The area (in sq. units) of the triangle formed by the straight line x+y=3 and the angular bisectors of the pair of straight lines x^2-y^2+2 y=1 , is

Step-by-step solution

pair of straight line is aligned & x^2-y^2+2 y=1 & x^2- (y^2-2 y+1 )=0 & x^2-(y-1)^2=0 & (x-y+1)(x+y-1)=0 & aligned So, lines are x+y-1=0 and x-y+1=0 equation of angle bisector is array rlrl & & A₁ x+B₁ y+c₁ A₁^2+B₁^2 & = A₂ x+B₂ y+c₂ A₂^2+B₂^2 & & x+y-1 2 & = x-y+1 2 & So, & x+y-1 2 & = x-y+1 2 & and & x+y-1 2 & = -(x-y+1) 2 & x+y-1 & =x-y+1 array and aligned & x+y-1=-x+y-1 & 2 y=2 and x=0 & y-1=0 y=1 aligned Now, vertices of formed by x=0, y=1 and x+y=3 are (0,1),(0,3) and (2,1) Area of = 1 2 (2)(2)=2 sq. units.