1. Problem: Find the image coordinates of points P(6, -2), Q(-9, -3), and R(-2, 5) after translations: a) by T(5, 1) b) by T(-3, 2) 2. Problem: The image of point 2 under translation T(5, 7) is (-3, 4). Find the original coordinates of point 2. 3. Problem: The image of point G(3, -5) under a translation is G'(-9, 0). Find the translation vector components. 4. Problem: Given a rhombus with points K(4, 4), L(5, 0), N(5, 8), find the image of point M under translation T(-2, 1) followed by T(5, 9). 5. Problem: Under translation T(3, a) followed by T(b, 5), point P(6, -18) maps to P'(2, -9). Find values of a and b. --- 1. a) Translation formula: $$P' = (x + t_x, y + t_y)$$ For P(6, -2) and T(5, 1): $$P' = (6 + 5, -2 + 1) = (11, -1)$$ For Q(-9, -3): $$Q' = (-9 + 5, -3 + 1) = (-4, -2)$$ For R(-2, 5): $$R' = (-2 + 5, 5 + 1) = (3, 6)$$ b) For T(-3, 2): For P(6, -2): $$P' = (6 - 3, -2 + 2) = (3, 0)$$ For Q(-9, -3): $$Q' = (-9 - 3, -3 + 2) = (-12, -1)$$ For R(-2, 5): $$R' = (-2 - 3, 5 + 2) = (-5, 7)$$ 2. Given image point 2' = (-3, 4) and translation T(5, 7), original point 2 = (x, y). Using translation formula: $$2' = (x + 5, y + 7) = (-3, 4)$$ Solve for x and y: $$x + 5 = -3 \Rightarrow x = -3 - 5 = -8$$ $$y + 7 = 4 \Rightarrow y = 4 - 7 = -3$$ Original point 2 is $$(-8, -3)$$. 3. Given G(3, -5) and G'(-9, 0), find translation vector $$T = (t_x, t_y)$$. Using translation formula: $$G' = (3 + t_x, -5 + t_y) = (-9, 0)$$ Solve for $$t_x$$ and $$t_y$$: $$3 + t_x = -9 \Rightarrow t_x = -9 - 3 = -12$$ $$-5 + t_y = 0 \Rightarrow t_y = 0 + 5 = 5$$ Translation vector is $$T = (-12, 5)$$. 4. Given rhombus points K(4, 4), L(5, 0), N(5, 8), find M and its image after translations. Since KLMN is a rhombus, M can be found using the midpoint formula: Midpoint of diagonal KN equals midpoint of diagonal LM. Midpoint KN: $$\left(\frac{4 + 5}{2}, \frac{4 + 8}{2}\right) = \left(\frac{9}{2}, 6\right) = (4.5, 6)$$ Let M = (x, y), midpoint LM: $$\left(\frac{5 + x}{2}, \frac{0 + y}{2}\right) = (4.5, 6)$$ Solve: $$\frac{5 + x}{2} = 4.5 \Rightarrow 5 + x = 9 \Rightarrow x = 4$$ $$\frac{0 + y}{2} = 6 \Rightarrow y = 12$$ So, $$M = (4, 12)$$. Apply translation T(-2, 1): $$M' = (4 - 2, 12 + 1) = (2, 13)$$ Then translation T(5, 9): $$M'' = (2 + 5, 13 + 9) = (7, 22)$$ Final image of M is $$M'' = (7, 22)$$. 5. Given translations T(3, a) and T(b, 5), and P(6, -18) maps to P'(2, -9). Combined translation vector: $$T_{total} = (3 + b, a + 5)$$ Using translation formula: $$P' = (6 + 3 + b, -18 + a + 5) = (2, -9)$$ Solve for a and b: $$6 + 3 + b = 2 \Rightarrow 9 + b = 2 \Rightarrow b = 2 - 9 = -7$$ $$-18 + a + 5 = -9 \Rightarrow a - 13 = -9 \Rightarrow a = -9 + 13 = 4$$ Values are $$a = 4$$ and $$b = -7$$. --- Final answers: 1. a) P'(11, -1), Q'(-4, -2), R'(3, 6) b) P'(3, 0), Q'(-12, -1), R'(-5, 7) 2. Point 2 original coordinates: (-8, -3) 3. Translation vector: (-12, 5) 4. M = (4, 12), M'' = (7, 22) 5. a = 4, b = -7