If two sides of a triangle are given by 3x^2- | vedclass

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(D) The given pair of equations of two sides of the triangle is $3x^2-5xy+2y^2=0$. Factoring this,we get $(3x-2y)(x-y)=0$. So,the equations of the two

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$10x+5y=21$

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(D) The given pair of equations of two sides of the triangle is $3x^2-5xy+2y^2=0$. Factoring this,we get $(3x-2y)(x-y)=0$. So,the equations of the two sides are $L_1: 3x-2y=0$ and $L_2: x-y=0$. The orthocentre $H$ is $(2,1)$. The altitude from the vertex (intersection of $L_1$ and $L_2$,which is $(0,0)$) to the third side is perpendicular to the third side. Alternatively,the altitude from a vertex to the opposite side passes through the orthocentre. The altitude from the vertex $(0,0)$ to the side $L_1$ is perpendicular to $L_1: 3x-2y=0$. Its equation is $2x+3y=0$. The altitude from the vertex $(0,0)$ to the side $L_2$ is perpendicular to $L_2: x-y=0$. Its equation is $x+y=0$. Let the third side be $L_3$. The altitude from the vertex where $L_1$ and $L_2$ meet (i.e.,$(0,0)$) is not useful here. Instead,consider the vertices $A, B, C$. Let $A=(0,0)$. $AB$ is $3x-2y=0$ and $AC$ is $x-y=0$. The altitude from $B$ to $AC$ passes through $H(2,1)$ and is perpendicular to $AC: x-y=0$. The line perpendicular to $x-y=0$ is $x+y+k=0$. Since it passes through $(2,1)$,$2+1+k=0 \Rightarrow k=-3$. So,$x+y-3=0$. The intersection of $x+y-3=0$ and $3x-2y=0$ gives vertex $B$: $3x-2(3-x)=0$ $\Rightarrow 5x=6$ $\Rightarrow x=6/5, y=9/5$. The altitude from $C$ to $AB$ passes through $H(2,1)$ and is perpendicular to $AB: 3x-2y=0$. The line perpendicular to $3x-2y=0$ is $2x+3y+k=0$. Since it passes through $(2,1)$,$4+3+k=0 \Rightarrow k=-7$. So,$2x+3y-7=0$. The intersection of $2x+3y-7=0$ and $x-y=0$ gives vertex $C$: $2x+3x-7=0$ $\Rightarrow 5x=7$ $\Rightarrow x=7/5, y=7/5$. The third side $BC$ passes through $(6/5, 9/5)$ and $(7/5, 7/5)$. The slope $m = \frac{7/5-9/5}{7/5-6/5} = \frac{-2/5}{1/5} = -2$. The equation is $y-7/5 = -2(x-7/5)$ $\Rightarrow 5y-7 = -10x+14$ $\Rightarrow 10x+5y=21$.

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

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