1. **State the problem:** We have points A(0,8) and B(16,0). Point D divides segment AB in ratio 1:3. Line L passes through D with gradient $\sqrt{3}$ and also passes through $(-2,f)$. We need to show that $f < -4$. 2. **Find coordinates of D:** Since D divides AB in ratio 1:3, let $AD = x$ and $DB = 3x$. Using section formula for internal division, coordinates of D are: $$D = \left( \frac{1 \times 16 + 3 \times 0}{1+3}, \frac{1 \times 0 + 3 \times 8}{1+3} \right) = \left( \frac{16}{4}, \frac{24}{4} \right) = (4,6)$$ 3. **Equation of line L:** Line L passes through D(4,6) with gradient $m = \sqrt{3}$. Using point-slope form: $$y - 6 = \sqrt{3}(x - 4)$$ Simplify: $$y = \sqrt{3}x - 4\sqrt{3} + 6$$ 4. **Find $f$ at $x = -2$:** Substitute $x = -2$ into line L: $$f = \sqrt{3}(-2) - 4\sqrt{3} + 6 = -2\sqrt{3} - 4\sqrt{3} + 6 = -6\sqrt{3} + 6$$ 5. **Show $f < -4$:** Calculate approximate value: $$\sqrt{3} \approx 1.732$$ So, $$f \approx -6 \times 1.732 + 6 = -10.392 + 6 = -4.392$$ Since $-4.392 < -4$, we have shown that $f < -4$. **Final answer:** $f < -4$ is true.