If 2x^2-5xy+2y^2=0 represents two sides of a | vedclass

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(C) The given equation is $2x^2-5xy+2y^2=0$. Factoring the equation: $(2x-y)(x-2y)=0$. Thus,the two sides are $y=2x$ and $y=\frac{1}{2}x$. Let the ver

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$x+y-3=0$

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(C) The given equation is $2x^2-5xy+2y^2=0$. Factoring the equation: $(2x-y)(x-2y)=0$. Thus,the two sides are $y=2x$ and $y=\frac{1}{2}x$. Let the vertices of the triangle be $O(0,0)$,$A(a, 2a)$,and $B(2b, b)$. The centroid is given as $(1,1)$. Using the centroid formula: $\frac{0+a+2b}{3}=1 \Rightarrow a+2b=3$ $(i)$ and $\frac{0+2a+b}{3}=1 \Rightarrow 2a+b=3$ (ii). Solving equations $(i)$ and (ii),we get $a=1$ and $b=1$. So,the vertices are $O(0,0)$,$A(1,2)$,and $B(2,1)$. The third side passes through $A(1,2)$ and $B(2,1)$. The equation of the line passing through $(1,2)$ and $(2,1)$ is $y-2 = \frac{1-2}{2-1}(x-1)$,which simplifies to $y-2 = -1(x-1) \Rightarrow x+y-3=0$.

If $2x^2-5xy+2y^2=0$ represents two sides of a triangle whose centroid is $(1,1)$,then the equation of the third side is

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