# A triangle has corners at (2, 1 ), ( 1, 3 ), and (5 , 5 ). If the triangle is dilated by  3 x around (1, 6), what will the new coordinates of its corners be?

May 10, 2017

$\left(2 , 1\right) \to \left(4 , - 9\right)$
$\left(1 , 3\right) \to \left(1 , - 3\right)$
$\left(5 , 5\right) \to \left(13 , 3\right)$

#### Explanation:

The vector equation of a line from $\left(1 , 6\right)$ to $\left(2 , 1\right)$ is:

$\left(x , y\right) = \left(1 , 6\right) + t \left(\left(2 - 1\right) \hat{i} + \left(1 - 6\right) \hat{j}\right)$

Perform the subtraction within the vector:

$\left(x , y\right) = \left(1 , 6\right) + t \left(\hat{i} - 5 \hat{j}\right) \text{ [1]}$

The vector equation of a line from $\left(1 , 6\right)$ to $\left(1 , 3\right)$ is:

$\left(x , y\right) = \left(1 , 6\right) + t \left(\left(1 - 1\right) \hat{i} + \left(3 - 6\right) \hat{j}\right)$

Perform the subtraction within the vector:

$\left(x , y\right) = \left(1 , 6\right) + t \left(- 3 \hat{j}\right) \text{ [2]}$

The vector equation of a line from $\left(1 , 6\right)$ to $\left(5 , 5\right)$ is:

$\left(x , y\right) = \left(1 , 6\right) + t \left(\left(5 - 1\right) \hat{i} + \left(5 - 6\right) \hat{j}\right)$

Perform the subtraction within the vector:

$\left(x , y\right) = \left(1 , 6\right) + t \left(4 \hat{i} - \hat{j}\right) \text{ [3]}$

Here are the 3 vector equations:

$\left(x , y\right) = \left(1 , 6\right) + t \left(\hat{i} - 5 \hat{j}\right) \text{ [1]}$
$\left(x , y\right) = \left(1 , 6\right) + t \left(- 3 \hat{j}\right) \text{ [2]}$
$\left(x , y\right) = \left(1 , 6\right) + t \left(4 \hat{i} - \hat{j}\right) \text{ [3]}$

To obtain the new points, evaluate equations [1], [2], and [3] at t = 3

$\left(x , y\right) = \left(1 , 6\right) + 3 \left(\hat{i} - 5 \hat{j}\right)$
$\left(x , y\right) = \left(1 , 6\right) + 3 \left(- 3 \hat{j}\right)$
$\left(x , y\right) = \left(1 , 6\right) + 3 \left(4 \hat{i} - \hat{j}\right)$

$\left(x , y\right) = \left(4 , - 9\right)$
$\left(x , y\right) = \left(1 , - 3\right)$
$\left(x , y\right) = \left(13 , 3\right)$

May 10, 2017

$\left(4 , - 9\right) , \left(1 , - 3\right) , \left(13 , 3\right)$

#### Explanation:

$\text{let " A=(2,1),B=(1,3),C=(5,5)" and } D = \left(1 , 6\right)$

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

$\vec{D A} = \underline{a} - \underline{d} = \left(\begin{matrix}2 \\ 1\end{matrix}\right) - \left(\begin{matrix}1 \\ 6\end{matrix}\right) = \left(\begin{matrix}1 \\ - 5\end{matrix}\right)$

$\Rightarrow \vec{D A '} = 3 \left(\begin{matrix}1 \\ - 5\end{matrix}\right) = \left(\begin{matrix}3 \\ - 15\end{matrix}\right)$

$\Rightarrow A ' = \left(1 + 3 , 6 - 15\right) = \left(4 , - 9\right)$
$\textcolor{b l u e}{\text{--------------------------------------------------}}$

$\vec{D B} = \underline{b} - \underline{d} = \left(\begin{matrix}1 \\ 3\end{matrix}\right) - \left(\begin{matrix}1 \\ 6\end{matrix}\right) = \left(\begin{matrix}0 \\ - 3\end{matrix}\right)$

$\Rightarrow \vec{D B '} = 3 \left(\begin{matrix}0 \\ - 3\end{matrix}\right) = \left(\begin{matrix}0 \\ - 9\end{matrix}\right)$

$\Rightarrow B ' = \left(1 + 0 , 6 - 9\right) = \left(1 , - 3\right)$
$\textcolor{b l u e}{\text{--------------------------------------------------}}$

$\vec{D C} = \underline{c} - \underline{d} = \left(\begin{matrix}5 \\ 5\end{matrix}\right) - \left(\begin{matrix}1 \\ 6\end{matrix}\right) = \left(\begin{matrix}4 \\ - 1\end{matrix}\right)$

$\Rightarrow \vec{D C '} = 3 \left(\begin{matrix}4 \\ - 1\end{matrix}\right) = \left(\begin{matrix}12 \\ - 3\end{matrix}\right)$

$\Rightarrow C ' = \left(1 + 12 , 6 - 3\right) = \left(13 , 3\right)$
$\textcolor{b l u e}{\text{-------------------------------------------------}}$