areageometry
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 $= \sqrt{s(s-a)(s-b)(s-c)}$
Draw a common chord of circles $G$ and $F$. The plan is to calculate the areas of segments $CEDB$ and $CEDO$ separately.
Let the distance of this chord from centre of $G$ be $x$. Upon using pythagorean theorem on $\triangle CEO$ and $\triangle CEA$, we will get $x = 2$.
To calculate the area of segment $CEDB$, we subtract the area of triangle $COD$ from sector $CODB$. Note that $CE = \sqrt{36-x^2} = 4\sqrt{2}$. So angle $COD$ can be written as $2 \arctan (\tfrac{CE}{OE}) = 2 \arctan{2\sqrt2}$.
Area of sector $CODB$ is simply $\frac{2\arctan(2\sqrt 2)}{2\pi} \pi \cdot 6^2 = 36 \arctan(2\sqrt 2)$. The area of triangle $COD$ is $8\sqrt 2$.
Therefore area of segment $CEDB$ is $\color{red}{36 \arctan(2\sqrt 2)-8\sqrt 2}$
Repeat the calculations to get area of segment $CEDO$. That is subtract area of triangle $CDA$ from area of sector $CODA$. I get upon calculation, area of segment $CEDO $ as $ \color{red}{81\arctan(\tfrac{4\sqrt 2}{ 7})-28\sqrt 2}$.
Now area of the shaded region in question is simply the area of circle $F$ minus the area of segments $CEDB$ and $CEDO$. So the required answer is:
$$81 \pi-36\arctan(2\sqrt2) - 81 \arctan(\tfrac{4\sqrt2}{7}) + 36\sqrt 2$$
Best Answer
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.