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

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$\left(\frac{1}{2}, \frac{1}{2}\right)$

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(C) The given equation of the pair of lines is $2x^2 - 3xy + y^2 = 0$. Factoring this,we get $(2x - y)(x - y) = 0$. So,the two lines are $L_1: y = 2x$ and $L_2: y = x$. The third line is $L_3: x + y = 1$. The vertices of the triangle are the intersection points of these lines: $1$. Intersection of $L_1$ and $L_2$ is $(0, 0)$. $2$. Intersection of $L_1$ and $L_3$: $x + 2x = 1 \implies 3x = 1 \implies x = 1/3, y = 2/3$. Vertex is $(1/3, 2/3)$. $3$. Intersection of $L_2$ and $L_3$: $x + x = 1 \implies 2x = 1 \implies x = 1/2, y = 1/2$. Vertex is $(1/2, 1/2)$. Let the vertices be $A(0, 0)$,$B(1/3, 2/3)$,and $C(1/2, 1/2)$. The altitude from $B$ to $AC$ (line $y=x$) has slope $-1$ and passes through $(1/3, 2/3)$: $y - 2/3 = -1(x - 1/3) \implies y = -x + 1$. The altitude from $C$ to $AB$ (line $y=2x$) has slope $-1/2$ and passes through $(1/2, 1/2)$: $y - 1/2 = -1/2(x - 1/2) \implies y = -1/2x + 3/4$. Solving these: $-x + 1 = -1/2x + 3/4 \implies 1/2x = 1/4 \implies x = 1/2$. Then $y = -1/2 + 1 = 1/2$. The orthocentre is $(1/2, 1/2)$.

The orthocentre of the triangle formed by the lines $2x^2 - 3xy + y^2 = 0$ and $x + y = 1$ is

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