1. **State the problem:** Determine if the point $\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ lies on the unit circle. 2. **Recall the unit circle equation:** The unit circle is defined by the equation $$x^2 + y^2 = 1$$ where the radius is 1 and the center is at the origin $(0,0)$. 3. **Substitute the point coordinates into the equation:** $$\left(-\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = ?$$ 4. **Calculate each term:** $$\left(-\frac{\sqrt{2}}{2}\right)^2 = \frac{(\sqrt{2})^2}{2^2} = \frac{2}{4} = \frac{1}{2}$$ $$\left(\frac{\sqrt{2}}{2}\right)^2 = \frac{(\sqrt{2})^2}{2^2} = \frac{2}{4} = \frac{1}{2}$$ 5. **Sum the squares:** $$\frac{1}{2} + \frac{1}{2} = 1$$ 6. **Interpretation:** Since the sum equals 1, the point satisfies the unit circle equation. 7. **Conclusion:** The point $\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ lies on the unit circle. **Answer:** B) Yes, because the coordinates $\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ satisfy the equation $x^2 + y^2 = 1$.