The distance of the origin from the centroid | vedclass

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(C) Let the sides be $AB: x - 2y + 1 = 0$ and $AC: 2x - y - 1 = 0$. Solving these,we get vertex $A(1, 1)$.The altitude from vertex $B$ passes through

Vedclass · Accepted Answer

$2\sqrt{2}$

Vedclass · Answer

(C) Let the sides be $AB: x - 2y + 1 = 0$ and $AC: 2x - y - 1 = 0$. Solving these,we get vertex $A(1, 1)$.The altitude from vertex $B$ passes through $H\left(\frac{7}{3}, \frac{7}{3}ight)$ and is perpendicular to $AC$. The slope of $AC$ is $2$,so the slope of altitude $BH$ is $-\frac{1}{2}$. The equation of $BH$ is $y - \frac{7}{3} = -\frac{1}{2}\left(x - \frac{7}{3}ight)$,which simplifies to $x + 2y - 7 = 0$.Vertex $B$ is the intersection of $AB$ and $BH$: $x - 2y = -1$ and $x + 2y = 7$. Adding gives $2x = 6 \Rightarrow x = 3$,so $y = 2$. Thus,$B(3, 2)$.The altitude from vertex $C$ passes through $H\left(\frac{7}{3}, \frac{7}{3}ight)$ and is perpendicular to $AB$. The slope of $AB$ is $\frac{1}{2}$,so the slope of altitude $CH$ is $-2$. The equation of $CH$ is $y - \frac{7}{3} = -2\left(x - \frac{7}{3}ight)$,which simplifies to $2x + y - 7 = 0$.Vertex $C$ is the intersection of $AC$ and $CH$: $2x - y = 1$ and $2x + y = 7$. Adding gives $4x = 8 \Rightarrow x = 2$,so $y = 3$. Thus,$C(2, 3)$.The centroid $G$ of $	riangle ABC$ with vertices $A(1, 1), B(3, 2), C(2, 3)$ is $\left(\frac{1+3+2}{3}, \frac{1+2+3}{3}ight) = (2, 2)$.The distance of the centroid $(2, 2)$ from the origin $(0, 0)$ is $\sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.

The distance of the origin from the centroid of the triangle whose two sides have the equations $x - 2y + 1 = 0$ and $2x - y - 1 = 0$ and whose orthocenter is $\left(\frac{7}{3}, \frac{7}{3}\right)$ is

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