle be}$

$A \left(1 , 2\right) , B \left(2 , 3\right) \text{ and } C \left(1 , 5\right)$

$\text{and A',B',C' be the images of A,B and C respectively}$
$\text{under the dilatation}$

$\text{let the centre of dilatation be } D \left(\textcolor{m a \ge n t a}{3} , \textcolor{b l u e}{1}\right)$

• vec(DA)=ula-uld=((1),(2))-((3),(1))=((-2),(1))

$\Rightarrow \vec{D A '} = \textcolor{red}{4} \vec{D A} = \textcolor{red}{4} \left(\begin{matrix}- 2 \\ 1\end{matrix}\right) = \left(\begin{matrix}- 8 \\ 4\end{matrix}\right)$

$\Rightarrow A ' = \left(\textcolor{m a \ge n t a}{3} - 8 , \textcolor{b l u e}{1} + 4\right) = \left(- 5 , 5\right)$

• vec(DB)=ulb-uld=((2),(3))-((3),(1))=((-1),(2))

$\Rightarrow \vec{D B '} = \textcolor{red}{4} \vec{D B} = \textcolor{red}{4} \left(\begin{matrix}- 1 \\ 2\end{matrix}\right) = \left(\begin{matrix}- 4 \\ 8\end{matrix}\right)$

$\Rightarrow B ' = \left(\textcolor{m a \ge n t a}{3} - 4 , \textcolor{b l u e}{1} + 8\right) = \left(- 1 , 9\right)$

• vec(DC)=ulc-uld=((1),(5))-((3),(1))=((-2),(4))

$\Rightarrow \vec{D C '} = \textcolor{red}{4} \vec{D C} = \textcolor{red}{4} \left(\begin{matrix}- 2 \\ 4\end{matrix}\right) = \left(\begin{matrix}- 8 \\ 16\end{matrix}\right)$

rArrC'=(color(magenta)(3)-8,color(blue)(1)+16))=(-5,17)