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

• Correct answer
  A

  3
• B

  $3.5$
• C

  4
• D

  $4.5$

Solution

area of $P Q C=\frac{1}{2}$ area of $A B C$ $$ \begin{array}{l} \frac{1}{2}(9-a)\left(1-\frac{a}{9}\right)=\frac{1}{2} \times \frac{1}{2}(8 \times 1) \\ (9-a)^{2}=36 \Rightarrow a=3 \end{array} $$

Hence, the correct answer is option: 'A'.