The centroid of the triangle formed by the li | vedclass

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(A) To find the vertices of the triangle,we solve the equations of the lines in pairs:$1$. Intersection of $x+y-1=0$ and $x-y-1=0$:Adding the two equa

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$\left(\frac{4}{3}, 1\right)$

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(A) To find the vertices of the triangle,we solve the equations of the lines in pairs:$1$. Intersection of $x+y-1=0$ and $x-y-1=0$:Adding the two equations gives $2x-2=0$,so $x=1$. Substituting $x=1$ into $x+y-1=0$ gives $y=0$. Thus,vertex $A$ is $(1, 0)$.$2$. Intersection of $x-y-1=0$ and $x-3y+3=0$:Subtracting the second from the first gives $2y-4=0$,so $y=2$. Substituting $y=2$ into $x-y-1=0$ gives $x=3$. Thus,vertex $B$ is $(3, 2)$.$3$. Intersection of $x-3y+3=0$ and $x+y-1=0$:Subtracting the second from the first gives $-4y+4=0$,so $y=1$. Substituting $y=1$ into $x+y-1=0$ gives $x=0$. Thus,vertex $C$ is $(0, 1)$.The centroid $(G)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.Substituting the coordinates of $A, B, C$:$G = \left(\frac{1+3+0}{3}, \frac{0+2+1}{3}\right) = \left(\frac{4}{3}, \frac{3}{3}\right) = \left(\frac{4}{3}, 1\right)$.

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

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