If $2 x^2-5 x y+2 y^2=0$ represents two sides of a triangle whose centroid is $(1,1)$, then the equation of…
If $2 x^2-5 x y+2 y^2=0$ represents two sides of a triangle whose centroid is $(1,1)$, then the equation of the third side is
$x+y+3=0$
$x-y-3=0$
$x+y-3=0$
$x-y+3=0$
Solution
Given equation are $2 x^2-5 x y+2 y^2=0$
$\Rightarrow(x-2 y)(2 x-y)=0 \Rightarrow x=2 y \& y=2 x$
Let $A(a, 2 a) \& B(2 b, b)$ and one point $(0,0)$ so
$\frac{a+2 b+0}{3}=1 \Rightarrow a+2 b=3$ ... (i)
$\& \frac{2 a+b+0}{3}=1 \Rightarrow 2 a+b=3$ ... (ii)
on solving we get, $\mathrm{a}=1, \mathrm{~b}=1$
So equation of line passing through $\mathrm{A}(1,2)$ \& $\mathrm{B}(2,1)$ is
$\begin{aligned}
& y-2=\frac{1-2}{2-1}(x-1) \\
& \Rightarrow y-2=-x+1 \Rightarrow x+y-3=0
\end{aligned}$