1. Given points A(-2t, h), B(pt, t), and C(2p, 2t) lie on a straight line, and B divides AC such that $AB = \frac{1}{3} AC$. 2. To find $p$ in terms of $h$, use the section formula for point B dividing AC in ratio 1:2 (since $AB = \frac{1}{3} AC$ means B divides AC into 1 part and 2 parts). 3. Coordinates of A: $(-2t, h)$, C: $(2p, 2t)$. 4. Using section formula, B divides AC in ratio 1:2, so $$x_B = \frac{1 \cdot 2p + 2 \cdot (-2t)}{1+2} = \frac{2p - 4t}{3}$$ $$y_B = \frac{1 \cdot 2t + 2 \cdot h}{3} = \frac{2t + 2h}{3}$$ 5. But B is given as $(pt, t)$, so equate: $$pt = \frac{2p - 4t}{3}$$ $$t = \frac{2t + 2h}{3}$$ 6. From second equation: $$3t = 2t + 2h \implies t = 2h$$ 7. Substitute $t=2h$ into first equation: $$p(2h) = \frac{2p - 4(2h)}{3} \implies 2ph = \frac{2p - 8h}{3}$$ 8. Multiply both sides by 3: $$6ph = 2p - 8h$$ 9. Rearrange: $$2p - 6ph = 8h \implies 2p = 6ph + 8h = h(6p + 8)$$ 10. Divide both sides by $h$ (assuming $h \neq 0$): $$\frac{2p}{h} = 6p + 8$$ 11. Rearrange to isolate $p$: $$2p = h(6p + 8) \implies 2p = 6ph + 8h$$ 12. Bring terms involving $p$ to one side: $$2p - 6ph = 8h \implies p(2 - 6h) = 8h$$ 13. Finally, $$p = \frac{8h}{2 - 6h}$$ --- 14. For problem 2, points A(1,4), B(0,6), and point P moves such that $PA : PB = 2 : 3$. 15. The locus of P is the set of points where the ratio of distances to A and B is constant. 16. Using the Apollonius circle formula: $$\frac{PA}{PB} = \frac{2}{3} \implies \frac{PA^2}{PB^2} = \frac{4}{9}$$ 17. Let P be $(x,y)$, then: $$PA^2 = (x-1)^2 + (y-4)^2$$ $$PB^2 = (x-0)^2 + (y-6)^2 = x^2 + (y-6)^2$$ 18. Set up equation: $$\frac{(x-1)^2 + (y-4)^2}{x^2 + (y-6)^2} = \frac{4}{9}$$ 19. Cross multiply and simplify: $$9[(x-1)^2 + (y-4)^2] = 4[x^2 + (y-6)^2]$$ 20. Expand: $$9(x^2 - 2x + 1 + y^2 - 8y + 16) = 4(x^2 + y^2 - 12y + 36)$$ 21. Simplify: $$9x^2 - 18x + 9 + 9y^2 - 72y + 144 = 4x^2 + 4y^2 - 48y + 144$$ 22. Bring all terms to one side: $$9x^2 - 18x + 9y^2 - 72y + 153 = 4x^2 + 4y^2 - 48y + 144$$ 23. Subtract right side: $$(9x^2 - 4x^2) - 18x + (9y^2 - 4y^2) - 72y + 48y + (153 - 144) = 0$$ 24. Simplify: $$5x^2 - 18x + 5y^2 - 24y + 9 = 0$$ 25. This is the equation of the locus of P. --- 26. For problem 3, lines JK: $y = px - k$ and RT: $y = (k + 2)x - p$ are perpendicular. 27. Slopes are $m_1 = p$ and $m_2 = k + 2$. 28. Since lines are perpendicular: $$m_1 \times m_2 = -1 \implies p(k + 2) = -1$$ 29. Solve for $p$: $$p = \frac{-1}{k + 2}$$ --- 30. For problem 4(a), line passes through M(0,3) and N(5,0). 31. Find equation in form $\frac{x}{a} + \frac{y}{b} = 1$. 32. Intercepts are $a$ on x-axis and $b$ on y-axis. 33. Since line passes through N(5,0), $a=5$. 34. Since line passes through M(0,3), $b=3$. 35. Equation is: $$\frac{x}{5} + \frac{y}{3} = 1$$ --- 36. For problem 4(b), point P moves such that distance from N(5,0) is always 3. 37. Distance formula: $$\sqrt{(x-5)^2 + (y-0)^2} = 3$$ 38. Square both sides: $$(x-5)^2 + y^2 = 9$$ 39. This is the equation of the locus of P. Final answers: 1. $$p = \frac{8h}{2 - 6h}$$ 2. $$5x^2 - 18x + 5y^2 - 24y + 9 = 0$$ 3. $$p = \frac{-1}{k + 2}$$ 4(a). $$\frac{x}{5} + \frac{y}{3} = 1$$ 4(b). $$(x-5)^2 + y^2 = 9$$