Rewritten on 2016-11-12. The OP raised very good questions in the comments. Note that this is not intended as an exhaustive answer (as one might expect from, say, a mathematician?), but more like observations from someone who routinely uses bilinear interpolation for numerical data.
How can a bilinear interpolation be defined for an arbitrary quadrilateral,
i.e. without running into singularities?
Bilinear interpolation is usually defined as
$$f(u,v) = (1-u) (1-v) F_{00} + u (1 - v) F_{01} + (1-u) v F_{10} + u v F_{11}$$
where $0 \le u, v \le 1$ and
$$\begin{array}{}
f(0,0) = F_{00} \\
f(0,1) = F_{01} \\
f(1,0) = F_{10} \\
f(1,1) = F_{11} \\
f(\frac{1}{2},\frac{1}{2}) = \frac{F_{00}+F_{01}+F_{10}+F_{11}}{4}
\end{array}$$
If we use notation
$$p(t; p_0, p_1) = (1-t) p_0 + t p_1 = p_0 + t (p_1 - p_0)$$
for the simplest form of linear interpolation, with $0 \le t \le 1$, $p(0;p_0,p_1) = p_0$, $p(1;p_0,p_1) = p_1$, then bilinear interpolation can be written as
$$f(u,v) = p(u; p(v; F_{00}, F_{01}), p(v; F_{10}, F_{11}))$$
so this simply extends the single-variable linear interpolation to two variables and $2^2 = 4$ samples.
Bilinear interpolation is not often used for arbitrary quadrilaterals. After pondering the questions OP posed in the comments, I realized that the typical form used for interpolation,
$$\begin{cases}
x(u,v) = x_{00} + u ( x_{10} - x_{00}) + v ( x_{01} - x_{00} ) \\
y(u,v) = y_{00} + u ( y_{10} - y_{00}) + v ( y_{01} - y_{00} ) \\
f(u,v) = (1-v) \left ( (1-u) f_{00} + u f_{10} \right ) + (v) \left ( (1-u) f_{10} + u f_{11} \right )
\end{cases}$$
is not applicable to arbitrary quadrilaterals, as it assumes it to be a parallelogram, i.e. with
$$\begin{cases}
x_{11} = x_{10} + x_{01} - x_{00} \\
y_{11} = y_{10} + y_{01} - y_{00}
\end{cases}$$
Solving $x = x(u,v)$, $y = y(u,v)$ for $u$ and $v$ yields
$$\begin{cases}
A = x_{00} (y_{01} - y_{10}) + x_{01} (y_{10} - y_{00}) + x_{10} (y_{00} - y_{01}) \\
u = \frac{ (x_{01} - x_{00}) y - (y_{01} - y_{00}) x + x_{00} y_{01} - y_{00} x_{01} }{A} \\
v = \frac{ (x_{00} - x_{10}) y - (y_{00} - y_{10}) x - x_{00} y_{10} + y_{00} x_{10} }{A}
\end{cases}$$
where $$A = \left(\vec{p}_{10} - \vec{p}_{00}\right) \times \left(\vec{p}_{01} - \vec{p}_{00}\right)$$
where $\times$ signifies the 2D analog of vector cross product, so $\lvert A \rvert$ is the area of the parallelogram. Thus, exactly one solution exists for all non-degenerate parallelograms.
For the most common use case, a regular rectangular axis-aligned grid of samples $p_{ji}$, $0 \le j, i \in \mathbb{Z}$, we have
$$\begin{cases}
x = a_x + b_x i \\
y = a_y + b_y j
\end{cases}$$
with $b_x \ne 0$, $b_y \ne 0$, corresponding to interpolation parameters
$$\begin{cases}
i = \left\lfloor \frac{x - a_x}{b_x} \right\rfloor \\
j = \left\lfloor \frac{y - a_y}{b_y} \right\rfloor \\
u = \frac{x - a_x}{b_x} - i \\
v = \frac{y - a_y}{b_y} - j
\end{cases}$$
so that
$$p(x,y) = (1-v) \left ( (1-u) p_{j,i} + (u) p_{j,i+1} \right ) + (v) \left ( (1-u) p_{j+1,i} + (u) p_{j+1,i+1} \right )$$
To apply bilinear interpolation to an arbitrary quadrilateral, we need to use
$$\begin{cases}
x(u,v) = (1-u)(1-v) x_{00} + (u)(1-v) x_{10} + (1-u)(v) x_{01} + (u)(v) x_{11} \\
y(u,v) = (1-u)(1-v) y_{00} + (u)(1-v) y_{10} + (1-u)(v) y_{01} + (u)(v) y_{11} \\
f(u,v) = (1-u)(1-v) f_{00} + (u)(1-v) f_{10} + (1-u)(v) f_{01} + (u)(v) f_{11}
\end{cases}$$
In some cases it is sufficient to produce additional samples, for example so that each quadrilateral can be split into four sub-quadrilaterals, doubling the resolution. Then, we do not need to solve for $x$ and $y$, and only need to compute
$$\begin{array}{cc}
x\left(\frac{1}{2},0\right), & y\left(\frac{1}{2},0\right), & f\left(\frac{1}{2},0\right) \\
x\left(\frac{1}{2},1\right), & y\left(\frac{1}{2},1\right), & f\left(\frac{1}{2},1\right) \\
x\left(0,\frac{1}{2}\right), & y\left(0,\frac{1}{2}\right), & f\left(0,\frac{1}{2}\right) \\
x\left(1,\frac{1}{2}\right), & y\left(1,\frac{1}{2}\right), & f\left(1,\frac{1}{2}\right)
\end{array}$$
However, solving $(u,v)$ for some specific $(x,y)$ is quite complicated. Indeed, I was surprised how complicated it turns out to be! (I apologize for misrepresenting this case as "easy" in a previous edit. Mea culpa.)
In practice, we first try to solve $u$ or $v$, and then the other by substituting into one of the equations above. If we decide we wish to solve $u$ first, we need to solve
$$U_2 u^2 + U_1 u + U_0 = 0$$
where
$$\begin{cases}
U_2 = (y_{00}-y_{01}) (x_{10}-x_{11}) - (x_{00}-x_{01}) (y_{10}-y_{11}) \\
U_1 = (y_{00}-y_{01}-y_{10}+y_{11}) x - (x_{00}-x_{01}-x_{10}+x_{11}) y + (x_{11}-2 x_{10}) y_{00} + (2 x_{00}-x_{01}) y_{10} + y_{01} x_{10} - y_{11} x_{00} \\
U_0 = (y_{10}-y_{00}) x - (x_{10}-x_{00}) y + y_{00} x_{10} - x_{00} y_{10}
\end{cases}$$
The possible solutions are
$$\begin{cases}
u = \frac{-U_1 \pm \sqrt{ U_1^2 - 4 U_2 U_0}}{2 U_2}, & U_2 \ne 0 \\
u = \frac{-U_0}{U_1}, & U_2 = 0, U_1 \ne 0 \\
u = 0, & U_2 = 0, U_0 = 0
\end{cases}$$
If we find $0 \le u \le 1$, we solve for $v$ by substituting into $X(u,v) = x$,
$$v = \frac{ (y_{00} - y_{10}) u + y - y_{00} }{ (y_{00} - y_{01} - y_{10} + y_{11}) u - y_{00} + y_{01} }$$
or into $Y(u,v) = y$,
$$v = \frac{ (x_{00} - x_{10}) u + x - x_{00} }{ (x_{00} - x_{01} - x_{10} + x_{11}) u - x_{00} + x_{01} }$$
If we find no solutions, we try to solve for $v$ in
$$V_2 v^2 + V_1 v + V_0 = 0$$
where
$$\begin{cases}
V_2 = (x_{00}-x_{01}) (y_{10}-y_{11}) - (y_{00}-y_{01}) (x_{10}-x_{11}) \\
V_1 = (x_{00}-x_{01}-x_{10}+x_{11}) y - (y_{00}-y_{01}-y_{10}+y_{11}) x + (y_{11}-2 y_{10}) x_{00} + (2 y_{00}-y_{01}) x_{10} + x_{01} y_{10} - x_{11} y_{00} \\
V_0 = (x_{10}-x_{00}) y - (y_{10}-y_{00}) x + x_{00} y_{10} - y_{00} x_{10}
\end{cases}$$
The possible solutions are similar to those for $u$:
$$\begin{cases}
v = \frac{-V_1 \pm \sqrt{ V_1^2 - 4 V_2 V_0}}{2 V_2}, & V_2 \ne 0 \\
v = \frac{-V_0}{V_1}, & V_2 = 0, V_1 \ne 0 \\
v = 0, & V_2 = 0, V_0 = 0
\end{cases}$$
If you find $0 \le v \le 1$, you solve for $u$ by substituting into $X(u,v) = x$,
$$u = \frac{(x_{00} - x_{01}) v + x - x_{00} }{ (x_{00} - x_{01} - x_{10} + x_{11}) v - x_{00} + x_{10} }$$
or into $Y(u,v) = y$,
$$u = \frac{(y_{00} - y_{01}) v + y - y_{00} }{ (y_{00} - y_{01} - y_{10} + y_{11}) v - y_{00} + y_{10} }$$
It is also possible to solve $(u,v)$ numerically, by calculating $X(u,v)$ and $Y(u,v)$ repeatedly with different $u$, $v$, until $\lvert X(u,v) - x \rvert \le \epsilon$ and $\lvert Y(u,v) - y \rvert \le \epsilon$, where $\epsilon$ is the maximum acceptable error in $x$ and $y$ (maximum distance to correct $(x,y)$ being $\sqrt{2}\epsilon$).
There are a number of different methods for the numerical search. Some of the following observations may come in handy, when implementing a numerical search:
$$\begin{array}{rl}
\frac{d \, X(u,v)}{d\,u} = & x_{10} - x_{00} + v ( x_{11} - x_{01} - x_{10} + x_{00} ) \\
\frac{d \, X(u,v)}{d\,v} = & x_{01} - x_{00} + u ( x_{11} - x_{01} - x_{10} + x_{00} ) \\
\frac{d \, Y(u,v)}{d\,u} = & y_{10} - y_{00} + v ( y_{11} - y_{01} - y_{10} + y_{00} ) \\
\frac{d \, Y(u,v)}{d\,v} = & y_{01} - y_{00} + u ( y_{11} - y_{01} - y_{10} + y_{00} ) \\
X(u + du, v) - X(u, v) = & du \left ( x_{10} - x_{00} + v ( x_{11} - x_{01} - x_{10} + x_{00} ) \right ) \\
X(u, v + dv) - X(u, v) = & dv \left ( x_{01} - x_{00} + u ( x_{11} - x_{01} - x_{10} + x_{00} ) \right ) \\
Y(u + du, v) - Y(u, v) = & du \left ( y_{10} - y_{00} + v ( y_{11} - y_{01} - y_{10} + y_{00} ) \right ) \\
Y(u, v + dv) - Y(u, v) = & dv \left ( y_{01} - y_{00} + u ( y_{11} - y_{01} - y_{10} + y_{00} ) \right )
\end{array}$$
In other words, it is true that the bilinear interpolation is quite difficult for arbitrary quadrilaterals, and very problematic for self-intersecting quadrilaterals. However, the most common quadrilateral types -- rectangles and parallelograms -- are easy, and even the general case is solvable at least numerically, even in the presence of singularities.
Why bilinear interpolation with quadrilaterals?
As I've shown above, for the rectangles and parallelograms -- the only quadrilaterals I've used bilinear interpolation with in real solutions --, bilinear interpolation is easy and simple.
Indeed, the emphasis on quadrilaterals (in the sense of arbitrary quadrilaterals) seems incorrect, as bilinear interpolation is mostly used with rectangles or parallelograms.
Perhaps the emphasis should be on that bilinear interpolation uses two variables to interpolate between four known values; or more generally, $k$-linear interpolation uses $k$ variables to interpolate between $2^k$ values. Trilinear interpolation is similarly common for cuboids with vertices
$$\begin{cases}
\vec{p}_{011} = \vec{p}_{010} + \vec{p}_{001} - \vec{p}_{000} \\
\vec{p}_{101} = \vec{p}_{100} + \vec{p}_{001} - \vec{p}_{000} \\
\vec{p}_{110} = \vec{p}_{100} + \vec{p}_{010} - \vec{p}_{000} \\
\vec{p}_{111} = \vec{p}_{100} + \vec{p}_{010} + \vec{p}_{001} - 2 \vec{p}_{000}
\end{cases}$$
i.e. cuboids defined by one vertex and three edge vectors.
Regular grids are ubiquitous, and linear mapping is the simplest interpolation method, with easy properties. Cubic interpolation and other interpolation methods do produce better results, but are computationally more expensive, and the properties may produce unwanted behaviour: most typically, the interpolated value is no longer guaranteed to reside within the range spanned by the constants.
Best Answer
I would say this. Your points $P_1$, $P_2$ and $P_3$ form a triangle with area $A_0$. Now place the point $P_4$. Now, assuming there are no self-intersections, $$\operatorname{Area}(P_1 P_2 P_3 P_4) = A_0 \pm \operatorname{Area}(P_1 P_4 P_3),$$ with the last area counted negatively if $P_4$ and $P_2$ are on the same side of $P_1 P_3$, and positively if they're on opposite sides.
So this is really a problem about the area of a certain triangle $P_1 P_4 P_3$, taking into account the sign convention I mentioned. Now using the formula $(1/2) \times$ base $\times$ height, you'll see that there is a line of points $P_4$ that work. You then need to take the intersection of that line with the other line you mentioned. The point $P_4$ may or may not satisfy the condition of the quadrilateral not having self-intersections.