The circumcentre of the triangle formed by th | vedclass

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(B) The given lines are:$x+y+2=0 \quad \dots(i)$$2x+y+8=0 \quad \dots(ii)$$x-y-2=0 \quad \dots(iii)$Solving $(i)$ and $(ii)$ by subtracting $(i)$ from

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$(-4,0)$

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(B) The given lines are:$x+y+2=0 \quad \dots(i)$$2x+y+8=0 \quad \dots(ii)$$x-y-2=0 \quad \dots(iii)$Solving $(i)$ and $(ii)$ by subtracting $(i)$ from $(ii)$:$(2x+y+8) - (x+y+2) = 0 \implies x+6=0 \implies x=-6$Substituting $x=-6$ in $(i)$: $-6+y+2=0 \implies y=4$. So,vertex is $(-6, 4)$.Solving $(i)$ and $(iii)$ by adding them:$(x+y+2) + (x-y-2) = 0 \implies 2x=0 \implies x=0$Substituting $x=0$ in $(i)$: $0+y+2=0 \implies y=-2$. So,vertex is $(0, -2)$.Solving $(ii)$ and $(iii)$ by adding them:$(2x+y+8) + (x-y-2) = 0 \implies 3x+6=0 \implies x=-2$Substituting $x=-2$ in $(iii)$: $-2-y-2=0 \implies y=-4$. So,vertex is $(-2, -4)$.The vertices are $A(-6, 4), B(0, -2), C(-2, -4)$.Check for right-angled triangle:Slope of $AB = \frac{-2-4}{0-(-6)} = \frac{-6}{6} = -1$Slope of $BC = \frac{-4-(-2)}{-2-0} = \frac{-2}{-2} = 1$Since (Slope of $AB$) $	imes$ (Slope of $BC$) $= -1 	imes 1 = -1$,the triangle is right-angled at $B(0, -2)$.The circumcentre of a right-angled triangle is the midpoint of the hypotenuse $AC$.Circumcentre $= \left(\frac{-6+(-2)}{2}, \frac{4+(-4)}{2}ight) = \left(\frac{-8}{2}, \frac{0}{2}ight) = (-4, 0)$.

The circumcentre of the triangle formed by the lines $x+y+2=0, 2x+y+8=0$ and $x-y-2=0$ is

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