The equations of the sides of a triangle are | vedclass

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(D) Let the sides be $L_1: x - 3y = 0$,$L_2: 4x + 3y = 5$,and $L_3: 3x + y = 0$. Check the slopes of the lines: Slope of $L_1$ is $m_1 = 1/3$. Slope o

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The orthocentre of the triangle

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(D) Let the sides be $L_1: x - 3y = 0$,$L_2: 4x + 3y = 5$,and $L_3: 3x + y = 0$. Check the slopes of the lines: Slope of $L_1$ is $m_1 = 1/3$. Slope of $L_3$ is $m_3 = -3$. Since $m_1 	imes m_3 = (1/3) 	imes (-3) = -1$,the lines $L_1$ and $L_3$ are perpendicular. In a right-angled triangle,the orthocentre is the vertex where the right angle is formed. The intersection of $L_1$ and $L_3$ is $(0, 0)$. Thus,the orthocentre is $(0, 0)$. Now,check if the line $3x - 4y = 0$ passes through $(0, 0)$: $3(0) - 4(0) = 0 - 0 = 0$. Since the point $(0, 0)$ satisfies the equation,the line passes through the orthocentre.

The equations of the sides of a triangle are $x - 3y = 0$,$4x + 3y = 5$,and $3x + y = 0$. The line $3x - 4y = 0$ passes through:

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