Question

If $2x^2$ $-5xy+2y^2=0$ represents two sides of a triangle whose centroid is $(2,3)$, then the area of that triangle (in sq. units) is

• A. $12$
• B. $6$
• C. $4$
• D. $8$

Answer 1

Answer: (B)

$6$

The given pair of lines is,

$2x^2-5xy+2y^2=0$
Put $t=\frac{x}{y},$
$2t^2-5t+2=0$
$t=\frac{5\pm \sqrt{25-16}}{4}$
$=\frac{5\pm 3}{6}$
$t=\frac{4}{2},\frac{1}{2}$
$\therefore y=\frac{x}{2}$ & $y=2x$
One of the vertices will be $(0,0)$. Let the other two vertices be $(h,\frac{h}{2})$ and $(k,2k)$
$\therefore \frac{h+k}{3}=2\Rightarrow h+k=6$ ------(1)
and $2k+\frac{h}{2}=9$ -----(2)
from 1 & 2,
$k=4$ & $h=2$
So, area of $\Delta ABC=\frac{1}{2}(8\times 2-4\times 1)$
$=\frac{1}{2}[16-4]$
$=\frac{12}{2}$
Area $=6$