A triangle's vertices are: $A(0 ; 0), B(4 ; 1), C(4 ; 0)$. Let's write the equation of the line that passes through the point $D(0 ;-1)$ and bisects the area of the triangle $ABC$.
From the coordinates, we can read that the triangle $ABC$ is a right-angled triangle, and furthermore, that the quadrilateral $ABCD$ is a parallelogram, with two sides being parallel and of equal length. It follows that the diagonals of $ABCD$ bisect each other at point $F$. The line $DB=e$ thus bisects the area of triangle $ABC$. The equation of line $e$ can be easily written; its slope is: $m=\frac{1}{2}$; it passes through point $D$, so its equation is: $y=\frac{1}{2} x-1$.

Question: Can there exist another line that passes through point $D$ and bisects the area of triangle $ABC$?

If such a line existed, it would intersect either side $AB$ or side $BC$ of the triangle. Let $f$ be a line passing through $D$ that intersects the side $AB$ of the triangle.

The line $f$ divides the triangle $ABF$ into two parts, one of which (which we have shaded on the diagram) connects to the triangle $BFC$, thereby increasing its area, which means that line $f$ cannot divide the triangle $ABC$ into two equal areas. Similarly, it can be shown that no line starting from $D$ exists that intersects the side $BC$ of the triangle and bisects the area of triangle $ABC$.