Area of the triangle determined by the line $x+y=3$ and the bisector of angle between the lines $x^2-y^2+2y=1$ What is the area of the triangle formed by the lines $x+y=3$ and angle bisectors of pair of straight lines $x^2-y^2+2y=1$ . I found the intersection point of these equations $(1,2)$ but not getting any idea of how to find the area . Any way to solve this problem. Thanks!
 A: First, observe that just like @Nicholas said, the equation $\,x^2-y^2+2y=1\,$ defines two lines:
\begin{align}
x^2-y^2+2y=1 \iff x^2 = (y-1)^2 \implies
\begin{cases} l_1: & y = x + 1 \\ l_2: & y = -x + 1 \end{cases}
\end{align}
The slope of the first one is $\,\dfrac{\pi}{4} = 45º,\,$ and for the second one the slope is $\,\dfrac{3\pi}{4} = 135º,\,$ i.e. they are perpendicular.
Equations of bisecting lines can be found by adding and subtracting equations of original lines:
\begin{align}
\begin{cases}b_1:&x=0\\b_2:&y=1\end{cases}
\end{align}
Points of intersection of these lines with each other and with the line $\,l_0: \ \; y = 3-x\,$ are
\begin{align}
A & := l_0 \cap b_1 &\iff& &&\begin{cases}y = 3-x\\ x = 0\end{cases} 
&\implies&& A &= \big(\,0,\,3\,\big) \\
B & := b_1 \cap b_2 &\iff& &&\begin{cases}x = 0\\y=1\end{cases} 
&\implies&& B &= \big(\,0,\,1\,\big) \\
C & := l_0 \cap b_1 &\iff& &&\begin{cases}y = 3-x\\ y=1\end{cases} 
&\implies&& C &= \big(\,2,\,1\,\big) \\
\end{align}
Observe that $\;b_1\perp b_2\;$ so that $\;\triangle\, ABC\;$ is right triangle.
Therefore the area $\,S\,$ of $\;\triangle\, ABC\;$ is just a half of product of legs $\, AB\,$ and $\,BC\,$:
\begin{align}
S_{\triangle ABC} = \dfrac{1}{2}\,\big\|\left(0,\,1\right) - \left(0,\,3\right)\big\|\, \big\|\left(2,\,1\right) - \left(0,\,1\right)\big\| = 2
\end{align}

