1 Answer

Krishwader · Answer 1 · 2023-03-18T12:24:40+0000

Given the coordinates of the triangle END

$\begin{gathered} E\rightarrow(-5,-1)_ \\ N\operatorname{\rightarrow}(-3,-2) \\ \begin{equation*} D\rightarrow(-3,2) \end{equation*} \end{gathered}$

Firstly the triangle was translated 2units horizontally to the right then shifted downward by 3units.

$\begin{gathered} E\rightarrow(-5+2,-1-3)=(-3,-4) \\ N\rightarrow(-3+2,-2-3)=(-1,-5) \\ D\rightarrow(-3+2,2-3)=(-1,-1) \end{gathered}$

The transformed image was then rotated 90degrees counterclockwise about the origin

When rotating a point 90 degrees counterclockwise about the origin our point A(x,y) becomes A'(-y,x).

Hence,

$\begin{gathered} E^(\prime)\rightarrow(-(-4),-3)=(4,-3) \\ N\rightarrow(-(-5),-1)=(5,-1) \\ D\rightarrow(-(-1),-1)=(1,-1) \end{gathered}$

Therefore, the coordinates of the transformed image is

$\begin{gathered} E^(\prime)\rightarrow(4,-3) \\ N\rightarrow(5,-1) \\ D\rightarrow(1,-1) \end{gathered}$

The graph of the image will be shown below

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