Rhombus Coordinates 933068

1. **State the problem:** We are given a rhombus ABCD with points $B(2, 10)$ and $D(6, 2)$, and $A$ lies on the x-axis. We need to find the coordinates of the midpoint $M$ of $BD$, and the coordinates of points $A$ and $C$. 2. **Recall properties of a rhombus:** - All sides are equal in length. - The diagonals bisect each other at right angles. - The midpoint of diagonal $BD$ is $M$, which is also the midpoint of diagonal $AC$. 3. **Find midpoint $M$ of $BD$:** $$M = \left(\frac{x_B + x_D}{2}, \frac{y_B + y_D}{2}\right) = \left(\frac{2 + 6}{2}, \frac{10 + 2}{2}\right) = (4, 6)$$ 4. **Find coordinates of $A$:** Since $A$ lies on the x-axis, its $y$-coordinate is 0. Let $A = (x_A, 0)$. 5. **Use midpoint property of diagonals:** Since $M$ is midpoint of $AC$, $$M = \left(\frac{x_A + x_C}{2}, \frac{y_A + y_C}{2}\right) = (4, 6)$$ 6. **Express $C$ in terms of $A$:** From midpoint formula: $$\frac{x_A + x_C}{2} = 4 \implies x_C = 8 - x_A$$ $$\frac{0 + y_C}{2} = 6 \implies y_C = 12$$ So, $C = (8 - x_A, 12)$. 7. **Use equal side lengths:** Calculate length $BD$: $$BD = \sqrt{(6 - 2)^2 + (2 - 10)^2} = \sqrt{4^2 + (-8)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$$ Since all sides are equal, $AB = BD = 4\sqrt{5}$. 8. **Express $AB$ in terms of $x_A$:** $$AB = \sqrt{(2 - x_A)^2 + (10 - 0)^2} = \sqrt{(2 - x_A)^2 + 100}$$ Set equal to $4\sqrt{5}$: $$\sqrt{(2 - x_A)^2 + 100} = 4\sqrt{5}$$ Square both sides: $$(2 - x_A)^2 + 100 = 16 \times 5 = 80$$ Simplify: $$(2 - x_A)^2 = 80 - 100 = -20$$ This is impossible (negative number under square), so check calculation. 9. **Recalculate $BD$ length:** $$BD = \sqrt{(6 - 2)^2 + (2 - 10)^2} = \sqrt{4^2 + (-8)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$$ Correct. 10. **Re-express $AB$ length:** $$AB = \sqrt{(2 - x_A)^2 + (10 - 0)^2} = \sqrt{(2 - x_A)^2 + 100}$$ Set equal to $4\sqrt{5}$: $$\sqrt{(2 - x_A)^2 + 100} = 4\sqrt{5}$$ Square both sides: $$(2 - x_A)^2 + 100 = 80$$ $$(2 - x_A)^2 = -20$$ Again negative, so no real solution. 11. **Check if $A$ lies on x-axis to the left of y-axis:** Try $A = (x_A, 0)$ with $x_A < 0$. 12. **Use vector approach:** The diagonals bisect each other at $M$ and are perpendicular. Vector $\overrightarrow{BD} = (6 - 2, 2 - 10) = (4, -8)$. Since diagonals are perpendicular: $$\overrightarrow{AC} \cdot \overrightarrow{BD} = 0$$ Let $A = (x_A, 0)$ and $C = (8 - x_A, 12)$. Vector $\overrightarrow{AC} = (8 - x_A - x_A, 12 - 0) = (8 - 2x_A, 12)$. Dot product: $$(8 - 2x_A, 12) \cdot (4, -8) = 0$$ Calculate: $$(8 - 2x_A) \times 4 + 12 \times (-8) = 0$$ $$32 - 8x_A - 96 = 0$$ $$-8x_A - 64 = 0$$ $$-8x_A = 64$$ $$x_A = -8$$ 13. **Find $A$ and $C$ coordinates:** $$A = (-8, 0)$$ $$C = (8 - (-8), 12) = (16, 12)$$ 14. **Verify side length $AB$:** $$AB = \sqrt{(2 - (-8))^2 + (10 - 0)^2} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}$$ 15. **Verify side length $BD$:** $$BD = 4\sqrt{5} \approx 8.944$$ 16. **Verify side length $AB$ and $BD$ are equal:** $10\sqrt{2} \approx 14.142$, not equal to $8.944$. 17. **Check if rhombus condition is met:** Since $AB$ and $BD$ are not equal, $BD$ is a diagonal, not a side. 18. **Recall rhombus diagonals bisect each other and are perpendicular, but sides are equal.** We must find $A$ such that $AB = BC = CD = DA$. 19. **Use midpoint $M$ and vector approach:** Since $M$ is midpoint of $BD$ and $AC$, $$M = \frac{A + C}{2} = (4, 6)$$ So, $$C = 2M - A = (8 - x_A, 12)$$ 20. **Use vector $AB$ and $BC$ equal length:** $$AB = \sqrt{(2 - x_A)^2 + (10 - 0)^2}$$ $$BC = \sqrt{(8 - x_A - 2)^2 + (12 - 10)^2} = \sqrt{(6 - x_A)^2 + 2^2}$$ Set equal: $$(2 - x_A)^2 + 100 = (6 - x_A)^2 + 4$$ Expand: $$4 - 4x_A + x_A^2 + 100 = 36 - 12x_A + x_A^2 + 4$$ Simplify: $$104 - 4x_A = 40 - 12x_A$$ $$104 - 4x_A - 40 + 12x_A = 0$$ $$64 + 8x_A = 0$$ $$8x_A = -64$$ $$x_A = -8$$ 21. **Coordinates:** $$A = (-8, 0)$$ $$C = (8 - (-8), 12) = (16, 12)$$ 22. **Check side lengths:** $$AB = \sqrt{(2 - (-8))^2 + (10 - 0)^2} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}$$ $$BC = \sqrt{(16 - 2)^2 + (12 - 10)^2} = \sqrt{14^2 + 2^2} = \sqrt{196 + 4} = \sqrt{200} = 10\sqrt{2}$$ 23. **Conclusion:** - Midpoint $M = (4, 6)$ - $A = (-8, 0)$ - $C = (16, 12)$ These satisfy the rhombus properties. **Final answer:** $$M = (4, 6), \quad A = (-8, 0), \quad C = (16, 12)$$