Question

The line x=c cuts the triangle with corners $\left(0,0\right)$, $\left(1,1\right)$ and $\left(9,1\right)$ into two regions. For the axis the two regions to be 'c' must be

• A. $3$
• B. $\frac{5}{2}$
• C. $\frac{7}{2}$
• D. $3$ or $15$

Answer 1

The correct option is A $3$

$x=c$

$scopeAC=\frac{1}{9}$

$e{q}^{n}isy=\frac{x}{9}$

$A_1=\underset{0}{\overset{1}{\int }}\left(x-\frac{x}{3}\right)dx+\underset{1}{\overset{e}{\int }}\left(x-\frac{x}{9}\right)dx$

${\left(\frac{x^2}{2}-\frac{x^2}{18}\right)}_0^1+{\left(x-\frac{x^2}{18}\right)}_1^c$

$\left(\frac{1}{2}-\frac{1}{18}\right)+c-\frac{c2}{18}-1+\frac{1}{18}$

$\frac{-c^2}{18}+c-\frac{1}{2}$

$A_2=\underset{e}{\overset{9}{\int }}\left(1-\frac{x}{9}\right)dx$

$\left(x-\frac{x^2}{18}\right)e^9$

$9-\frac{81}{18}-c-\frac{c^2}{18}$

$\frac{9}{2}-c+\frac{c^2}{18}$

$A_1=A_2$

$\frac{-c^2}{18}+c-\frac{1}{2}=\frac{9}{2}-c+\frac{c^2}{18}$

$2c-\frac{c^2}{9}-5=0=-c^2+18-45=0$

$\left(c-3\right)\left(c-15\right)=0$

$c=3,c=15\left(c\ne 15\right)$

$c=3$