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