1. **Problem:** Find the point of intersection between Route A passing through (2,5) and (8,17) and Route B passing through (3,20) and (9,2). 2. **Step 1: Find the equations of the lines for Route A and Route B.** - The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For Route A: $$m_A = \frac{17 - 5}{8 - 2} = \frac{12}{6} = 2$$ Using point-slope form with point (2,5): $$y - 5 = 2(x - 2)$$ Simplify: $$y = 2x - 4 + 5 = 2x + 1$$ For Route B: $$m_B = \frac{2 - 20}{9 - 3} = \frac{-18}{6} = -3$$ Using point-slope form with point (3,20): $$y - 20 = -3(x - 3)$$ Simplify: $$y = -3x + 9 + 20 = -3x + 29$$ 3. **Step 2: Find the intersection point by solving the system:** $$\begin{cases} y = 2x + 1 \\ y = -3x + 29 \end{cases}$$ Set equal: $$2x + 1 = -3x + 29$$ $$2x + 3x = 29 - 1$$ $$5x = 28$$ $$x = \frac{28}{5} = 5.6$$ Substitute $x=5.6$ into $y=2x+1$: $$y = 2(5.6) + 1 = 11.2 + 1 = 12.2$$ 4. **Answer:** The point of intersection is $\boxed{\left(5.6, 12.2\right)}$. This point is the most favorable location for the major bus stop where the two routes intersect.