1. **Problem statement:** Determine if the points are collinear by checking if the gradients (slopes) between pairs of points are equal. 2. **Formula for gradient:** The gradient $m$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ If three points are collinear, the gradients between each pair must be equal. --- ### 7a. Points A(-3,-5), B(1,5), C(9,25) 3. Calculate gradient $m_{AB}$: $$m_{AB} = \frac{5 - (-5)}{1 - (-3)} = \frac{10}{4} = \frac{5}{2}$$ 4. Calculate gradient $m_{BC}$: $$m_{BC} = \frac{25 - 5}{9 - 1} = \frac{20}{8} = \frac{5}{2}$$ 5. Calculate gradient $m_{AC}$: $$m_{AC} = \frac{25 - (-5)}{9 - (-3)} = \frac{30}{12} = \frac{5}{2}$$ 6. Since $m_{AB} = m_{BC} = m_{AC} = \frac{5}{2}$, the points are collinear. --- ### 7b. Points P(2,-3), Q(1,-7), R(5,11) 7. Calculate gradient $m_{PQ}$: $$m_{PQ} = \frac{-7 - (-3)}{1 - 2} = \frac{-4}{-1} = 4$$ 8. Calculate gradient $m_{QR}$: $$m_{QR} = \frac{11 - (-7)}{5 - 1} = \frac{18}{4} = \frac{9}{2}$$ 9. Calculate gradient $m_{PR}$: $$m_{PR} = \frac{11 - (-3)}{5 - 2} = \frac{14}{3}$$ 10. Since $m_{PQ} = 4$, $m_{QR} = \frac{9}{2}$, and $m_{PR} = \frac{14}{3}$ are not equal, the points are not collinear. --- ### 8. Points S(3,-6), T(k,1), U(9,0) are collinear. Find $k$. 11. Calculate gradient $m_{ST}$: $$m_{ST} = \frac{1 - (-6)}{k - 3} = \frac{7}{k - 3}$$ 12. Calculate gradient $m_{TU}$: $$m_{TU} = \frac{0 - 1}{9 - k} = \frac{-1}{9 - k}$$ 13. Since points are collinear, gradients are equal: $$\frac{7}{k - 3} = \frac{-1}{9 - k}$$ 14. Cross multiply: $$7(9 - k) = -1(k - 3)$$ $$63 - 7k = -k + 3$$ 15. Rearrange terms: $$63 - 7k + k = 3$$ $$63 - 6k = 3$$ 16. Subtract 63 from both sides: $$-6k = 3 - 63$$ $$-6k = -60$$ 17. Divide both sides by -6: $$k = \frac{-60}{-6} = 10$$ --- ### 9. Points A(a,a), B(7,10), C(-1,5) are collinear. Find $a$. 18. Calculate gradient $m_{AB}$: $$m_{AB} = \frac{10 - a}{7 - a}$$ 19. Calculate gradient $m_{BC}$: $$m_{BC} = \frac{5 - 10}{-1 - 7} = \frac{-5}{-8} = \frac{5}{8}$$ 20. Since points are collinear, gradients are equal: $$\frac{10 - a}{7 - a} = \frac{5}{8}$$ 21. Cross multiply: $$8(10 - a) = 5(7 - a)$$ $$80 - 8a = 35 - 5a$$ 22. Rearrange terms: $$80 - 8a + 5a = 35$$ $$80 - 3a = 35$$ 23. Subtract 80 from both sides: $$-3a = 35 - 80$$ $$-3a = -45$$ 24. Divide both sides by -3: $$a = \frac{-45}{-3} = 15$$ --- **Final answers:** - 7a: Points are collinear. - 7b: Points are not collinear. - 8: $k = 10$ - 9: $a = 15$