Find the circumcentre of Delta ABC with verti | vedclass

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(C) Let the circumcentre be $O(x, y)$. The circumcentre is equidistant from the vertices $A, B$,and $C$. Therefore,$OA^2 = OB^2 = OC^2$.Given $A(-1, 1

Vedclass · Accepted Answer

$(-3, -2)$

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(C) Let the circumcentre be $O(x, y)$. The circumcentre is equidistant from the vertices $A, B$,and $C$. Therefore,$OA^2 = OB^2 = OC^2$.Given $A(-1, 1)$,$B(0, -4)$,and $C(-1, -5)$.$OA^2 = (x + 1)^2 + (y - 1)^2 = x^2 + 2x + 1 + y^2 - 2y + 1 = x^2 + y^2 + 2x - 2y + 2$.$OB^2 = (x - 0)^2 + (y + 4)^2 = x^2 + y^2 + 8y + 16$.$OC^2 = (x + 1)^2 + (y + 5)^2 = x^2 + 2x + 1 + y^2 + 10y + 25 = x^2 + y^2 + 2x + 10y + 26$.Equating $OA^2 = OC^2$:$x^2 + y^2 + 2x - 2y + 2 = x^2 + y^2 + 2x + 10y + 26$$-2y + 2 = 10y + 26$$-12y = 24 \implies y = -2$.Equating $OA^2 = OB^2$:$x^2 + y^2 + 2x - 2y + 2 = x^2 + y^2 + 8y + 16$$2x - 2y + 2 = 8y + 16$$2x - 10y = 14$.Substituting $y = -2$:$2x - 10(-2) = 14$$2x + 20 = 14$$2x = -6 \implies x = -3$.Thus,the circumcentre is $(-3, -2)$.

Find the circumcentre of $\Delta ABC$ with vertices $A(-1, 1)$,$B(0, -4)$,and $C(-1, -5)$.

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