1. Problem: Find the angle of rotation about the origin that maps point A(4, 5) to A'(-4, -5). 2. Explanation: Rotation about the origin by angle $\theta$ transforms point $(x, y)$ to $(x', y')$ by $$x' = x\cos\theta - y\sin\theta, \quad y' = x\sin\theta + y\cos\theta.$$ 3. Substitute $x=4$, $y=5$, $x'=-4$, $y'=-5$: $$-4 = 4\cos\theta - 5\sin\theta,$$ $$-5 = 4\sin\theta + 5\cos\theta.$$ 4. Multiply first equation by 5 and second by 4: $$-20 = 20\cos\theta - 25\sin\theta,$$ $$-20 = 16\sin\theta + 20\cos\theta.$$ 5. Subtract second from first: $$0 = 4\cos\theta - 41\sin\theta \implies 4\cos\theta = 41\sin\theta \implies \tan\theta = \frac{4}{41}.$$ 6. Check consistency with original equations; however, since $A'$ is the negative of $A$, this corresponds to rotation by $\pm 180^\circ$. 7. Final answer: The angle of rotation is $\pm 180^\circ$. --- 8. Problem: Find the image of point $P(x, y)$ under translation vector $(m, n)$. 9. Explanation: Translation moves every point by adding the vector components: $$P'(x', y') = (x + m, y + n).$$ 10. Final answer: The image is $P'(x + m, y + n)$. --- 11. Problem: Given image $P'(m, n)$ of $P(x, y)$ under translation, find the translation vector. 12. Explanation: Translation vector is the difference between image and original point: $$\text{Translation vector} = (m - x, n - y).$$ 13. Final answer: Translation vector is $(m - x, n - y)$. --- 14. Problem: Find translation vector if point $(4, 5)$ maps to $(6, 8)$ under translation. 15. Explanation: Translation vector = $(6 - 4, 8 - 5) = (2, 3)$. 16. Final answer: Translation vector is $(2, 3)$. --- 17. Problem: Find coordinates of image of point $A(x, y)$ enlarged about origin with scale factor $K$. 18. Explanation: Enlargement about origin scales coordinates: $$A'(x', y') = (Kx, Ky).$$ 19. Final answer: Coordinates of image are $(Kx, Ky)$. --- 20. Problem: Find image of $A(4, 5)$ enlarged by $E[(0, 1), 2]$. 21. Explanation: Enlargement about center $(0, 1)$ with scale factor 2: $$A' = (0 + 2(4 - 0), 1 + 2(5 - 1)) = (8, 9).$$ 22. Final answer: Image coordinates are $(8, 9)$. --- 23. Problem: If $A = \begin{pmatrix}2 & 1 \\ 3 & 4\end{pmatrix}$, find matrix $B$ such that $AB = I_2$. 24. Explanation: $B = A^{-1}$, inverse of $A$. 25. Calculate determinant: $$\det A = 2 \times 4 - 1 \times 3 = 8 - 3 = 5.$$ 26. Inverse: $$B = \frac{1}{5} \begin{pmatrix}4 & -1 \\ -3 & 2\end{pmatrix} = \begin{pmatrix} \frac{4}{5} & -\frac{1}{5} \\ -\frac{3}{5} & \frac{2}{5} \end{pmatrix}.$$ 27. Final answer: $B = \begin{pmatrix} \frac{4}{5} & -\frac{1}{5} \\ -\frac{3}{5} & \frac{2}{5} \end{pmatrix}$. --- 28. Problem: Given $A = \begin{pmatrix}2 & 1 \\ 5 & 3\end{pmatrix}$, $B = \begin{pmatrix}3 & 4 \\ 2 & 3\end{pmatrix}$, find $C$ such that $AC = B$. 29. Explanation: Multiply both sides by $A^{-1}$: $$C = A^{-1}B.$$ 30. Calculate $\det A = 2 \times 3 - 1 \times 5 = 6 - 5 = 1.$ 31. Inverse: $$A^{-1} = \begin{pmatrix}3 & -1 \\ -5 & 2\end{pmatrix}.$$ 32. Compute $C = A^{-1}B$: $$C = \begin{pmatrix}3 & -1 \\ -5 & 2\end{pmatrix} \begin{pmatrix}3 & 4 \\ 2 & 3\end{pmatrix} = \begin{pmatrix}7 & 9 \\ -11 & -14\end{pmatrix}.$$ 33. Final answer: $C = \begin{pmatrix}7 & 9 \\ -11 & -14\end{pmatrix}$. --- 34. Problem: Show that for $A = \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix}$, $A^2 - 2A$ is the null matrix. 35. Explanation: Compute $A^2$: $$A^2 = A \times A = \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix} \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix} = \begin{pmatrix}2 & 2 \\ 2 & 2\end{pmatrix}.$$ 36. Compute $A^2 - 2A$: $$\begin{pmatrix}2 & 2 \\ 2 & 2\end{pmatrix} - 2 \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix} = \begin{pmatrix}2 & 2 \\ 2 & 2\end{pmatrix} - \begin{pmatrix}2 & 2 \\ 2 & 2\end{pmatrix} = \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}.$$ 37. Final answer: $A^2 - 2A$ is the null matrix. --- 38. Problem: Prove for $A = \begin{pmatrix}1 & 1 \\ 8 & 3\end{pmatrix}$ that $A^2 - 4A = 5I$. 39. Explanation: Compute $A^2$: $$A^2 = \begin{pmatrix}1 & 1 \\ 8 & 3\end{pmatrix} \begin{pmatrix}1 & 1 \\ 8 & 3\end{pmatrix} = \begin{pmatrix}9 & 4 \\ 32 & 17\end{pmatrix}.$$ 40. Compute $A^2 - 4A$: $$\begin{pmatrix}9 & 4 \\ 32 & 17\end{pmatrix} - 4 \begin{pmatrix}1 & 1 \\ 8 & 3\end{pmatrix} = \begin{pmatrix}9 & 4 \\ 32 & 17\end{pmatrix} - \begin{pmatrix}4 & 4 \\ 32 & 12\end{pmatrix} = \begin{pmatrix}5 & 0 \\ 0 & 5\end{pmatrix} = 5I.$$ 41. Final answer: $A^2 - 4A = 5I$ is proven. --- 42. Problem: If $A = \begin{pmatrix}x & 5 \\ 3 & y\end{pmatrix}$, $B = \begin{pmatrix}4 & 3 \\ 5 & -2\end{pmatrix}$ and $A^t = B$, find $x$ and $y$. 43. Explanation: Transpose $A$ is $$A^t = \begin{pmatrix}x & 3 \\ 5 & y\end{pmatrix} = B = \begin{pmatrix}4 & 3 \\ 5 & -2\end{pmatrix}.$$ 44. Equate corresponding elements: $$x = 4, \quad 3 = 3, \quad 5 = 5, \quad y = -2.$$ 45. Final answer: $x = 4$, $y = -2$. --- 46. Problem: If $(a \, b; c \, d) - (1 \, 2; 3 \, 4) = (5 \, 6; 6 \, 8)$, find $a, b, c, d$. 47. Explanation: Add matrices on right side: $$(a - 1, b - 2; c - 3, d - 4) = (5, 6; 6, 8).$$ 48. Equate elements: $$a - 1 = 5 \Rightarrow a = 6,$$ $$b - 2 = 6 \Rightarrow b = 8,$$ $$c - 3 = 6 \Rightarrow c = 9,$$ $$d - 4 = 8 \Rightarrow d = 12.$$ 49. Final answer: $a=6, b=8, c=9, d=12$. --- 50. Problem: If $(a \, b; c \, d) - (2 \, 4; 6 \, 8) = (10 \, 12; 12 \, 16)$, find $a, b, c, d$. 51. Explanation: $$(a - 2, b - 4; c - 6, d - 8) = (10, 12; 12, 16).$$ 52. Equate elements: $$a = 12, b = 16, c = 18, d = 24.$$ 53. Final answer: $a=12, b=16, c=18, d=24$. --- 54. Problem: If $A = \begin{pmatrix}3 \\ 2\end{pmatrix}$, $B = \begin{pmatrix}-1 \\ 4\end{pmatrix}$, $C = \begin{pmatrix}-3 \\ -1\end{pmatrix}$, find $A + 2B - 3C$. 55. Explanation: $$A + 2B - 3C = \begin{pmatrix}3 \\ 2\end{pmatrix} + 2 \begin{pmatrix}-1 \\ 4\end{pmatrix} - 3 \begin{pmatrix}-3 \\ -1\end{pmatrix} = \begin{pmatrix}3 \\ 2\end{pmatrix} + \begin{pmatrix}-2 \\ 8\end{pmatrix} + \begin{pmatrix}9 \\ 3\end{pmatrix} = \begin{pmatrix}10 \\ 13\end{pmatrix}.$$ 56. Final answer: $\begin{pmatrix}10 \\ 13\end{pmatrix}$. --- 57. Problem: Find values of $a$ and $b$ if midpoint of $(a+1, 1)$ and $(6, b+2)$ is $(4, 2)$. 58. Explanation: Midpoint formula: $$\left(\frac{a+1+6}{2}, \frac{1 + b + 2}{2}\right) = (4, 2).$$ 59. Equate components: $$\frac{a + 7}{2} = 4 \Rightarrow a + 7 = 8 \Rightarrow a = 1,$$ $$\frac{b + 3}{2} = 2 \Rightarrow b + 3 = 4 \Rightarrow b = 1.$$ 60. Final answer: $a=1$, $b=1$. --- 61. Problem: Find other end of line segment if one end is $(4, 4)$ and midpoint is $(-2, 2)$. 62. Explanation: Let other end be $(x, y)$. Midpoint formula: $$\left(\frac{4 + x}{2}, \frac{4 + y}{2}\right) = (-2, 2).$$ 63. Equate components: $$\frac{4 + x}{2} = -2 \Rightarrow 4 + x = -4 \Rightarrow x = -8,$$ $$\frac{4 + y}{2} = 2 \Rightarrow 4 + y = 4 \Rightarrow y = 0.$$ 64. Final answer: Other end is $(-8, 0)$. --- 65. Problem: Find other end of line segment if one end is $(2, 1)$ and midpoint is $(4, 2)$. 66. Explanation: Let other end be $(x, y)$. $$\left(\frac{2 + x}{2}, \frac{1 + y}{2}\right) = (4, 2).$$ 67. Equate components: $$2 + x = 8 \Rightarrow x = 6,$$ $$1 + y = 4 \Rightarrow y = 3.$$ 68. Final answer: Other end is $(6, 3)$. --- 69. Problem: Find ratio in which point $(5, 3)$ divides line segment joining $(2, 3)$ and $(7, 3)$. 70. Explanation: Since $y$ coordinates are same, ratio $m:n$ satisfies $$x = \frac{m \times 7 + n \times 2}{m + n} = 5.$$ 71. Let ratio be $m:n = k:1$: $$5 = \frac{7k + 2}{k + 1} \Rightarrow 5(k + 1) = 7k + 2 \Rightarrow 5k + 5 = 7k + 2 \Rightarrow 2k = 3 \Rightarrow k = \frac{3}{2}.$$ 72. Final answer: Ratio is $3:2$. --- 73. Problem: Find ratio in which point $(2, b)$ divides line segment joining $(-4, 3)$ and $(6, 3)$. 74. Explanation: $y$ coordinates are same, so $b=3$. 75. Let ratio be $m:n = k:1$: $$2 = \frac{6k - 4}{k + 1} \Rightarrow 2(k + 1) = 6k - 4 \Rightarrow 2k + 2 = 6k - 4 \Rightarrow 4k = 6 \Rightarrow k = \frac{3}{2}.$$ 76. Final answer: Ratio is $3:2$. --- 77. Problem: Find ratio in which X-axis divides line joining points $(2, -4)$ and $(-3, 6)$. 78. Explanation: Let ratio be $m:n = k:1$ and point on X-axis has $y=0$. 79. Using section formula for $y$ coordinate: $$0 = \frac{-4k + 6}{k + 1} \Rightarrow -4k + 6 = 0 \Rightarrow 4k = 6 \Rightarrow k = \frac{3}{2}.$$ 80. Final answer: Ratio is $3:2$. --- 81. Problem: Find equation of line with slope $-3$ passing through $(2, -2)$. 82. Explanation: Use point-slope form: $$y - y_1 = m(x - x_1) \Rightarrow y + 2 = -3(x - 2).$$ 83. Simplify: $$y + 2 = -3x + 6 \Rightarrow 3x + y = 4.$$ 84. Final answer: Equation is $3x + y = 4$. --- 85. Problem: Find coordinates where line $4x + 5y = 20$ intersects X-axis and Y-axis. 86. Explanation: - At X-axis, $y=0$: $$4x = 20 \Rightarrow x = 5.$$ - At Y-axis, $x=0$: $$5y = 20 \Rightarrow y = 4.$$ 87. Final answer: Intercepts are $(5, 0)$ and $(0, 4)$. --- 88. Problem: Given $AP = BP$ and $P(4, 3)$ lies on line segment $AB$ with $A$ on X-axis and $B$ on Y-axis, find $A$ and $B$. 89. Explanation: Let $A = (a, 0)$ and $B = (0, b)$. 90. Since $P$ is midpoint (because $AP = BP$), $$P = \left(\frac{a + 0}{2}, \frac{0 + b}{2}\right) = (4, 3).$$ 91. Equate components: $$\frac{a}{2} = 4 \Rightarrow a = 8,$$ $$\frac{b}{2} = 3 \Rightarrow b = 6.$$ 92. Final answer: $A = (8, 0)$, $B = (0, 6)$. --- 93. Problem: Find $p$ if distance from point $(2, -3)$ to line $px - 4y + 7 = 0$ is 5 units. 94. Explanation: Distance formula from point $(x_0, y_0)$ to line $Ax + By + C = 0$ is $$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}.$$ 95. Substitute: $$5 = \frac{|p \times 2 - 4 \times (-3) + 7|}{\sqrt{p^2 + 16}} = \frac{|2p + 12 + 7|}{\sqrt{p^2 + 16}} = \frac{|2p + 19|}{\sqrt{p^2 + 16}}.$$ 96. Square both sides: $$25 = \frac{(2p + 19)^2}{p^2 + 16} \Rightarrow 25(p^2 + 16) = (2p + 19)^2.$$ 97. Expand: $$25p^2 + 400 = 4p^2 + 76p + 361.$$ 98. Rearrange: $$25p^2 - 4p^2 - 76p + 400 - 361 = 0 \Rightarrow 21p^2 - 76p + 39 = 0.$$ 99. Solve quadratic: $$p = \frac{76 \pm \sqrt{(-76)^2 - 4 \times 21 \times 39}}{2 \times 21} = \frac{76 \pm \sqrt{5776 - 3276}}{42} = \frac{76 \pm \sqrt{2500}}{42} = \frac{76 \pm 50}{42}.$$ 100. Solutions: $$p = \frac{76 + 50}{42} = \frac{126}{42} = 3,$$ $$p = \frac{76 - 50}{42} = \frac{26}{42} = \frac{13}{21}.$$ 101. Final answer: $p = 3$ or $p = \frac{13}{21}$.