Question

find the area of a triangle whose three sides are having the equations x+y=2 , x-y=0 , x+2y-6=0.​

Answer 1

X+y=2

x-y=0

x+2y-6=0

first find points of intersection of these lines

x+y=2

x-y=0 solve this we find (1,1)

x-y=0

x+2y=6 solve then (2,2)

x+2y=6

x+y=2 solve then (-2,4)

now (1,1) , (2,2) and (-2,4) are the vertices of triangle .

so area of triangle=1/2 {1 (2-4) +2 (4-1)-2 (1-2)}=1/2 {-2 +6 +2}=3 square unit