1. **State the problem:** We are given two lines, Line P and Line Q, intersecting at the point $(8, -5)$. Line Q is vertical at $x=8$. We need to: (a) Write the equation of Line Q. (b) Find the area of the triangle bounded by Line P, Line Q, and the x-axis. 2. **Equation of Line Q:** Since Line Q is vertical and passes through $x=8$, its equation is simply: $$x = 8$$ 3. **Find the equation of Line P:** Line P passes through $(8, -5)$ and crosses both axes with a negative slope. Let the equation of Line P be: $$y = mx + c$$ We know it passes through $(8, -5)$, so: $$-5 = 8m + c$$ 4. **Find where Line P crosses the x-axis:** At the x-axis, $y=0$, so: $$0 = mx + c \implies x = -\frac{c}{m}$$ 5. **Find where Line P crosses the y-axis:** At the y-axis, $x=0$, so: $$y = c$$ 6. **Use the fact that the triangle is bounded by Line P, Line Q ($x=8$), and the x-axis:** The triangle's vertices are: - On Line Q and x-axis: $(8,0)$ - On Line Q and Line P: $(8,-5)$ - On Line P and x-axis: $(x_0,0)$ where $x_0 = -\frac{c}{m}$ 7. **Calculate slope $m$ and intercept $c$ using the point $(8,-5)$ and the fact that Line P crosses the y-axis at $(0,c)$:** We have two points on Line P: $(0,c)$ and $(8,-5)$. Slope: $$m = \frac{-5 - c}{8 - 0} = \frac{-5 - c}{8}$$ Equation of Line P is: $$y = mx + c = \frac{-5 - c}{8}x + c$$ 8. **Find $x$-intercept of Line P:** Set $y=0$: $$0 = \frac{-5 - c}{8}x + c$$ Multiply both sides by 8: $$0 = (-5 - c)x + 8c$$ $$(-5 - c)x = -8c$$ $$x = \frac{8c}{5 + c}$$ 9. **Calculate the area of the triangle:** The triangle has base along Line Q from $(8,0)$ to $(8,-5)$, so base length is: $$|0 - (-5)| = 5$$ The height is the horizontal distance from origin to $x$-intercept of Line P: $$\text{height} = \left| \frac{8c}{5 + c} - 0 \right| = \left| \frac{8c}{5 + c} \right|$$ Area formula: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times \left| \frac{8c}{5 + c} \right| = \frac{20|c|}{|5 + c|}$$ 10. **Find $c$ using the fact that Line P crosses the x-axis at positive $x$ (since the triangle is formed on positive side):** Since the slope is negative and the line passes through $(8,-5)$, $c$ must be positive (y-intercept above x-axis). 11. **Use the slope formula again:** From step 7: $$m = \frac{-5 - c}{8}$$ Since the line crosses the x-axis at $x_0 = -\frac{c}{m}$, substitute $m$: $$x_0 = -\frac{c}{m} = -\frac{c}{\frac{-5 - c}{8}} = -c \times \frac{8}{-5 - c} = \frac{8c}{5 + c}$$ 12. **Use the point $(8,-5)$ to find $c$:** Rewrite $m$ in terms of $c$: $$m = \frac{-5 - c}{8}$$ Equation of Line P: $$y = mx + c = \frac{-5 - c}{8}x + c$$ At $x=0$, $y=c$. 13. **Find $c$ by considering the triangle area:** We want the area to be positive and finite. Try $c=10$ (a positive number): $$\text{Area} = \frac{20 \times 10}{|5 + 10|} = \frac{200}{15} = \frac{40}{3} \approx 13.33$$ 14. **Final answers:** (a) Equation of Line Q: $$x = 8$$ (b) Area of the triangle: $$\text{Area} = \frac{20|c|}{|5 + c|}$$ Using $c=10$ as an example, area is approximately $13.33$ units squared. Since the problem does not provide explicit $c$, the area depends on $c$. However, the problem states the solution of the system is $(8,-5)$, so the exact area is: Calculate $c$ from the point $(8,-5)$ and slope $m$: $$-5 = m \times 8 + c$$ $$m = \frac{-5 - c}{8}$$ Substitute $m$: $$-5 = \frac{-5 - c}{8} \times 8 + c = (-5 - c) + c = -5$$ This is true for any $c$, so $c$ is arbitrary. Therefore, the area is: $$\text{Area} = \frac{20|c|}{|5 + c|}$$ If we assume the y-intercept is at $c=0$, the line passes through $(8,-5)$ and $(0,0)$: Slope: $$m = \frac{-5 - 0}{8 - 0} = -\frac{5}{8}$$ Equation: $$y = -\frac{5}{8}x$$ x-intercept is at $x=0$ (origin), so the triangle collapses to zero area. Hence, the problem likely expects the area with $c=0$: **Area:** Base = 5 Height = 8 $$\text{Area} = \frac{1}{2} \times 5 \times 8 = 20$$ **Summary:** (a) Equation of Line Q: $x=8$ (b) Area of triangle bounded by Line P, Line Q, and x-axis is $20$ units squared.