Question

Let $O$ be an interior point of $△ABC$ such that $\overline{OA}+2\overline{AB}+3\overline{OC}=\overline{O}$, then the ratio of $△ABC$ to area of $△AOC$ is

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

Answer 1

Answer: (A)

$2$

Refer to the figure. Let O be the origin.

Let $\overline{OA}=\overline{a}$

$\overline{OB}=\overline{b}$

$\overline{OC}=\overline{c}$

Given, $\overline{OA}+2\overline{OB}+3\overline{OC}=0$

$\therefore \overline{a}+2\overline{b}+3\overline{c}=0$

$\therefore \overline{a}=-2\overline{b}-3\overline{c}$ (1)

1) Area of triangle ABC is given by,

$A\left(△ABC\right)=\frac{1}{2}|\overline{AB}\times \overline{BC}|$

$\therefore A\left(△ABC\right)=\frac{1}{2}|(\overline{b}-\overline{a})\times (\overline{c}-\overline{b})|$

From equation (1),

$A\left(△ABC\right)=\frac{1}{2}|(\overline{b}-(-2\overline{b}-3\overline{c}))\times (\overline{c}-\overline{b})|$

$A\left(△ABC\right)=\frac{1}{2}|(\overline{b}+2\overline{b}+3\overline{c})\times (\overline{c}-\overline{b})|$

$A\left(△ABC\right)=\frac{1}{2}|(3\overline{b}+3\overline{c})\times (\overline{c}-\overline{b})|$

$\therefore A\left(△ABC\right)=\frac{3}{2}|(\overline{b}+\overline{c})\times (\overline{c}-\overline{b})|$

$\therefore A\left(△ABC\right)=\frac{3}{2}|(\overline{b}\times \overline{c})-(\overline{b}\times \overline{b})+(\overline{c}\times \overline{c})-(\overline{c}\times \overline{b})|$

But, $\overline{b}\times \overline{b}=0$ and $\overline{c}\times \overline{c}=0$

$\therefore A\left(△ABC\right)=\frac{3}{2}|(\overline{b}\times \overline{c})-(\overline{c}\times \overline{b})|$

$\therefore A\left(△ABC\right)=\frac{3}{2}|(\overline{b}\times \overline{c})+(\overline{b}\times \overline{c})|$

$(\because (\overline{a}\times \overline{b})=-(\overline{b}\times \overline{a}))$

$\therefore A\left(△ABC\right)=\frac{3}{2}\left|2(\overline{b}\times \overline{c})\right|$

$\therefore A\left(△ABC\right)=3\left|(\overline{b}\times \overline{c})\right|$ (2)

2) Area of triangle AOC is given by,

$\therefore \left(△AOC\right)=\frac{1}{2}|\overline{OA}\times \overline{OC}|$

$\therefore \left(△AOC\right)=\frac{1}{2}|\overline{a}\times \overline{c}|$

From equation (1),

$A\left(△AOC\right)=\frac{1}{2}|(-2\overline{b}-3\overline{c})\times \overline{c}|$

$\therefore A\left(△AOC\right)=\frac{1}{2}|-2(\overline{b}\times \overline{c})-3(\overline{c}\times \overline{c})|$

$\therefore A\left(△AOC\right)=\frac{2}{2}\left|(\overline{b}\times \overline{c})\right|$

$\therefore A\left(△AOC\right)=\left|(\overline{b}\times \overline{c})\right|$ (3)

Now, $\frac{A\left(△ABC\right)}{A\left(△AOC\right)}=\frac{3\left|(\overline{b}\times \overline{c})\right|}{\left|(\overline{b}\times \overline{c})\right|}$

$\frac{A\left(△ABC\right)}{A\left(△AOC\right)}=3$