Triangle Dilation 366389

1. **State the problem:** We need to find the image of triangle $\triangle ABC$ after a dilation centered at point $P$ with a scale factor of $\frac{1}{2}$. 2. **Recall the dilation formula:** If $P$ is the center of dilation and $k$ is the scale factor, then for any point $X$, the image $X'$ is given by: $$X' = P + k(X - P)$$ This means we take the vector from $P$ to $X$, scale it by $k$, and then add it back to $P$. 3. **Apply the formula to each vertex:** Let the coordinates of $A$, $B$, $C$, and $P$ be known or given. Since the problem provides approximate segment lengths but no exact coordinates, the key is understanding the scale effect: - Each segment length from $P$ to a vertex is multiplied by $\frac{1}{2}$. - The image triangle $\triangle A'B'C'$ is similar to $\triangle ABC$ with side lengths half as long. 4. **Interpretation of the image:** - The image triangle is smaller, centered around $P$. - Each vertex $A'$, $B'$, and $C'$ lies on the line segment from $P$ to the original vertex, exactly halfway between $P$ and that vertex. 5. **Summary:** The dilation with center $P$ and scale factor $\frac{1}{2}$ maps each vertex $X$ of $\triangle ABC$ to $X'$ such that: $$X' = P + \frac{1}{2}(X - P)$$ This produces a smaller triangle $\triangle A'B'C'$ similar to $\triangle ABC$ with side lengths half the original. **Final answer:** The image of $\triangle ABC$ under the dilation centered at $P$ with scale factor $\frac{1}{2}$ is a triangle $\triangle A'B'C'$ where each vertex is the midpoint of the segment from $P$ to the corresponding vertex of $\triangle ABC$.