The circumcentre of the triangle with vertice | vedclass

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(A) Let the vertices of the triangle be $A(-2, 3)$,$B(2, -1)$,and $C(4, 0)$.Let $O(x, y)$ be the circumcentre of the triangle.By definition,the circum

Vedclass · Accepted Answer

$\left(\frac{3}{2}, \frac{5}{2}\right)$

Vedclass · Answer

(A) Let the vertices of the triangle be $A(-2, 3)$,$B(2, -1)$,and $C(4, 0)$.Let $O(x, y)$ be the circumcentre of the triangle.By definition,the circumcentre is equidistant from all vertices,so $OA = OB = OC$,which implies $OA^2 = OB^2 = OC^2$.First,set $OA^2 = OB^2$:$(x + 2)^2 + (y - 3)^2 = (x - 2)^2 + (y + 1)^2$$x^2 + 4x + 4 + y^2 - 6y + 9 = x^2 - 4x + 4 + y^2 + 2y + 1$$8x - 8y = -8 \Rightarrow x - y = -1$ ... $(i)$Next,set $OB^2 = OC^2$:$(x - 2)^2 + (y + 1)^2 = (x - 4)^2 + (y - 0)^2$$x^2 - 4x + 4 + y^2 + 2y + 1 = x^2 - 8x + 16 + y^2$$4x + 2y = 11$ ... $(ii)$Solving equations $(i)$ and $(ii)$:From $(i)$,$y = x + 1$.Substitute into $(ii)$: $4x + 2(x + 1) = 11$ $\Rightarrow 6x = 9$ $\Rightarrow x = \frac{3}{2}$.Then $y = \frac{3}{2} + 1 = \frac{5}{2}$.Thus,the circumcentre is $\left(\frac{3}{2}, \frac{5}{2}\right)$.

The circumcentre of the triangle with vertices $(-2, 3)$,$(2, -1)$,and $(4, 0)$ is:

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