The sides of a triangle are x = 2,y + 1 = 0,a | vedclass

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(A) The given lines are $L_1: x = 2$,$L_2: y = -1$,and $L_3: x + 2y = 4$.First,find the vertices of the triangle by solving the equations of the lines

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$(4, 0)$

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(A) The given lines are $L_1: x = 2$,$L_2: y = -1$,and $L_3: x + 2y = 4$.First,find the vertices of the triangle by solving the equations of the lines pairwise:$1$. Intersection of $L_1$ and $L_2$: $x = 2, y = -1 \implies B(2, -1)$.$2$. Intersection of $L_1$ and $L_3$: $x = 2 \implies 2 + 2y = 4 \implies 2y = 2 \implies y = 1 \implies A(2, 1)$.$3$. Intersection of $L_2$ and $L_3$: $y = -1 \implies x + 2(-1) = 4 \implies x - 2 = 4 \implies x = 6 \implies C(6, -1)$.The slopes of the sides are: $m_{AB} = 	ext{undefined}$ (vertical line),$m_{BC} = 0$ (horizontal line),and $m_{AC} = \frac{-1 - 1}{6 - 2} = \frac{-2}{4} = -\frac{1}{2}$.Since $AB \perp BC$,the triangle is a right-angled triangle with the right angle at vertex $B(2, -1)$.For a right-angled triangle,the circumcenter is the midpoint of the hypotenuse $AC$.The coordinates of the midpoint of $AC$ are $(\frac{2 + 6}{2}, \frac{1 - 1}{2}) = (4, 0)$.

The sides of a triangle are $x = 2$,$y + 1 = 0$,and $x + 2y = 4$. Find its circumcenter.

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