Example 6 As shown in Figure 4, the line $y=-\frac{\sqrt{3}}{3} x +1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. An isosceles right triangle $\triangle ABC$ is constructed in the first quadrant with $AB$ as one of the legs, and $\angle BAC=90^{\circ}$. If there is a point $P\left(a, \frac{1}{2}\right)$ in the second quadrant, and the area of $\triangle ABP$ is equal to the area of the right triangle $\triangle ABC$, find the value of $a$.
(2001, National Junior High School Mathematics Competition, Tianjin Preliminary Contest)
Solution: From the given, we have $A(\sqrt{3}, 0) 、 B(0,1), O A=$ $\sqrt{3} 、 O B=1$.
Thus, $A B=\sqrt{O A^{2}+O B^{2}}=2$.
Therefore, $S_{\triangle A B C}=\frac{1}{2} \times 2 \times 2=2$.
Connect $P O$. Then
$$
\begin{array}{l}
S_{\triangle A B P}=S_{\triangle P B O}+S_{\triangle A B O}-S_{\triangle A P O} \\
=\frac{1}{2} \times 1 \times|a|+\frac{1}{2} \times 1 \times \sqrt{3}-\frac{1}{2} \times \sqrt{3} \times \frac{1}{2} \\
=-\frac{a}{2}+\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{4} .
\end{array}
$$

Since $S_{\triangle A B P}=S_{\triangle B B C}$,
we have $-\frac{a}{2}+\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{4}=2$. Therefore, $a=\frac{\sqrt{3}-8}{2}$.