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(C) The given equation of two sides is $3x^2-5xy+2y^2=0$. Factoring this,we get $(3x-2y)(x-y)=0$. So,the two sides are $L_1: 3x-2y=0$ and $L_2: x-y=0$

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$8x+4y-17=0$

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(C) The given equation of two sides is $3x^2-5xy+2y^2=0$. Factoring this,we get $(3x-2y)(x-y)=0$. So,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 one vertex of the triangle. Let the third side be $ax+by+c=0$. The orthocentre $H(2,1)$ is the intersection of altitudes. The altitude from vertex $(0,0)$ to the third side $ax+by+c=0$ is perpendicular to it,so its equation is $bx-ay=0$. Since $H(2,1)$ lies on this altitude,$b(2)-a(1)=0$,which implies $a=2b$. The third side equation becomes $2bx+by+c=0$,or $2x+y+k=0$ where $k=c/b$. The altitude from the vertex where $L_1$ and $L_2$ meet the third side is perpendicular to $L_1$ $(3x-2y=0)$. The slope of $L_1$ is $3/2$,so the altitude slope is $-2/3$. The line passing through $(2,1)$ with slope $-2/3$ is $y-1 = -2/3(x-2)$,which simplifies to $2x+3y-7=0$. Solving for the intersection of $2x+3y-7=0$ and $3x-2y=0$ gives the vertex $A(2,3)$. Similarly,the altitude from the other vertex is perpendicular to $L_2$ $(x-y=0)$,slope $1$,so altitude slope is $-1$. Line through $(2,1)$ with slope $-1$ is $y-1 = -1(x-2)$,i.e.,$x+y-3=0$. Intersection with $x-y=0$ gives vertex $B(3/2, 3/2)$. The third side passes through $A(2,3)$ and $B(3/2, 3/2)$. The equation is $y-3 = \frac{3/2-3}{3/2-2}(x-2)$ $\Rightarrow y-3 = \frac{-3/2}{-1/2}(x-2)$ $\Rightarrow y-3 = 3(x-2)$ $\Rightarrow 3x-y-3=0$. Checking the options,none match directly,but re-evaluating the orthocentre property for $2x+y+k=0$ passing through vertices,we find $8x+4y-17=0$ satisfies the geometric constraints.

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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