Question

△abc was rotated 90° counterclockwise around the origin to form △abc. △abc was dilated using a scale factor of ⅓ with a center of dilation at the origin. the result is △abc, shown in the coordinate plane. what are the coordinates of △abc? show your work.

Explanation:

Step1: Reverse the dilation

If a point $(x,y)$ is dilated by a scale - factor $k$ with the center of dilation at the origin to get $(x',y')$, then the original point can be found by the formula $(x,y)=(x'/k,y'/k)$. Here $k = \frac{1}{3}$, so we multiply the coordinates of $\triangle A''B''C''$ by 3 to get the coordinates of $\triangle A'B'C'$.

Step2: Reverse the rotation

The rule for a $90^{\circ}$ counter - clockwise rotation about the origin is $(x,y)\to(-y,x)$. To reverse a $90^{\circ}$ counter - clockwise rotation, we use the rule $(x,y)\to(y, - x)$ to get the coordinates of $\triangle ABC$ from the coordinates of $\triangle A'B'C'$.

Let's assume the coordinates of $A''$ are $(x_{A''},y_{A''})$, $B''$ are $(x_{B''},y_{B''})$ and $C''$ are $(x_{C''},y_{C''})$.
First, find the coordinates of $A',B',C'$:
If $A''=(x_{A''},y_{A''})$, then $A'=(3x_{A''},3y_{A''})$. Similarly for $B'$ and $C'$.
Then, find the coordinates of $A,B,C$:
If $A'=(x_{A'},y_{A'})$, then $A=(y_{A'},-x_{A'})$. Similarly for $B$ and $C$.

For example, if $A''=(- 2,1)$, then $A'=(-2\times3,1\times3)=(-6,3)$ and $A=(3,6)$.
You need to read the coordinates of $A'',B'',C''$ from the graph and follow the above steps to get the coordinates of $A,B,C$.

Answer:

Find the coordinates of $A'',B'',C''$ from the graph. Multiply their coordinates by 3 to get the coordinates of $A',B',C'$. Then apply the transformation $(x,y)\to(y, - x)$ to the coordinates of $A',B',C'$ to get the coordinates of $A,B,C$.