Thank you for reporting, we will resolve it shortly
Q.
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
Straight Lines
Solution:
Area of $\triangle O A B=\frac{1}{2}(1)(8)=4$ sq. units
The equation of $O B$ is
$y=\frac{1}{9} x$
Hence, the point $E$ is $(C, C / 9)$.
Now, the area of $\triangle B D E$ is $2 sq$. 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 $C=3$
