The orthocenter of the triangle formed by the | vedclass

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(D) Let the lines be $L_1: 4x - 7y + 10 = 0$,$L_2: x + y - 5 = 0$,and $L_3: 7x + 4y - 15 = 0$. First,check the slopes of the lines. The slope of $L_1$

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$(1, 2)$

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(D) Let the lines be $L_1: 4x - 7y + 10 = 0$,$L_2: x + y - 5 = 0$,and $L_3: 7x + 4y - 15 = 0$. First,check the slopes of the lines. The slope of $L_1$ is $m_1 = 4/7$. The slope of $L_3$ is $m_3 = -7/4$. Since $m_1 	imes m_3 = (4/7) 	imes (-7/4) = -1$,the lines $L_1$ and $L_3$ are perpendicular to each other. In a right-angled triangle,the orthocenter is the vertex where the right angle is formed. To find the vertex,solve $L_1$ and $L_3$: $4x - 7y = -10$ $7x + 4y = 15$ Multiplying $L_1$ by $4$ and $L_3$ by $7$: $16x - 28y = -40$ $49x + 28y = 105$ Adding these: $65x = 65 \implies x = 1$. Substituting $x = 1$ into $L_1$: $4(1) - 7y = -10 \implies -7y = -14 \implies y = 2$. Thus,the orthocenter is $(1, 2)$.

The orthocenter of the triangle formed by the lines $4x - 7y + 10 = 0$,$x + y - 5 = 0$,and $7x + 4y - 15 = 0$ is:

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