Question

The line $x=c$ cuts the triangle with corners $(0,0);(1,1)$ and $(9,1)$ into two regions. For the area of the two regions to be the same $c$ must be equal to

• A. $512$
• B. $3$
• C. $7/2$
• D. $3$ or $15$

Answer 1

Answer: (D)

$3$ or $15$

$\begin{array}{c}SlopeAC=\frac{1}{9}slope\\ y=\frac{1}{9}x\\ AB\to y=x\\ A_1={\int }_0^1\left(x-\frac{x}{9}\right)dx+{\int }_1^c\left(1-\frac{x}{9}\right)dx\\ {\left[\frac{x^2}{2}-\frac{x^2}{10}\right]}_0^1+{\left[x-\frac{x^2}{18}\right]}_1^c\\ ={\left[\frac{8x^2}{18}\right]}_0^1+{\left[x-\frac{x^2}{10}\right]}_1^c\\ ={\left[\frac{4x^2}{9}\right]}_0^1+{\left[x-\frac{x^2}{10}\right]}_1^c\\ =\frac{4}{9}+c-\frac{c^2}{10}-1+\frac{1}{18}\\ =\frac{8}{16}+\frac{1}{18}-1+c-\frac{c^2}{18}\\ =c-\frac{c^2}{18}-\frac{1}{2}\\ A_2={\int }_c^9\left(1-\frac{x}{9}\right)dx={\left[x-\frac{x^2}{10}\right]}_c^9\\ =9-\frac{81}{18}-c+\frac{c^2}{18}=9-\frac{9}{2}+\frac{c^2}{18}-c\\ 2x-\frac{2c^2}{18}=\frac{9}{2}+\frac{1}{2}=5\\ c^2-18c+45=0\\ c^2-15c-3c+45=0\\ c=3,15\end{array}$