1. **State the problem:** Find the points of intersection A and B between the line $y=2$ and the parabola $y=\frac{1}{2}x^2$. 2. **Find points of intersection:** Set the two equations equal to find $x$: $$2 = \frac{1}{2}x^2$$ Multiply both sides by 2: $$4 = x^2$$ Take the square root: $$x = \pm 2$$ So the points are: $$A = (2, 2), \quad B = (-2, 2)$$ 3. **Find lengths of segments OA, OB, and AB:** - $O = (0,0)$ - Length $OA = \sqrt{(2-0)^2 + (2-0)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ - Length $OB = \sqrt{(-2-0)^2 + (2-0)^2} = \sqrt{4 + 4} = 2\sqrt{2}$ - Length $AB = \sqrt{(2 - (-2))^2 + (2-2)^2} = \sqrt{4^2 + 0} = 4$ 4. **Find area $S$ of triangle $OAB$:** Use the formula for area of triangle with vertices $O(0,0)$, $A(2,2)$, $B(-2,2)$: $$S = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ Substitute: $$S = \frac{1}{2} |0(2-2) + 2(2-0) + (-2)(0-2)| = \frac{1}{2} |0 + 4 + 4| = \frac{1}{2} \times 8 = 4$$ **Final answers:** - Points of intersection: $A(2,2)$ and $B(-2,2)$ - Lengths: $OA = OB = 2\sqrt{2}$, $AB = 4$ - Area of triangle $OAB = 4$