1. **State the problem:** We are given a rectangle ABCD with points A(3,4) and B(5,4). One of the diameters of the circle circumscribing this rectangle is given by the line $4y = x + 7$. We need to find the area of the rectangle. 2. **Understand the problem:** The circle circumscribing the rectangle is the circle passing through all four vertices of the rectangle. The diameter of this circle is a line given by $4y = x + 7$. 3. **Find the center of the circle:** The diameter line $4y = x + 7$ can be rewritten as $x - 4y + 7 = 0$. The center of the circle lies on this diameter line and is the midpoint of the diagonal of the rectangle. 4. **Find coordinates of points A and B:** Given A(3,4) and B(5,4). 5. **Find the length and direction of AB:** Since both points have the same y-coordinate, AB is horizontal with length $|5 - 3| = 2$. 6. **Find the coordinates of point D:** Since ABCD is a rectangle, the diagonal AC is perpendicular to AB. The diagonal AC is the diameter of the circle. 7. **Find midpoint M of diagonal AC:** The midpoint M lies on the diameter line $4y = x + 7$. 8. **Find coordinates of C and D:** Since AB is horizontal, the other side AD is vertical. The y-coordinate of D is different from A and B. 9. **Calculate the length of AD:** Let the length of AD be $h$. Then the area of the rectangle is $2 imes h$. 10. **Find the center M:** The midpoint of diagonal AC is the center of the circle and lies on $4y = x + 7$. 11. **Calculate the coordinates of C:** Since ABCD is a rectangle, vector AC is diagonal. Vector AC = vector AB + vector AD. 12. **Use the midpoint formula:** If A is (3,4) and C is (x_c,y_c), then midpoint M is $\left(\frac{3+x_c}{2}, \frac{4+y_c}{2}\right)$. 13. **Since M lies on $4y = x + 7$, substitute:** $$4 \times \frac{4 + y_c}{2} = \frac{3 + x_c}{2} + 7$$ $$2(4 + y_c) = \frac{3 + x_c}{2} + 7$$ 14. **Simplify:** $$8 + 2y_c = \frac{3 + x_c}{2} + 7$$ $$8 + 2y_c - 7 = \frac{3 + x_c}{2}$$ $$1 + 2y_c = \frac{3 + x_c}{2}$$ 15. **Multiply both sides by 2:** $$2 + 4y_c = 3 + x_c$$ $$x_c = 4y_c - 1$$ 16. **Since ABCD is a rectangle, vector AC is perpendicular to vector BD. But since AB is horizontal, AD is vertical, so C has coordinates (5, y_c) or (3, y_c + h). We can find C by using the fact that the diagonal AC is the diameter.** 17. **Calculate length of diagonal AC:** $$AC = \sqrt{(x_c - 3)^2 + (y_c - 4)^2}$$ 18. **The center M is midpoint of AC:** $$M = \left(\frac{3 + x_c}{2}, \frac{4 + y_c}{2}\right)$$ 19. **Since M lies on $4y = x + 7$, substitute $x = \frac{3 + x_c}{2}$ and $y = \frac{4 + y_c}{2}$:** $$4 \times \frac{4 + y_c}{2} = \frac{3 + x_c}{2} + 7$$ 20. **From step 15, we have $x_c = 4y_c - 1$. Substitute into the midpoint formula:** $$M_x = \frac{3 + 4y_c - 1}{2} = \frac{2 + 4y_c}{2} = 1 + 2y_c$$ $$M_y = \frac{4 + y_c}{2}$$ 21. **Substitute into the diameter line equation:** $$4 M_y = M_x + 7$$ $$4 \times \frac{4 + y_c}{2} = 1 + 2y_c + 7$$ $$2(4 + y_c) = 8 + 2y_c$$ $$8 + 2y_c = 8 + 2y_c$$ 22. **This is true for all $y_c$, so we need another condition. Since ABCD is a rectangle, the sides are perpendicular. AB is horizontal, so AD is vertical. So the x-coordinate of D is the same as A, and the y-coordinate of D is different.** 23. **Coordinates of D are (3, y_d). Since B is (5,4), and AB is horizontal, AD is vertical. So the length of AD is $|y_d - 4|$.** 24. **The diagonal BD is also a diameter of the circle. The midpoint of BD is also the center M.** 25. **Calculate midpoint of BD:** $$M = \left(\frac{5 + 3}{2}, \frac{4 + y_d}{2}\right) = (4, \frac{4 + y_d}{2})$$ 26. **Since M lies on $4y = x + 7$, substitute:** $$4 \times \frac{4 + y_d}{2} = 4 + 7$$ $$2(4 + y_d) = 11$$ $$8 + 2 y_d = 11$$ $$2 y_d = 3$$ $$y_d = 1.5$$ 27. **Length of AD:** $$|y_d - 4| = |1.5 - 4| = 2.5$$ 28. **Length of AB:** $$|5 - 3| = 2$$ 29. **Area of rectangle:** $$\text{Area} = \text{length} \times \text{width} = 2 \times 2.5 = 5$$ **Final answer:** The area of the rectangle is $5$ square units.