The coordinates of the vertices of a quadrilateral are $A(0 ; 0), B(16 ; 0), C(8 ; 8), D(0,8)$. Write the equation of the line parallel to $AC$ that divides the area of the quadrilateral in half.
Solution. The quadrilateral is a right-angled trapezoid, whose area is

$$
T=\frac{8 \cdot(16+8)}{2}=96 \text { square units. }
$$

The slope of line $AC$ is 1. The area of the triangle cut off by line $AC$ is: $\frac{16 \cdot 8}{2}=64$ square units; this is larger than half the area of the trapezoid, which means that the area-bisecting line passes "below" $AC$, and cuts a triangle from the trapezoid.

![](https://cdn.mathpix.com/cropped/2024_05_02_c227b760fb9c7ba91740g-1.jpg?height=344&width=566&top_left_y=481&top_left_x=774)

The vertices of the cut-off triangle are $A^{\prime}, B, C^{\prime}$, let $A^{\prime} B=x$, and the height corresponding to side $A^{\prime} B$ be $m=y$. The area of $\triangle A^{\prime} B C^{\prime}$ is $\frac{x y}{2}=48$, according to the condition. From the similarity of triangles $ABC$ and $A^{\prime} B C^{\prime}$, we have $x=2 y$, and from the area, $y=4 \sqrt{3}$ and $x=8 \sqrt{3}$. The slope of the line $A^{\prime} C^{\prime}$ is 1, as it is parallel to $AC$. The coordinates of point $A^{\prime}$ are $(16-8 \sqrt{3} ; 0)$. The equation of line $AC^{\prime}$ is: $y=x+8 \sqrt{3}-16$.