1. **State the problem:** We have two similar triangles on the coordinate plane, triangle A with vertices at approximately $(-8,6)$, $(-2,6)$, $(-5,7.5)$ and triangle A' with vertices at $(-10,7)$, $(2,7)$, $(-4,9)$. We want to find the scale factor and the center of enlargement that maps triangle A to triangle A'. 2. **Formula and rules:** - The scale factor $k$ is the ratio of corresponding side lengths or distances between corresponding points. - The center of enlargement $C=(x_c,y_c)$ satisfies the relation for each vertex $P$ of the original figure and its image $P'$: $$P' = C + k(P - C)$$ which can be rearranged to find $C$ if $k$ is known. 3. **Calculate side lengths of triangle A:** - Side between $(-8,6)$ and $(-2,6)$: $$d = \sqrt{(-2 + 8)^2 + (6 - 6)^2} = \sqrt{6^2 + 0} = 6$$ - Side between $(-2,6)$ and $(-5,7.5)$: $$d = \sqrt{(-5 + 2)^2 + (7.5 - 6)^2} = \sqrt{(-3)^2 + 1.5^2} = \sqrt{9 + 2.25} = \sqrt{11.25}$$ - Side between $(-5,7.5)$ and $(-8,6)$: $$d = \sqrt{(-8 + 5)^2 + (6 - 7.5)^2} = \sqrt{(-3)^2 + (-1.5)^2} = \sqrt{9 + 2.25} = \sqrt{11.25}$$ 4. **Calculate side lengths of triangle A':** - Side between $(-10,7)$ and $(2,7)$: $$d' = \sqrt{(2 + 10)^2 + (7 - 7)^2} = \sqrt{12^2 + 0} = 12$$ - Side between $(2,7)$ and $(-4,9)$: $$d' = \sqrt{(-4 - 2)^2 + (9 - 7)^2} = \sqrt{(-6)^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40}$$ - Side between $(-4,9)$ and $(-10,7)$: $$d' = \sqrt{(-10 + 4)^2 + (7 - 9)^2} = \sqrt{(-6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40}$$ 5. **Calculate scale factor $k$:** - Using the horizontal side: $$k = \frac{12}{6} = 2$$ - Using the other sides: $$k = \frac{\sqrt{40}}{\sqrt{11.25}} = \frac{\sqrt{40}}{\sqrt{11.25}} = \sqrt{\frac{40}{11.25}} = \sqrt{3.555\ldots} \approx 1.886$$ Since the sides are similar but not exactly matching, we use the horizontal side for exact scale factor $k=2$ (likely due to rounding in coordinates). 6. **Find center of enlargement $C=(x_c,y_c)$:** - Use the formula for one pair of corresponding points, for example $P=(-8,6)$ and $P'=(-10,7)$: $$P' = C + k(P - C)$$ $$\Rightarrow P' = kP + (1 - k)C$$ Rearranged: $$C = \frac{P' - kP}{1 - k}$$ - Calculate $x_c$: $$x_c = \frac{-10 - 2(-8)}{1 - 2} = \frac{-10 + 16}{-1} = \frac{6}{-1} = -6$$ - Calculate $y_c$: $$y_c = \frac{7 - 2(6)}{1 - 2} = \frac{7 - 12}{-1} = \frac{-5}{-1} = 5$$ 7. **Verify center with another point:** - For $P=(-2,6)$ and $P'=(2,7)$: $$C = \frac{2 - 2(-2)}{1 - 2} = \frac{2 + 4}{-1} = -6$$ $$C = \frac{7 - 2(6)}{1 - 2} = 5$$ Matches previous result. **Final answer:** - Scale factor $k = 2$ - Center of enlargement $C = (-6, 5)$