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(C) Let the lines be $L_1: x - 2y = 0$,$L_2: 4x + 3y - 5 = 0$,and $L_3: 2x + y = 0$. First,find the vertices of the triangle by solving the pairs of e

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Orthocenter

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(C) Let the lines be $L_1: x - 2y = 0$,$L_2: 4x + 3y - 5 = 0$,and $L_3: 2x + y = 0$. First,find the vertices of the triangle by solving the pairs of equations: Vertex $A = L_1 \cap L_2$: Solving $x = 2y$ and $4(2y) + 3y = 5 \implies 11y = 5 \implies y = 5/11, x = 10/11$. So $A = (10/11, 5/11)$. Vertex $B = L_2 \cap L_3$: Solving $y = -2x$ and $4x + 3(-2x) = 5 \implies -2x = 5 \implies x = -5/2, y = 5$. So $B = (-5/2, 5)$. Vertex $C = L_3 \cap L_1$: Solving $x = 2y$ and $2(2y) + y = 0 \implies 5y = 0 \implies y = 0, x = 0$. So $C = (0, 0)$. The orthocenter $H(x, y)$ is the intersection of altitudes. Altitude from $A$ to $BC$: Slope of $BC$ is $m_{BC} = (5-0)/(-5/2-0) = -2$. The altitude is perpendicular to $BC$,so its slope is $1/2$. Equation: $y - 5/11 = 1/2(x - 10/11) \implies 2y - 10/11 = x - 10/11 \implies x - 2y = 0$. Altitude from $B$ to $AC$: Slope of $AC$ is $m_{AC} = (5/11-0)/(10/11-0) = 1/2$. The altitude is perpendicular to $AC$,so its slope is $-2$. Equation: $y - 5 = -2(x + 5/2) \implies y - 5 = -2x - 5 \implies 2x + y = 0$. Wait,the lines $L_1$ and $L_3$ are perpendicular $(m_1 = 1/2, m_3 = -2)$. Thus,the triangle is a right-angled triangle at vertex $C(0,0)$. In a right-angled triangle,the orthocenter is the vertex at the right angle,which is $C(0,0)$. The line $3y - 4x = 0$ passes through $(0,0)$. Therefore,the line passes through the orthocenter.

The equations of the sides of a triangle are $x - 2y = 0$,$4x + 3y = 5$,and $2x + y = 0$. Through which point does the line $3y - 4x = 0$ pass?

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