If A(-2, 1),B(0, -2),and C(1, 2) are the vert | vedclass

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(B) Let $O(h, k)$ be the circumcentre of $	riangle ABC$ with vertices $A(-2, 1)$,$B(0, -2)$,and $C(1, 2)$.Since $O$ is the circumcentre,$OA = OB = OC

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$\frac{3 \sqrt{17}}{22}$

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(B) Let $O(h, k)$ be the circumcentre of $	riangle ABC$ with vertices $A(-2, 1)$,$B(0, -2)$,and $C(1, 2)$.Since $O$ is the circumcentre,$OA = OB = OC$.$OA^2 = OB^2 \Rightarrow (h+2)^2 + (k-1)^2 = h^2 + (k+2)^2$$h^2 + 4h + 4 + k^2 - 2k + 1 = h^2 + k^2 + 4k + 4$$4h - 6k + 1 = 0$  --- $(i)$$OB^2 = OC^2 \Rightarrow h^2 + (k+2)^2 = (h-1)^2 + (k-2)^2$$h^2 + k^2 + 4k + 4 = h^2 - 2h + 1 + k^2 - 4k + 4$$2h + 8k - 1 = 0$  --- $(ii)$Solving equations $(i)$ and $(ii)$:Multiply $(ii)$ by $2$: $4h + 16k - 2 = 0$ --- $(iii)$Subtract $(i)$ from $(iii)$: $(4h + 16k - 2) - (4h - 6k + 1) = 0$$22k - 3 = 0 \Rightarrow k = \frac{3}{22}$Substitute $k$ in $(ii)$: $2h + 8(\frac{3}{22}) - 1 = 0$ $\Rightarrow 2h + \frac{12}{11} - 1 = 0$ $\Rightarrow 2h = -\frac{1}{11}$ $\Rightarrow h = -\frac{1}{22}$Equation of line $BC$ passing through $(0, -2)$ and $(1, 2)$:Slope $m = \frac{2 - (-2)}{1 - 0} = 4$$y - (-2) = 4(x - 0) \Rightarrow 4x - y - 2 = 0$Perpendicular distance $d$ from $O(-\frac{1}{22}, \frac{3}{22})$ to $4x - y - 2 = 0$ is:$d = \frac{|4(-\frac{1}{22}) - \frac{3}{22} - 2|}{\sqrt{4^2 + (-1)^2}} = \frac{|-\frac{4}{22} - \frac{3}{22} - \frac{44}{22}|}{\sqrt{17}} = \frac{|-\frac{51}{22}|}{\sqrt{17}} = \frac{51}{22\sqrt{17}} = \frac{3\sqrt{17}}{22}$

If $A(-2, 1)$,$B(0, -2)$,and $C(1, 2)$ are the vertices of a triangle $ABC$,then the perpendicular distance from its circumcentre to the side $BC$ is

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