area enclosed within 3 lines

Equations of lines $L1$ and $L2$ are $y = x − 2$ and $y = −2x − 2$. If $y = −x$ is the angle bisector of lines $L2$ and $L3$, then what is the area enclosed within the 3 lines $L1, L2$ and $L3$?

equation of line $L1$ is ==> $y = x-2$ Equation of line $L2$ is ==> $y = -2x-2$

!can I find the coordinates of intersection of three line, if yes how? If there intersection points are found then we can find length of sides and then apply hero's formula i.e Area $= \sqrt{s(s-a)(s-b)(s-c)}$

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One way: find points of intersection, find distances between points, find all angles, use one angle to get a height, use area formula. –  Addem Apr 7 '14 at 17:42
One point of intersection: x-2 = -2x-2 so x=0 and y=-2. –  Addem Apr 7 '14 at 17:44
Can you find $L3$? –  chenbai Apr 8 '14 at 8:23

The problem is artificially constructed in such a way that all calculated data are integers (or the square root of an integer). Referring to the diagram below:-

A. By solving L(1) and L(2), $A = (0, –2)$.

B. $L(1)$ cuts the x-axis at $B = (2, 0)$.

C. $L$ cuts $L(1)$ at $C = (1, –1)$.

D. By considering the slopes of $L(1)$ and $L$, $L \perp L(1)$.

E. $L(2)$ and $L$ intersect at $D = (-2, 2)$.

F. BY [$ASA$], $\triangle DAC$ must congruent to $\triangle DXC$ for some point $X$ on $AC$ produced.

G. $X$ happens to be $B$ because $\triangle OAB$, $\triangle OAC$ and $\triangle OCB$ are all $45^\circ-45^\circ-90^\circ$ right angled triangles.

The above provides all the necessary information to compute the required area.

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