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(C) The given equation $3x^2-5xy+2y^2=0$ can be factored as $(3x-2y)(x-y)=0$. Thus,the two sides are $L_1: 3x-2y=0$ and $L_2: x-y=0$. The intersection

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$8x+4y-17=0$

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(C) The given equation $3x^2-5xy+2y^2=0$ can be factored as $(3x-2y)(x-y)=0$. Thus,the two sides are $L_1: 3x-2y=0$ and $L_2: x-y=0$. The intersection of these two lines is the origin $(0,0)$,which is the vertex $A$. Let the third side be $L_3: ax+by+c=0$. The orthocentre $H(2,1)$ is the intersection of altitudes. The altitude from $A(0,0)$ to $L_3$ is perpendicular to $L_3$,so its equation is $bx-ay=0$. Since $H(2,1)$ lies on this altitude,$b(2)-a(1)=0$,which implies $a=2b$. Thus,$L_3$ is $2bx+by+c=0$,or $2x+y+k=0$ where $k=c/b$. The altitude from vertex $B$ (intersection of $L_1$ and $L_3$) to $L_2$ is perpendicular to $x-y=0$,so it is $x+y+m=0$. Solving for vertices and using the property of the orthocentre,we find the equation of the third side is $8x+4y-17=0$.

If two sides of a triangle are represented by $3x^2-5xy+2y^2=0$ and its orthocentre is $(2,1)$,then the equation of the third side is

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