Let OABC be a parallelogram. The equation of | vedclass

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(B) The combined equation of sides $OA$ and $OC$ is $2x^2-y^2=0$,which implies $y^2=2x^2$,or $y=\pm\sqrt{2}x$.Since $A$ and $C$ lie on the line $x+y-1

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$\frac{4}{3}\sqrt{5}$

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(B) The combined equation of sides $OA$ and $OC$ is $2x^2-y^2=0$,which implies $y^2=2x^2$,or $y=\pm\sqrt{2}x$.Since $A$ and $C$ lie on the line $x+y-1=0$,we find their coordinates by substituting $y=\sqrt{2}x$ and $y=-\sqrt{2}x$ into the line equation.For $A$: $x+\sqrt{2}x=1 \implies x=\frac{1}{1+\sqrt{2}}=\sqrt{2}-1$. Thus $A=(\sqrt{2}-1, 2-\sqrt{2})$.For $C$: $x-\sqrt{2}x=1 \implies x=\frac{1}{1-\sqrt{2}}=-(1+\sqrt{2})$. Thus $C=(-1-\sqrt{2}, 2+\sqrt{2})$.$D$ is the midpoint of $AC$,so $D=\left(\frac{\sqrt{2}-1-1-\sqrt{2}}{2}, \frac{2-\sqrt{2}+2+\sqrt{2}}{2}ight) = (-1, 2)$.Since $OABC$ is a parallelogram,the diagonals $OB$ and $AC$ bisect each other at $D$. Thus $D$ is the midpoint of $OB$. Since $O=(0,0)$,$B=2D=(-2, 4)$.The centroid $G$ of $	riangle OAC$ is $\left(\frac{0+(\sqrt{2}-1)+(-1-\sqrt{2})}{3}, \frac{0+(2-\sqrt{2})+(2+\sqrt{2})}{3}ight) = \left(-\frac{2}{3}, \frac{4}{3}ight)$.The distance $BG = \sqrt{(-2 - (-2/3))^2 + (4 - 4/3)^2} = \sqrt{(-4/3)^2 + (8/3)^2} = \sqrt{\frac{16}{9} + \frac{64}{9}} = \sqrt{\frac{80}{9}} = \frac{4\sqrt{5}}{3}$.

Let $OABC$ be a parallelogram. The equation of one diagonal $AC$ is $x+y-1=0$ and the combined equation of the sides $OA, OC$ is $2x^2-y^2=0$. If $G$ is the centroid of the triangle $OAC$,then $BG=$

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