A triangle has corners at (4, 5 ), ( 2, 3 ), and (2 , 2 ). If the triangle is dilated by  4 x around (0, 1), what will the new coordinates of its corners be?

May 31, 2018

color(blue)(A'=(16x,16x+1)

color(blue)(B'=(8x,8x+1)

color(blue)(C'=(8x,4x+1)

Explanation:

Let $A = \left(4 , 5\right) , B = \left(2 , 3\right) , C = \left(2 , 2\right)$

Let:

$\vec{O D} = \left(\begin{matrix}0 \\ 1\end{matrix}\right)$ be the position vector of the dilation point.

Let:

$\vec{D A} = \left(\begin{matrix}4 \\ 4\end{matrix}\right)$, $\vec{D B} = \left(\begin{matrix}2 \\ 2\end{matrix}\right)$, $\vec{D C} = \left(\begin{matrix}2 \\ 1\end{matrix}\right)$

Scale factor $4 x$

Then the position vectors of the images of A,B and C are:

$\vec{O A '} = \vec{O D} + 4 x \vec{D A} = \left(\begin{matrix}0 \\ 1\end{matrix}\right) + 4 x \left(\begin{matrix}4 \\ 4\end{matrix}\right) = \left(\begin{matrix}16 x \\ 16 x + 1\end{matrix}\right)$

$\vec{O B '} = \vec{O D} + 4 x \vec{D B} = \left(\begin{matrix}0 \\ 1\end{matrix}\right) + 4 x \left(\begin{matrix}2 \\ 2\end{matrix}\right) = \left(\begin{matrix}8 x \\ 8 x + 1\end{matrix}\right)$

$\vec{O C '} = \vec{O D} + 4 x \vec{D C} = \left(\begin{matrix}0 \\ 1\end{matrix}\right) + 4 x \left(\begin{matrix}2 \\ 1\end{matrix}\right) = \left(\begin{matrix}8 x \\ 4 x + 1\end{matrix}\right)$

color(blue)(A'=(16x,16x+1)

color(blue)(B'=(8x,8x+1)

color(blue)(C'=(8x,4x+1)