If the vertices of a triangle are (-2,3), (6, | vedclass

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(D) Let the vertices of the triangle be $A(-2,3)$,$B(6,-1)$,and $C(4,3)$. Let $F(x,y)$ be the circumcentre. The circumcentre is equidistant from the v

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$(1,-1)$

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(D) Let the vertices of the triangle be $A(-2,3)$,$B(6,-1)$,and $C(4,3)$. Let $F(x,y)$ be the circumcentre. The circumcentre is equidistant from the vertices,so $FA^2 = FB^2 = FC^2$.$FA^2 = (x+2)^2 + (y-3)^2$$FB^2 = (x-6)^2 + (y+1)^2$$FC^2 = (x-4)^2 + (y-3)^2$Equating $FA^2 = FC^2$:$(x+2)^2 + (y-3)^2 = (x-4)^2 + (y-3)^2$$(x+2)^2 = (x-4)^2$$x^2 + 4x + 4 = x^2 - 8x + 16$$12x = 12 \implies x = 1$Equating $FA^2 = FB^2$:$(1+2)^2 + (y-3)^2 = (1-6)^2 + (y+1)^2$$9 + y^2 - 6y + 9 = 25 + y^2 + 2y + 1$$18 - 6y = 26 + 2y$$-8 = 8y \implies y = -1$Thus,the coordinates of the circumcentre are $(1,-1)$.

If the vertices of a triangle are $(-2,3), (6,-1)$ and $(4,3)$,then the coordinates of the circumcentre of the triangle are

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