The orthocenter of the triangle whose sides a | vedclass

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(B) Let the lines be $L_1: x+y+10=0$,$L_2: x-y-2=0$,and $L_3: 2x+y-7=0$. First,check the slopes of the lines: Slope of $L_1$ is $m_1 = -1$. Slope of $

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$(-4,-6)$

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(B) Let the lines be $L_1: x+y+10=0$,$L_2: x-y-2=0$,and $L_3: 2x+y-7=0$. First,check the slopes of the lines: Slope of $L_1$ is $m_1 = -1$. Slope of $L_2$ is $m_2 = 1$. Since $m_1 	imes m_2 = (-1) 	imes (1) = -1$,the lines $L_1$ and $L_2$ are perpendicular to each other. In a right-angled triangle,the orthocenter is the vertex where the right angle is formed. Therefore,the orthocenter is the point of intersection of $L_1$ and $L_2$. Solving $x+y+10=0$ and $x-y-2=0$: Adding the two equations: $(x+y+10) + (x-y-2) = 0 \implies 2x + 8 = 0 \implies x = -4$. Substituting $x = -4$ into $x-y-2=0$: $-4 - y - 2 = 0 \implies y = -6$. Thus,the orthocenter is $(-4, -6)$.

The orthocenter of the triangle whose sides are given by $x+y+10=0$,$x-y-2=0$,and $2x+y-7=0$ is

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