Question: Dilate $\triangle ABC$ with a scale factor of $3$. Enter the new coordinates. $A = (1,2)$ $B = (3,-1)$ $C = (-4,-2)$ $A' = ( ?, )$ $B' = ( , )$ $C' = ( , )

1. **State the problem:** We need to find the coordinates of the dilated triangle $\triangle A'B'C'$ after applying a dilation with scale factor $3$ to the original triangle $\triangle ABC$ with vertices $A(1,2)$, $B(3,-1)$, and $C(-4,-2)$. 2. **Formula for dilation:** When a point $(x,y)$ is dilated from the origin by a scale factor $k$, the new coordinates $(x',y')$ are given by: $$ (x', y') = (kx, ky) $$ 3. **Apply the formula to each vertex:** - For $A(1,2)$: $$ A' = (3 \times 1, 3 \times 2) = (3, 6) $$ - For $B(3,-1)$: $$ B' = (3 \times 3, 3 \times (-1)) = (9, -3) $$ - For $C(-4,-2)$: $$ C' = (3 \times (-4), 3 \times (-2)) = (-12, -6) $$ 4. **Final answer:** $$ A' = (3, 6), \quad B' = (9, -3), \quad C' = (-12, -6) $$ These are the coordinates of the dilated triangle $\triangle A'B'C'$ with scale factor $3$.