Find the area of a triangle whose three sides are having the equation x+y=2,x-y=0 and x+2y-6=0.

please let me know if this falls under co-ordinate geometry or theoritical geometry and I need the solution too.

The given equations are :

x + y = 2   ..............1
x - y = 0    ..............2
x + 2y - 6 = 0
 x + 2y = 6   ..............3

Adding  1  and  2  we get,2x = 2 x = 1Put x = 1 in  1 we get,1 + y = 2 y = 2 - 1 = 1 Therefore, point of intersection of  1  and  2 is  1,1Subtracting 2  from  3  we get,3y = 6y = 2Put y = 2 in  2 we get,x - 2 = 0 x = 2 The point of intersection of  2  and  3 is 2,2Subtracting 1  from  3  we get,y = 4Put y = 4 in  1  we get,x + 4 = 2 x = 2 - 4 = -2The point of intersection of 1  and  3 is -2,4Then area of  formed by joining these points    1,1, 2,2  and -2,4 is given by :Area of  =   12 x1 y2 - y3 + x2 y3 - y1 + x3 y1 - y2                   =   12 1 2 - 4 + 2 4 - 1 +-2 1 - 2                   =   12-2 + 6 + 2                   =   12 × 6 = 3 square units.
 

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