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  4. The distance of the origin from the centroid of the triangle whose two sides have the equations x -2 y +1=0 and 2 x - y -1=0 and whose orthocenter is ((7/3), (7/3)) is:

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

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Solution:

$AB \equiv x -2 y +1=0$
$AC \equiv 2 x - y -1=0$
So $A (1,1)$
Altitude from $B$ is $BH = x +2 y -7=0 \Rightarrow B (3,2)$
Altitude from $C$ is $CH =2 x + y -7=0 \Rightarrow C (2,3)$
Centroid of $\triangle ABC = E (2,2) OE =2 \sqrt{2}$