1. The problem asks to write a proportion showing that the slope of segment JL in triangle JKL is equal to the slope of segment MP in triangle MNP, given that the triangles are similar right triangles. 2. Recall the formula for the slope between two points $(x_1,y_1)$ and $(x_2,y_2)$ is: $$\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}$$ 3. For segment JL, points are $J(3,3)$ and $L(2,-3)$: $$\text{slope}_{JL} = \frac{-3 - 3}{2 - 3} = \frac{-6}{-1} = 6$$ 4. For segment MP, points are $M(7,5)$ and $P(-2,-8)$: $$\text{slope}_{MP} = \frac{-8 - 5}{-2 - 7} = \frac{-13}{-9} = \frac{13}{9}$$ 5. Since the slopes are not equal numerically, we check the other given points from the second graph for MP and JL: - $M(3,2)$ and $P(-6,3)$: $$\text{slope}_{MP} = \frac{3 - 2}{-6 - 3} = \frac{1}{-9} = -\frac{1}{9}$$ - $J(1,2)$ and $L(2,0)$: $$\text{slope}_{JL} = \frac{0 - 2}{2 - 1} = \frac{-2}{1} = -2$$ 6. These slopes are also not equal, so the problem likely wants the proportion of the slopes expressed as a ratio of corresponding side lengths due to similarity. 7. Since triangles are similar, corresponding sides are proportional. The proportion showing the equality of slopes is: $$\frac{\text{slope}_{JL}}{\text{slope}_{MP}} = \frac{\frac{y_L - y_J}{x_L - x_J}}{\frac{y_P - y_M}{x_P - x_M}} = 1$$ 8. Writing this as a proportion: $$\frac{y_L - y_J}{x_L - x_J} = \frac{y_P - y_M}{x_P - x_M}$$ This proportion shows the slope of JL equals the slope of MP. Final answer: $$\boxed{\frac{y_L - y_J}{x_L - x_J} = \frac{y_P - y_M}{x_P - x_M}}$$