## Precalculus (10th Edition)

$\left(-\frac{1}{2},\frac{\sqrt3}{2}\right)$
If we reflect a point $(x,y)$ about the $y$-axis then, the reflected point is $(-x,y)$. Hence reflecting $\left(\frac{1}{2},\frac{\sqrt3}{2}\right)$ about the $y$-axis gives $\left(-\frac{1}{2},\frac{\sqrt3}{2}\right)$.