The circumcentre of the triangle with vertice | vedclass

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(B) The circumcentre of a triangle is equidistant from its vertices. Let $O(x, y)$ be the circumcentre of the triangle with vertices $A(-2, 3), B(1, -

Vedclass · Accepted Answer

$\left(-\frac{6}{7}, \frac{2}{7}\right)$

Vedclass · Answer

(B) The circumcentre of a triangle is equidistant from its vertices. Let $O(x, y)$ be the circumcentre of the triangle with vertices $A(-2, 3), B(1, -2)$,and $C(2, 1)$.Since $OA = OB = OC$,we have $OA^2 = OB^2 = OC^2$.$OA^2 = (x + 2)^2 + (y - 3)^2 = x^2 + 4x + 4 + y^2 - 6y + 9 = x^2 + y^2 + 4x - 6y + 13$$OB^2 = (x - 1)^2 + (y + 2)^2 = x^2 - 2x + 1 + y^2 + 4y + 4 = x^2 + y^2 - 2x + 4y + 5$$OC^2 = (x - 2)^2 + (y - 1)^2 = x^2 - 4x + 4 + y^2 - 2y + 1 = x^2 + y^2 - 4x - 2y + 5$Equating $OA^2 = OB^2$:$x^2 + y^2 + 4x - 6y + 13 = x^2 + y^2 - 2x + 4y + 5$$6x - 10y + 8 = 0 \Rightarrow 3x - 5y + 4 = 0 \dots (i)$Equating $OB^2 = OC^2$:$x^2 + y^2 - 2x + 4y + 5 = x^2 + y^2 - 4x - 2y + 5$$2x + 6y = 0$ $\Rightarrow x + 3y = 0$ $\Rightarrow x = -3y \dots (ii)$Substituting $(ii)$ into $(i)$:$3(-3y) - 5y + 4 = 0$$-9y - 5y + 4 = 0$ $\Rightarrow -14y = -4$ $\Rightarrow y = \frac{4}{14} = \frac{2}{7}$Now,$x = -3\left(\frac{2}{7}\right) = -\frac{6}{7}$.Thus,the circumcentre is $\left(-\frac{6}{7}, \frac{2}{7}\right)$.

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

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