1. **State the problem:** We have an isosceles triangle ABC with AB = BC and side AC parallel to the x-axis. Points B and C are given as B(10, 8) and C(16, 2). Point A lies on the x-axis, so its y-coordinate is 0. We need to find the coordinates of point A. 2. **Key properties and formula:** Since AC is parallel to the x-axis, points A and C share the same y-coordinate. Given C(16, 2), and A lies on the x-axis, the y-coordinate of A is 0, so A must be at (x, 0). 3. **Isosceles triangle property:** AB = BC means the distance from A to B equals the distance from B to C. 4. **Calculate BC:** Use the distance formula: $$BC = \sqrt{(16 - 10)^2 + (2 - 8)^2} = \sqrt{6^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$$ 5. **Set AB equal to BC:** Let A = (x, 0). Then, $$AB = \sqrt{(x - 10)^2 + (0 - 8)^2} = \sqrt{(x - 10)^2 + 64}$$ Set equal to BC: $$\sqrt{(x - 10)^2 + 64} = 6\sqrt{2}$$ 6. **Square both sides:** $$(x - 10)^2 + 64 = 72$$ 7. **Simplify:** $$(x - 10)^2 = 72 - 64 = 8$$ 8. **Solve for x:** $$x - 10 = \pm \sqrt{8} = \pm 2\sqrt{2}$$ So, $$x = 10 \pm 2\sqrt{2}$$ 9. **Determine the correct x-coordinate:** Since C is at (16, 2) and A is on the x-axis forming base AC parallel to x-axis, A should be horizontally aligned with C, so x-coordinate of A is 16. But from the isosceles condition, we have two possible x values for A: approximately $10 + 2.828 = 12.828$ or $10 - 2.828 = 7.172$. Since AC is parallel to the x-axis, A and C must have the same y-coordinate, but A lies on the x-axis (y=0) and C is at y=2, so this contradicts the initial assumption. The problem states AC is parallel to the x-axis, so A and C must have the same y-coordinate. Given that, A must be at (16, 0) because it lies on the x-axis. 10. **Check if AB = BC with A(16, 0):** $$AB = \sqrt{(16 - 10)^2 + (0 - 8)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$ $$BC = 6\sqrt{2} \approx 8.485$$ AB is not equal to BC, so A(16, 0) does not satisfy the isosceles condition. 11. **Conclusion:** The only way for AC to be parallel to the x-axis and for AB = BC is for A to have y=0 and x-coordinate such that AB = BC. From step 8, the x-coordinates are $10 \pm 2\sqrt{2}$. Since C is at (16, 2), AC is parallel to the x-axis only if A and C have the same y-coordinate, but A is on the x-axis (y=0), so AC is not parallel to the x-axis if A is at y=0 and C at y=2. Therefore, the problem's description implies A lies on the x-axis (y=0), AC is parallel to the x-axis (so y-coordinates of A and C are equal), so C must also be on the x-axis (y=0). Given C(16, 2), this is contradictory. Assuming the problem means AC is horizontal (parallel to x-axis), then A and C must have the same y-coordinate, so A is at (x, 2). Given that, and A lies on the x-axis, the only way is that y=0 for A and y=2 for C, so AC is not parallel to x-axis. Hence, the problem likely means A lies on the x-axis, and AC is parallel to the x-axis, so A and C have the same y-coordinate 0. Therefore, C must be at (16, 0) instead of (16, 2) for the problem to be consistent. If C is at (16, 0), then: - BC distance: $$BC = \sqrt{(16 - 10)^2 + (0 - 8)^2} = \sqrt{6^2 + (-8)^2} = 10$$ - AB distance with A at (x, 0): $$AB = \sqrt{(x - 10)^2 + (0 - 8)^2} = \sqrt{(x - 10)^2 + 64}$$ Set AB = BC = 10: $$(x - 10)^2 + 64 = 100$$ $$(x - 10)^2 = 36$$ $$x - 10 = \pm 6$$ $$x = 16 \text{ or } 4$$ Since C is at (16, 0), A can be at (4, 0) or (16, 0). But A and C cannot be the same point, so A is at (4, 0). **Final answer:** The coordinates of point A are $(4, 0)$.