The triangle formed by the lines x+y−3=0,2x+y−4=0,3x−7y+11=0

Given lines are x+y−3=0-----12x+y−4=0-----23x−7y+11=0----3To get the points of intersection of lines (1) and (2), eliminate  y To eliminate y, subtract (1) from (2) 2x+y−4−x−y+3=0x−1=0x=1 Substitute x=1 in the equation (1)1+y−3=0y=2Hence the point of intersection of two lines (1) and (2) is 1,2 To get the points of intersection of lines (1) and (3), eliminate y   To eliminate y , add equation (1) after multiplying with 7 from (3) 3x−7y+11+7x+7y−21=010x−10=0x=1 Substitute x=1 in the equation (3) 31−7y+11=0−7y=−14y=2Hence the point of intersection of two lines (1) and (3) is 1,2All three lines are passing through the same point 1,2, it means the given three lines are concurrent. No triangle will be formed.