The centroid of the triangle formed by the lines $x+y-1=0, x-y-1=0, x-3 y+3=0$ is
The centroid of the triangle formed by the lines $x+y-1=0, x-y-1=0, x-3 y+3=0$ is
$\left(\frac{4}{3}, 1\right)$
$\left(\frac{-4}{3}, 1\right)$
$\left(\frac{8}{3}, 3\right)$
$\left(\frac{-8}{3}, 3\right)$
Solution
Point of intersection of lines $x+y-1=0$ and $x-y-1=0$ is $A(1,0)$.
Similarly, point of intersection of lines $x-y-1=0$ and $x-3 y+3=0$ is $B(3,2)$, and point of intersection of lines $x-3 y+3=0$ and $x+y-1=0$ is $C(0,1)$.
Now, centroid of $\triangle A B C$ is $\left(\frac{1+3+0}{3}, \frac{0+2+1}{3}\right)$
$
=\left(\frac{4}{3}, 1\right)
$
