Geometry problem dealing with trapezoid and a quadrilateral

Question

Vertex $A$ of a quadrilateral $ABCD$ lies on the $y$-axis and the rest of the points are $B(-5, -11), C(4, 0)$ and $D(7, 12)$. Define the vertex $A$ such that the quadrilateral $ABCD$ is a trapezoid.

I found this one to be very tricky. I tried to define $A(0, y)$, but I'm not getting anywhere. Any help would be appreciated.

Answer 1

For a trapezium, the two bases are parallel. Meaning AB is parallel to DC. The gradient of DC is $\frac{12-0}{7-4}=4$. The gradient of AB is $\frac{y+11}{5}$

Setting these two expression equal to each other and solving for $y$, should give you the answer (I’ll let you do that bit!)

Answer 2

The problem is not tricky. It is just not well-posed. There are two cases: Either $\overline{AB}$ is parallel to $\overline{CD}$ (in which case $y=4$) or $\overline{AD}$ is parallel to $\overline{BC}$ (in which case $y=31/9$).