Question

The line $x=C$ cuts the triangle with vertices $(0,0),(1,1)$, and $(9,1)$ into two regions. For the areas of the two regions to be the same, $C$ must be equal to

Answer 1

Answer: 3

Area of $△OAB=\frac{1}{2}(1)(8)=4$ sq. units

The equation of $OB$ is
$y=\frac{1}{9}x$
Hence, the point $E$ is $(C,C/9)$.
Now, the area of $△BDE$ is $2sq$. units. Therefore, or $\frac{1}{2}\left(1-\frac{C}{9}\right)(9-C)=2$
or $(9-C{)}^2=36$
or $9-C=\pm 6$
or $C=3$