The lines $x-y-2=0, x+y-4=0$ and $x+3 y=6$ meet in the common point :
The lines $x-y-2=0, x+y-4=0$ and $x+3 y=6$ meet in the common point :
$(1,2)$
$(2,2)$
$(3,1)$
$(1,1)$
Solution
The equation of lines are
$x-y-2=0$ ...(i)
$x+y-4=0$ ...(ii)
and $\quad x+3 y=6$ ...(iii)
On solving Eqs. (i) and (ii), we get
$x=3, y=1$
On putting these value in Eq. (iii)
$3+3=6$
which is satisfied. Thus the required point is $(3,1)$.