1. **Problem statement:** We have an isosceles triangle ABC where sides AB and BC are equal in length, and side AC is parallel to the x-axis. We need to find the coordinates of point A. 2. **Given information:** - AB = BC (isosceles triangle condition) - AC is parallel to the x-axis, so points A and C have the same y-coordinate. 3. **Approach:** - Let the coordinates of points B and C be known or assumed as $B(x_B,y_B)$ and $C(x_C,y_C)$. - Since AC is parallel to the x-axis, $y_A = y_C$. - Use the distance formula for AB and BC: $$AB = BC \implies \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2} = \sqrt{(x_B - x_C)^2 + (y_B - y_C)^2}$$ 4. **Substitute $y_A = y_C$ into the equation:** $$\sqrt{(x_A - x_B)^2 + (y_C - y_B)^2} = \sqrt{(x_B - x_C)^2 + (y_B - y_C)^2}$$ 5. **Square both sides to eliminate the square roots:** $$(x_A - x_B)^2 + (y_C - y_B)^2 = (x_B - x_C)^2 + (y_B - y_C)^2$$ 6. **Simplify:** $$(x_A - x_B)^2 = (x_B - x_C)^2$$ 7. **Take the square root:** $$x_A - x_B = \pm (x_B - x_C)$$ 8. **Solve for $x_A$:** $$x_A = x_B \pm (x_B - x_C)$$ 9. **Interpretation:** - There are two possible values for $x_A$ depending on the sign. - Choose the value that fits the triangle's orientation or additional information if given. **Final answer:** $$\boxed{(x_A, y_A) = \left(x_B \pm (x_B - x_C), y_C\right)}$$ This gives the coordinates of point A based on the coordinates of points B and C and the given conditions.