The circumcentre of the triangle with vertice | vedclass

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Question

(B) Let the vertices of $	riangle ABC$ be $A(-2, 3)$,$B(1, -2)$,and $C(2, 1)$.The circumcentre is the point of intersection of the perpendicular bise

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the vertices of $	riangle ABC$ be $A(-2, 3)$,$B(1, -2)$,and $C(2, 1)$.The circumcentre is the point of intersection of the perpendicular bisectors of the sides.Let $E$ be the circumcentre. Let $D$ be the midpoint of $BC$.$D = (\frac{1+2}{2}, \frac{-2+1}{2}) = (\frac{3}{2}, -\frac{1}{2})$.Slope of $BC = \frac{1 - (-2)}{2 - 1} = \frac{3}{1} = 3$.The slope of the perpendicular bisector $DE$ is $-\frac{1}{3}$.Equation of $DE$: $y - (-\frac{1}{2}) = -\frac{1}{3}(x - \frac{3}{2})$ $\Rightarrow y + \frac{1}{2} = -\frac{1}{3}x + \frac{1}{2}$ $\Rightarrow x + 3y = 0 \quad (i)$.Let $F$ be the midpoint of $AB$.$F = (\frac{-2+1}{2}, \frac{3-2}{2}) = (-\frac{1}{2}, \frac{1}{2})$.Slope of $AB = \frac{-2-3}{1-(-2)} = \frac{-5}{3}$.The slope of the perpendicular bisector $EF$ is $\frac{3}{5}$.Equation of $EF$: $y - \frac{1}{2} = \frac{3}{5}(x - (-\frac{1}{2}))$ $\Rightarrow y - \frac{1}{2} = \frac{3}{5}x + \frac{3}{10}$ $\Rightarrow 10y - 5 = 6x + 3$ $\Rightarrow 6x - 10y = -8$ $\Rightarrow 3x - 5y = -4 \quad (ii)$.Solving $(i)$ and $(ii)$: From $(i)$,$x = -3y$.Substitute into $(ii)$: $3(-3y) - 5y = -4$ $\Rightarrow -9y - 5y = -4$ $\Rightarrow -14y = -4$ $\Rightarrow y = \frac{2}{7}$.Then $x = -3(\frac{2}{7}) = -\frac{6}{7}$.Thus,the circumcentre is $(-\frac{6}{7}, \frac{2}{7})$.

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

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