This cannot be the median

Define points $A\left(0,0\right)$ , $B\left(1,1\right)$ and $C\left(9,1\right)$ . Let the equation be $x = a$ for some $a \in \left(1,9\right)$ . Also, define $P\left(a,1\right)$ and $Q\left(a,\dfrac{a}{9}\right)$ as the intersections of the line $x = a$ to the lines containing the points $B$ , $C$ , and $A$ , $C$ , respectively. Finally, define point $D\left(0,1\right)$ . Notice that $DPQA$ is an isosceles trapezoid since $DA \parallel PQ$ . Additionally, $\triangle DBA$ is an isosceles triangle with an area $\dfrac{1}{2}$ . Thus, the area of $BPQA$ is $\dfrac{a\left(1 + 1 - \dfrac{a}{9}\right)}{2} - \dfrac{1}{2} = \dfrac{a\left(2 - \dfrac{a}{9}\right)}{2} - \dfrac{1}{2}.$ On the other hand, the area of $\triangle PCQ$ is $\dfrac{\left(1 - \dfrac{a}{9}\right)\left(9 - a\right)}{2}.$ Solving for $a$ in the equation $\dfrac{a\left(2 - \dfrac{a}{9}\right)}{2} - \dfrac{1}{2} = \dfrac{\left(1 - \dfrac{a}{9}\right)\left(9 - a\right)}{2}$ gives $a = \lbrace3,15\rbrace$ and so the desired equation is $\boxed{x = 3}$ .