1. **State the problem:** We have three airplanes A, B, and C flying along paths represented by lines on a coordinate plane. - Airplane A's path passes through points $(2,-2)$ and $(4,1)$. - Airplane B's path is parallel to A's and passes through $(14,4)$. - Airplane C's path is perpendicular to A's and passes through $(7,-1)$. We need to determine if any of these paths intersect in the fourth quadrant (where $x>0$ and $y<0$). 2. **Find the equation of Airplane A's path:** The slope $m_A$ is given by $$m_A=\frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-2)}{4 - 2} = \frac{3}{2}.$$ Using point-slope form with point $(2,-2)$: $$y - (-2) = \frac{3}{2}(x - 2) \implies y + 2 = \frac{3}{2}x - 3 \implies y = \frac{3}{2}x - 5.$$ So Airplane A's path is: $$y = \frac{3}{2}x - 5.$$ 3. **Find the equation of Airplane B's path (parallel to A):** Parallel lines have the same slope, so $m_B = \frac{3}{2}$. Using point $(14,4)$: $$y - 4 = \frac{3}{2}(x - 14) \implies y = \frac{3}{2}x - 21 + 4 = \frac{3}{2}x - 17.$$ So Airplane B's path is: $$y = \frac{3}{2}x - 17.$$ 4. **Find the equation of Airplane C's path (perpendicular to A):** The slope of C, $m_C$, is the negative reciprocal of $m_A$: $$m_C = -\frac{2}{3}.$$ Using point $(7,-1)$: $$y - (-1) = -\frac{2}{3}(x - 7) \implies y + 1 = -\frac{2}{3}x + \frac{14}{3} \implies y = -\frac{2}{3}x + \frac{14}{3} - 1 = -\frac{2}{3}x + \frac{11}{3}.$$ So Airplane C's path is: $$y = -\frac{2}{3}x + \frac{11}{3}.$$ 5. **Check intersections between paths:** - **A and B:** Since they are parallel with different intercepts, they never intersect. - **A and C:** Solve $$\frac{3}{2}x - 5 = -\frac{2}{3}x + \frac{11}{3}.$$ Multiply both sides by 6 to clear denominators: $$6 \times \left(\frac{3}{2}x - 5\right) = 6 \times \left(-\frac{2}{3}x + \frac{11}{3}\right)$$ $$9x - 30 = -4x + 22$$ $$9x + 4x = 22 + 30$$ $$13x = 52 \implies x = 4.$$ Substitute back to find $y$: $$y = \frac{3}{2} \times 4 - 5 = 6 - 5 = 1.$$ Intersection point is $(4,1)$. - **B and C:** Solve $$\frac{3}{2}x - 17 = -\frac{2}{3}x + \frac{11}{3}.$$ Multiply both sides by 6: $$9x - 102 = -4x + 22$$ $$9x + 4x = 22 + 102$$ $$13x = 124 \implies x = \frac{124}{13} \approx 9.54.$$ Substitute back: $$y = \frac{3}{2} \times 9.54 - 17 = 14.31 - 17 = -2.69.$$ Intersection point is approximately $(9.54, -2.69)$. 6. **Determine if intersections are in the fourth quadrant ($x>0$, $y<0$):** - Intersection of A and C is at $(4,1)$: $y>0$, so **not** in the fourth quadrant. - Intersection of B and C is at $(9.54, -2.69)$: $x>0$ and $y<0$, so **this intersection is in the fourth quadrant**. 7. **Conclusion:** Only Airplane B and Airplane C's paths intersect in the air traffic controller's region (fourth quadrant).