If the equations of three sides of a triangle | vedclass

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(A) The equations of the three sides are $L_1: x = 2$,$L_2: y = -1$,and $L_3: x + 2y = 4$.$1$. Intersection of $L_1$ and $L_2$: $x = 2, y = -1 \implie

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$(4, 0)$

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(A) The equations of the three sides are $L_1: x = 2$,$L_2: y = -1$,and $L_3: x + 2y = 4$.$1$. Intersection of $L_1$ and $L_2$: $x = 2, y = -1 \implies A(2, -1)$.$2$. Intersection of $L_1$ and $L_3$: $x = 2, 2 + 2y = 4 \implies 2y = 2 \implies y = 1 \implies B(2, 1)$.$3$. Intersection of $L_2$ and $L_3$: $y = -1, x + 2(-1) = 4 \implies x - 2 = 4 \implies x = 6 \implies C(6, -1)$.Let the circumcentre be $O(x, y)$. The distance $OA^2 = OB^2 = OC^2$.$OA^2 = (x - 2)^2 + (y + 1)^2$$OB^2 = (x - 2)^2 + (y - 1)^2$$OC^2 = (x - 6)^2 + (y + 1)^2$Equating $OA^2 = OB^2$: $(x - 2)^2 + (y + 1)^2 = (x - 2)^2 + (y - 1)^2 \implies (y + 1)^2 = (y - 1)^2 \implies y^2 + 2y + 1 = y^2 - 2y + 1 \implies 4y = 0 \implies y = 0$.Equating $OA^2 = OC^2$ with $y = 0$: $(x - 2)^2 + (0 + 1)^2 = (x - 6)^2 + (0 + 1)^2 \implies (x - 2)^2 = (x - 6)^2 \implies x^2 - 4x + 4 = x^2 - 12x + 36 \implies 8x = 32 \implies x = 4$.Thus,the circumcentre is $(4, 0)$.

If the equations of three sides of a triangle are $x = 2$,$y + 1 = 0$,and $x + 2y = 4$,then the coordinates of the circumcentre of this triangle are:

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