The orthocentre of the triangle formed by lin | vedclass

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(A) Given lines are:$L_1: x+y+1=0$$L_2: x-y-1=0$$L_3: 3x+4y+5=0$Slope of $L_1$ $(m_1)$ is $-1$.Slope of $L_2$ $(m_2)$ is $1$.Since $m_1 	imes m_2 = (

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$(0,-1)$

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(A) Given lines are:$L_1: x+y+1=0$$L_2: x-y-1=0$$L_3: 3x+4y+5=0$Slope of $L_1$ $(m_1)$ is $-1$.Slope of $L_2$ $(m_2)$ is $1$.Since $m_1 	imes m_2 = (-1) 	imes (1) = -1$,the lines $L_1$ and $L_2$ are perpendicular to each other.Therefore,the triangle is a right-angled triangle,and the orthocentre of a right-angled triangle is the vertex where the right angle is formed.To find the vertex,solve $L_1$ and $L_2$:$x+y+1=0$$x-y-1=0$Adding the two equations: $2x = 0 \Rightarrow x = 0$.Substituting $x=0$ in $x+y+1=0$,we get $0+y+1=0 \Rightarrow y = -1$.Thus,the orthocentre is $(0, -1)$.

The orthocentre of the triangle formed by lines $x+y+1=0$,$x-y-1=0$,and $3x+4y+5=0$ is

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