Shortest distance from origin

Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = 32$. Base $BC = 8 \implies \frac{1}{2} \times 8 \times h = 32 \implies h = 8$.

Since side $BC$ lies on the vertical line $x = 4$, the perpendicular distance of $A(x_A, y_A)$ from the line $x = 4$ is 8. So $x_A = 4 + 8 = 12$ or $x_A = 4 - 8 = -4$.

The distance of $A(x_A, y_A)$ from origin $(0,0)$ is $\sqrt{x_A^2 + y_A^2}$. To minimize this distance, we choose $y_A = 0$ and $x_A = -4$ (which gives distance $|-4| = 4$). Thus, the shortest possible distance is 4 units.