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Consider a triangle $\mathrm{ABC}$ in the $\mathrm{xy}$-plane with vertices $\mathrm{A}=(0,0), \mathrm{B}=(1,1)$ and $\mathrm{C}=(9,1)$. If the line $\mathrm{x}=\mathrm{a}$ divides the triangle into two parts of equal area, then a equals
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The correct answer is:
3
area of $P Q C=\frac{1}{2}$ area of $A B C$
$$
\begin{array}{l}
\frac{1}{2}(9-a)\left(1-\frac{a}{9}\right)=\frac{1}{2} \times \frac{1}{2}(8 \times 1) \\
(9-a)^{2}=36 \Rightarrow a=3
\end{array}
$$
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