1. **Calculate the area of the rectangle with midpoints M(9,11) and N(5,8):** Given M and N are midpoints of two sides of the rectangle, and the sides are parallel to the axes. - M is midpoint of the top side, so its y-coordinate is the same as the top side's y. - N is midpoint of the left side, so its x-coordinate is the same as the left side's x. Let the rectangle's corners be (x_1,y_1), (x_2,y_1), (x_2,y_2), (x_1,y_2) with $x_1 < x_2$ and $y_1 < y_2$. Since M is midpoint of top side: $$M_x = \frac{x_1 + x_2}{2} = 9, \quad M_y = y_2 = 11$$ Since N is midpoint of left side: $$N_x = x_1 = 5, \quad N_y = \frac{y_1 + y_2}{2} = 8$$ From $N_x = x_1 = 5$ and $M_x = \frac{5 + x_2}{2} = 9$: $$\frac{5 + x_2}{2} = 9 \implies 5 + x_2 = 18 \implies x_2 = 13$$ From $M_y = y_2 = 11$ and $N_y = \frac{y_1 + 11}{2} = 8$: $$\frac{y_1 + 11}{2} = 8 \implies y_1 + 11 = 16 \implies y_1 = 5$$ Area of rectangle: $$\text{Area} = (x_2 - x_1)(y_2 - y_1) = (13 - 5)(11 - 5) = 8 \times 6 = 48$$ 2. **Calculate the area of the square with center (14,9) and vertex (16,11):** - The center is midpoint of the diagonal. - Distance from center to vertex is half the diagonal length. Calculate half diagonal length: $$d = \sqrt{(16 - 14)^2 + (11 - 9)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$$ Full diagonal length: $$D = 2d = 4\sqrt{2}$$ Area of square: $$\text{Area} = \left(\frac{D}{\sqrt{2}}\right)^2 = \left(\frac{4\sqrt{2}}{\sqrt{2}}\right)^2 = 4^2 = 16$$ 3. **Find coordinates of point A in isosceles triangle ABC with AB = BC and AC parallel to x-axis:** Given: - B(10,8), C(16,2) - AC parallel to x-axis means $y_A = y_C = 2$ Since AB = BC, distances are equal: $$AB = BC$$ Calculate $BC$: $$BC = \sqrt{(16 - 10)^2 + (2 - 8)^2} = \sqrt{6^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$$ Let $A = (x_A, 2)$. Calculate $AB$: $$AB = \sqrt{(x_A - 10)^2 + (2 - 8)^2} = \sqrt{(x_A - 10)^2 + (-6)^2} = \sqrt{(x_A - 10)^2 + 36}$$ Set equal to $BC$: $$\sqrt{(x_A - 10)^2 + 36} = 6\sqrt{2}$$ Square both sides: $$(x_A - 10)^2 + 36 = 72$$ Simplify: $$(x_A - 10)^2 = 36$$ Take square root: $$x_A - 10 = \pm 6$$ So: $$x_A = 10 + 6 = 16 \quad \text{or} \quad x_A = 10 - 6 = 4$$ Since $A$ and $C$ are distinct points on the same horizontal line, and $C$ is at (16,2), $A$ must be at (4,2). 4. **Find coordinates of G in rectangle DEFG with center C(6,7) and E(20,23):** Rectangle corners: D(top-left), E(top-right), F(bottom-right), G(bottom-left) Center $C$ is midpoint of diagonal $EG$: $$C = \left(\frac{x_E + x_G}{2}, \frac{y_E + y_G}{2}\right) = (6,7)$$ Given $E = (20,23)$, solve for $G = (x_G, y_G)$: $$\frac{20 + x_G}{2} = 6 \implies 20 + x_G = 12 \implies x_G = -8$$ $$\frac{23 + y_G}{2} = 7 \implies 23 + y_G = 14 \implies y_G = -9$$ So: $$G = (-8, -9)$$