If the circumcentre of the triangle with vert | vedclass

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(B) Let the vertices of the triangle be $A(a, -1)$,$B(6, -9)$,and $C(10, b)$. Let the circumcentre be $O(6, -5)$.Since the circumcentre is equidistant

Vedclass · Accepted Answer

$a = 6, b = -5$

Vedclass · Answer

(B) Let the vertices of the triangle be $A(a, -1)$,$B(6, -9)$,and $C(10, b)$. Let the circumcentre be $O(6, -5)$.Since the circumcentre is equidistant from all vertices,$OA = OB = OC$.First,calculate $OB^2 = (6 - 6)^2 + (-5 - (-9))^2 = 0^2 + 4^2 = 16$.Since $OA^2 = OB^2$,we have $(a - 6)^2 + (-1 - (-5))^2 = 16$.$(a - 6)^2 + 4^2 = 16 \implies (a - 6)^2 + 16 = 16 \implies (a - 6)^2 = 0 \implies a = 6$.Next,since $OC^2 = OB^2$,we have $(10 - 6)^2 + (b - (-5))^2 = 16$.$4^2 + (b + 5)^2 = 16 \implies 16 + (b + 5)^2 = 16 \implies (b + 5)^2 = 0 \implies b = -5$.Thus,the values are $a = 6$ and $b = -5$.

If the circumcentre of the triangle with vertices $(a, -1)$,$(6, -9)$,and $(10, b)$ is $(6, -5)$,then find the values of $a$ and $b$.

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