Area of a triangle

Question:
Area of triangle formed by the line
!!x+y!! !!=!! !!3!! and angle bisector of the pair of straight lines !!(x^2 -y^2 + 2y)!! !!=!! !!1!! is: !!a)" "1!!

!!b)" "2!!

!!c)" "4!!

!!d)" "8!!

Solution:
We have;
!!(x^2 -y^2 + 2y)!! !!=!! !!1!! or !!(x^2 - (y-1)^2)!! !!=!! !!0!!
!!x^2!! !!=!! !!(y-1)^2!! !!(x+y-1)!! !!=!! !!0!! or !!(x-y+1)!! !!=!! !!0!!

Here we can find the angle bisector using formula like below:

The equation of their angle bisectors will be:
!!Mod(x+y-1/sqrt(2))!! !!=!! !!Mod(x-y+1/sqrt(2))!!

Or we can see that the given lines have slope of 1, -1 respectively. So the angle bisector will have slope of !!0!!,!!oo!!. The point of intersection os !!(0,1)!!.

So we get:
!!x!! !!=!! !!0!! or !!y!! !!=!! !!1!!

The intersection of !!x!! !!=!! !!0!! !!y!! !!=!! !!1!! and !!(x+y)!! !!=!! !!3!! gives a right angled triangle with coordinates !!(0,3)!! !!(2,1)!! !!(0,1)!!
Therefore Area of the triangle is
!!(1/2).(2.2)!! !!=!! !!2!!

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