I tried using below stated amazing formula:
Then the area of the triangle that these lines will enclose is given by the magnitude of : $$\frac{det\begin{bmatrix}a_1 & b_1 & c_1\\a_2 & b_2 & c_2\\a_3 & b_3 & c_3\end{bmatrix}^2}{2C_1C_2C_3}$$ [Where $C_1,C_2,C_3$ are the co-factors of $c_1,c_2,c_3$ respectively in the above matrix.] ref.
But couldn't get the desired answer. System of linear questions are as follows.
$$2x+3y+-6=0$$ $$3x+2y+6=0$$ $$3x+-3y+6=0$$
I end up getting decimal answers!
Just solve those equations.
We can find the intersect points are $(0,2), (-2,0), (-6,6)$
Since it is so symmetry, you have plenty of method to find the area.
A simple method:
$Area = \frac{bh}{2} = \frac{2\sqrt{2} \times 5\sqrt{2}}{2} = \frac{10 \times 2}{2} =10$
Or subtraction area of two triangle with base on x-axis may be even easier.
$Area = \frac{5 \times 6}{2} -\frac{5 \times 2}{2} = 10$
Or use determinant formula. link
Basically, the first thing you need to do is find the length of each of the triangle's sides. You can do this by finding the intersection points of the lines, and the corresponding distances between them.
Let the lines be:
$l_1(x)=a_1x+b_1$
$l_2(x)=a_2x+b_2$
$l_3(x)=a_3x+b_3$
Your three points of intersection will then be:
$(\frac{b_2-b_1}{a_1-a_2}, (\frac{b_2-b_1}{a_1-a_2})x+b_1$)
$(\frac{b_3-b_2}{a_2-a_3}, (\frac{b_3-b_2}{a_2-a_3})x+b_2)$
$(\frac{b_1-b_3}{a_3-a_1}, (\frac{b_1-b_3}{a_3-a_1})x+b_3)$
Calculate the distance between each of these points using the formula $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$.
Once you've obtained that, let the three side lengths be $a,b,c$. Using Heron's formula, $s=\frac{a+b+c}{2}:A=\sqrt{s(s-a)(s-b)(s-c)}$.
