What?Area?

Let the line passing through B intersects AC at D. Then area of triangle ABD is (3/2)BD and triangle BCD is (1/2)BD. Therefore the area of the triangle ABC is the sum of these two, which is equal to 2BD. If the side length of the triangle ABC be a, then 2BD=(√3/4)a^2 or BD=(√3/8)a^2. If CD is of length x and angle CBD be α, then sin(α)=1/a, BD=(√3/2)(x/sinα). Also, x=(√3/2)cosα+(1/2)sinα. These yield x=(2a)/(1+√(3(a^2-1))), and BD=(√3)(a^2)/(1+√(3(a^2-1)))=(√3/8)a^2. Or, a^2=52/3 and the required area is (√3/4)(52/3)=13/(√3)=7.50555....

Suppose that the vertices of the triangle have coordinates $(0,0)$ , $(x,3)$ and $(y,-1)$ , where $x,y > 0$ . Then we need $x^2 + 9 \; = \; y^2 + 1 \; = \; (x-y)^2 + 16$ so that $y^2 = x^2 + 8$ and $2xy = x^2 + 15$ . Hence $\begin{aligned} 4x^2(x^2 + 8) & = \; (x^2 + 15)^2 \\ 3x^4 + 2x^2 - 225 & = \; 0 \\ (3x^2 - 25)(x^2 + 9) & = \; 0 \end{aligned}$ and hence $x = \tfrac{5}{\sqrt{3}}$ , and hence $y = \tfrac{7}{\sqrt{3}}$ . The triangle has area $\tfrac12(x + 3y) = \tfrac{13}{\sqrt{3}} = \boxed{7.505553499}$ .

Since we can have only 2 equations, we need to have only 2 independent variables. Since reflecting the triangle with respect to $y$ -axis or sliding the triangle back and forth along the y=3, 0 & -1 lines, have no effect on the area of the triangle. I elected to assign the $x$ -coordinate of the point on the $y=0$ line to be $0$ , i.e., at the origin and to use the most positive solutions. These are only arbitrary choices.

With those decisions made, solving $\sqrt{\left| x_3\right| ^2+9}=\sqrt{\left| x_3-x_{-1}\right| ^2+16}=\sqrt{\left| x_{-1}\right| ^2+1}$ gives $x$ on the $y=3$ line as $\frac{5}{\sqrt{3}}$ and $x$ on the $y=-1$ line as $\frac{7}{\sqrt{3}}$ . $\frac{1}{2} \left|\left| \begin{array}{ccc} 0 & 0 & 1 \\ \frac{7}{\sqrt{3}} & -1 & 1 \\ \frac{5}{\sqrt{3}} & 3 & 1 \\ \end{array} \right|\right| \Rightarrow \frac{13}{\sqrt{3}} \approx 7.50555349946514$

Please, understand the doubled vertical bars around the matrix of homogeneous coordinates as absolute value of the determinant.