Question

Let $D$ and $E$ be the midpoints of the sides $AC$ and $BC$ of a triangle $ABC$ respectively. If $O$ is an interior point of the triangle $ABC$ such that $OA+2OB+3OC=0$. then the area (in sq. units) of the triangle $ODE$ is

• A. 6
• B. 5
• C. $\frac{3}{4}$
• D. 0

Answer 1

Answer: (D)

Given,
$OA+2OB+3OC=O$

$\Rightarrow a+2b+3c=0\dots$(i)
Now, area of $\Delta ODE$
$=\frac{1}{2}|OD\times OE|=\frac{1}{2}\left|\left(\frac{a+c}{2}\right)\times \left(\frac{b+c}{2}\right)\right|$
$=\frac{1}{8}|a\times b+a\times c+c\times b+c\times c|$
$=\frac{1}{8}|a\times b+a\times c+c\times b|\dots$(ii)
Multiply Eq. (i) by (b),
$a\times b+2(b\times b)+3(c\times b)=0$
$\Rightarrow a\times b+3(c\times b)=0\dots$(iii)
Again multiply Eq. (i) by (c),
$a\times c+2(b\times c)+3(c\times c)=0$
$\Rightarrow a\times c+2(b\times c)=0\dots$ (iv)
Adding Eqs. (iii) and (iv), we get
$a\times b+3(c\times b)+a\times c+2(b\times c)=0$
$\Rightarrow a\times b+3(c\times b)+a\times c-2(c\times b)=0$
$\Rightarrow a\times b+a\times c+c\times b=0\dots$ (v)
$\therefore$ Area of $\Delta ODE=0$ [By Eq. (v)]