If the vertices of a triangle are (2, 1), (5, | vedclass

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(B) Let the circumcentre be $O(x, y)$ and the given vertices be $A(2, 1), B(5, 2),$ and $C(3, 4).$Since the circumcentre is equidistant from all verti

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$\left( \frac{13}{4}, \frac{9}{4} \right)$

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(B) Let the circumcentre be $O(x, y)$ and the given vertices be $A(2, 1), B(5, 2),$ and $C(3, 4).$Since the circumcentre is equidistant from all vertices,$OA^2 = OB^2 = OC^2.$From $OA^2 = OB^2: (x - 2)^2 + (y - 1)^2 = (x - 5)^2 + (y - 2)^2.$Expanding this: $x^2 - 4x + 4 + y^2 - 2y + 1 = x^2 - 10x + 25 + y^2 - 4y + 4.$Simplifying: $6x + 2y = 24 \implies 3x + y = 12$......$(i)$From $OA^2 = OC^2: (x - 2)^2 + (y - 1)^2 = (x - 3)^2 + (y - 4)^2.$Expanding this: $x^2 - 4x + 4 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 8y + 16.$Simplifying: $2x + 6y = 20 \implies x + 3y = 10$......(ii)Solving equations $(i)$ and (ii): Multiply $(i)$ by $3$: $9x + 3y = 36.$Subtracting (ii) from this: $(9x - x) + (3y - 3y) = 36 - 10 \implies 8x = 26 \implies x = \frac{13}{4}.$Substituting $x = \frac{13}{4}$ into $(i)$: $3(\frac{13}{4}) + y = 12 \implies y = 12 - \frac{39}{4} = \frac{48 - 39}{4} = \frac{9}{4}.$Thus,the circumcentre is $\left( \frac{13}{4}, \frac{9}{4} \right).$

If the vertices of a triangle are $(2, 1), (5, 2)$ and $(3, 4)$,then its circumcentre is

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