1. **Problem:** Show that the points (-2, 10), (3, -2), and (15, 3) are vertices of an isosceles triangle. 2. **Formula:** To check if a triangle is isosceles, calculate the distances between each pair of points using the distance formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. **Calculate distances:** - Between (-2, 10) and (3, -2): $$d_1 = \sqrt{(3 - (-2))^2 + (-2 - 10)^2} = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$ - Between (3, -2) and (15, 3): $$d_2 = \sqrt{(15 - 3)^2 + (3 - (-2))^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$$ - Between (-2, 10) and (15, 3): $$d_3 = \sqrt{(15 - (-2))^2 + (3 - 10)^2} = \sqrt{17^2 + (-7)^2} = \sqrt{289 + 49} = \sqrt{338}$$ 4. **Check equality:** We see that $d_1 = 13$ and $d_2 = 13$, so two sides are equal. 5. **Conclusion:** Since two sides are equal, the triangle with vertices (-2, 10), (3, -2), and (15, 3) is isosceles.