analytic geometrycoordinate systems
Given three lines $L_{1}\equiv a_1x+b_1y+c_1=0$ ,
$L_{2}\equiv a_2x+b_2y+c_2=0$, $L_{3}\equiv a_3x+b_3y+c_3=0$ where $c_1,c_2$ and $c_3$ are all positive.
Find the condition that the point $P\equiv(x_0,y_0)$ may lie inside the triangle formed by three given lines.
The answer given was $(a_1x_0+b_1y_0+c_1)(a_2x_0+b_0y+c_2)(a_3x_0+b_3y_0+c_3)<0$.
But I am not able to justify.
Clearly, we can scale the coefficients of a given linear equation by any (non-zero) constant and the result is unchanged. Therefore, by dividing-through by $\sqrt{a_i^2+b_i^2}$, we may assume our equations are in "normal form":
$$\begin{align} x \cos\theta + y \sin\theta - p &= 0 \\ x \cos\phi + y \sin\phi - q &= 0 \\ x \cos\psi + y \sin\psi - r &= 0 \end{align}$$
with $\theta$, $\phi$, $\psi$ and $p$, $q$, $r$ (and $A$, $B$, $C$ and $a$, $b$, $c$) as in the figure:
Then
$$C_1 = \left|\begin{array}{cc} \cos\phi & \sin\phi \\ \cos\psi & \sin\psi \end{array} \right| = \sin\psi\cos\phi - \cos\psi\sin\phi = \sin(\psi-\phi) = \sin \angle ROQ = \sin A$$ Likewise, $$C_2 = \sin B \qquad C_3 = \sin C$$
Moreover, $$D := \left|\begin{array}{ccc} \cos\theta & \sin\theta & - p \\ \cos\phi & \sin\phi & - q \\ \cos\psi & \sin\psi & - r \end{array}\right| = - \left( p C_1 + q C_2 + r C_3 \right) = - \left(\;p \sin A + q \sin B + r \sin C\;\right)$$
Writing $d$ for the circumdiameter of the triangle, the Law of Sines tells us that $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$
Therefore,
$$\begin{align} D &= - \left( \frac{ap}{d} + \frac{bq}{d} + \frac{cr}{d} \right) \\[4pt] &= -\frac{1}{d}\left(\;ap + b q + c r\;\right) \\[4pt] &= -\frac{1}{d}\left(\;2|\triangle COB| + 2|\triangle AOC| + 2|\triangle BOA| \;\right) \\[4pt] &= -\frac{2\;|\triangle ABC|}{d} \end{align}$$
Also, $$C_1 C_2 C_3 = \sin A \sin B \sin C = \frac{a}{d}\frac{b}{d}\sin C= \frac{2\;|\triangle ABC|}{d^2}$$
Finally: $$\frac{D^2}{2C_1C_2C_3} = \frac{4\;|\triangle ABC|^2/d^2}{4\;|\triangle ABC|/d^2} = |\triangle ABC|$$
If two lines do not intersect--i.e., they have no points in common--then the system of equations $$\begin{align*} a_1 x + b_1 y + c_1 &= 0 \\ a_2 x + b_2 y + c_2 &= 0 \end{align*}$$ will have no solution for $(x,y)$. Thus, if we solve the system, we find $$x = \frac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}, \quad y = \frac{a_2 c_1 - a_1 c_2}{a_1 b_2 - a_2 b_1}.$$ This solution does not exist or is indeterminate if $a_1 b_2 - a_2 b_1 = 0$.
However, some care is required: two lines coincide if $$(a_1, b_1, c_1) = k(a_2, b_2, c_2)$$ for some nonzero scalar constant $k$, and in this case, both the numerators and denominator of the aforementioned solution are zero, meaning that there are infinitely many points that the two equations share in common. So while it is a sufficient condition for $a_1 b_2 - a_2 b_1 \ne 0$ to imply that the lines intersect, it is not a strictly necessary condition, and for that reason, the question should have been better phrased by saying "two lines...intersect if..." rather than "only if"; alternatively, it might be phrased "two distinct lines...intersect only if...."
Best Answer
Hint: Observe that intersection point of 2 lines say $L_1,L_2$ lie on the same side of $L_3$ as point $P$ lies.
Repeating this for all combinations of lines gives the above result.