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

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Aicharya · Accepted Answer

Let vertices of $	riangle A B C$ is $A(-2,3), B(1,-2)$ and $C(2,1)$ $	herefore$ Circumcentre of triangle is a point where perpendicular bisector of sides meet. So, $D E$ is perpendicular bisector of $B C$. So, $D E$ is perpendicular bisector of $B C$. $D$ is mid-point of $B C$, so coordinate of $	herefore \quad D\left(\frac{1+2}{2}, \frac{-2+1}{2}ight)=D\left(\frac{3}{2}, \frac{-1}{2}ight)$ Slope of $B C=\frac{1+2}{2-1}=3$ Since $D E \perp B C$, slope of $D E=-\frac{1}{3}$ Now, equation of $D E$ is $ \begin{aligned} \left(y+\frac{1}{2}ight) & =\frac{-1}{3}\left(x-\frac{3}{2}ight) \ \Rightarrow \quad y+\frac{1}{2} & =-\frac{1}{3} x+\frac{1}{2} \Rightarrow 3 y=-x \end{aligned} $ Now, $F E$ is perpendicular bisector of $A B$ $	herefore$ Coordinate of $F\left(\frac{-1}{2}, \frac{1}{2}ight)$ Slope of $A B=\frac{3-(-2)}{-2-(1)}=\frac{-5}{3}$ So, slope of $E F=\frac{3}{5}$ $	herefore$ Equation of $F E, y-\frac{1}{2}=\frac{3}{5}(x+1 / 2)$ $D E$ and $E F$ intersect at circumcentre, so Multiplying by 3 in Eq. (i) and subtract then, we get $ \begin{aligned} & 3 x+9 y=0 \ & 3 x-5 y=-4 \ & -\quad+\quad+ \ & \hline \end{aligned} $ $ \begin{array}{r} 14 y=4 \ y=\frac{2}{7} \end{array} $ Put in Eq. (i), we get $ x=-3 	imes \frac{2}{7} \Rightarrow x=\frac{-6}{7} $ So, coordinate of circumcentre is $\left(\frac{-6}{7}, \frac{2}{7}ight)$

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

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