The centroid of the triangle formed by the li | vedclass

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(A) Let the intersection of lines $x-3y+3=0$ and $x+3y+3=0$ be $A$. Adding the equations: $2x+6=0 \implies x=-3$. Substituting $x=-3$ in $x-3y+3=0$,we

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$\left(0, -\frac{1}{3}\right)$

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(A) Let the intersection of lines $x-3y+3=0$ and $x+3y+3=0$ be $A$. Adding the equations: $2x+6=0 \implies x=-3$. Substituting $x=-3$ in $x-3y+3=0$,we get $y=0$. So,$A = (-3, 0)$.Let the intersection of lines $x+3y+3=0$ and $x+y-1=0$ be $B$. Subtracting the equations: $(x+3y+3) - (x+y-1) = 0 \implies 2y+4=0 \implies y=-2$. Substituting $y=-2$ in $x+y-1=0$,we get $x-2-1=0 \implies x=3$. So,$B = (3, -2)$.Let the intersection of lines $x-3y+3=0$ and $x+y-1=0$ be $C$. Subtracting the equations: $(x-3y+3) - (x+y-1) = 0 \implies -4y+4=0 \implies y=1$. Substituting $y=1$ in $x+y-1=0$,we get $x+1-1=0 \implies x=0$. So,$C = (0, 1)$.The centroid $G$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.$G = \left(\frac{-3+3+0}{3}, \frac{0-2+1}{3}\right) = \left(0, -\frac{1}{3}\right)$.

The centroid of the triangle formed by the lines $x-3y+3=0$,$x+3y+3=0$,and $x+y-1=0$ is

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