Find area of $\triangle ABC$ given point $C(-2,-1)$ and equations of the angle bisector $b_{a}$ and the height $h_{a}$ from the edge $A$ My diagram
The equation of the angle bisector $b_a$ is $2x+y-5=0$ and the equation of the height is $h_a$ is $x+y-5=0$, point $A=b_a \cap h_a$, point $C$ has coordinates $(-2,-1)$. What is the area of the triangle?
I have found the coordinates of point $A$: $A(x_A, y_A)=b_a \cap h_a \implies \begin{cases} 2x_A+y_A-5=0 \\ x_A+y_A-5=0 \end{cases} \implies A(0,5)$ but don't know how to continue.
 A: Guide: 
a) Reflect the $C$ across angle bisector $b_a$, name this point $C'$, then $B\in AC'$
(Solution: The line through $C$ perpendicular to $b_a$ is $y=-x/2$, so it cuts $b_a$ in $D(2,1)$. Since $D$ halves $CC'$ we have $C'(6,3)$ and so the line $AC'$ is $y=-x/3+5$)
b) Write perpendicular $\ell$ to $h_a$ thorugh $C$, then $B\in \ell$. 
So $B\in AC'\cap \ell$ so you can calculate $B$ and then area of $ABC$.
A: Idea:
Let the point $B$ be $(x,y)$. Then $BC$ is perpendicular to the altitude given $h_a$. So
$$m_{BC} \cdot m_{h_a}=-1 \implies \frac{y+1}{x+2}\cdot (-1)=-1 \implies y=x+1.$$
You have equation of line $AC$ (since both $A$ and $C$ are known). Suppose bisector $b_a$ meets the side $BC$ at $M$. Use the formula $\tan \alpha=\frac{m_1-m_2}{1+m_1m_2}$, to get 
$$\tan \left(\angle CAM\right)=\frac{m_{AC}-m_{b_a}}{1+m_{AC}m_{b_a}}=\frac{3-(-2)}{1+(-6)}=-1.$$
Likewise, 
$$\tan \left(\angle MAB\right)=\frac{m_{AB}-m_{b_a}}{1+m_{AB}m_{b_a}}=\frac{\frac{y-5}{x}-(-2)}{1-2(\frac{y-5}{x})}=\frac{y-5+2x}{x-2y+10}.$$
Then use the bisector condition $\angle CAM=\angle MAB$ to get another equation involving $x,y$ to determine point $B$. Once $B$ is known you can use the standard formula for the area.
