Let $\triangle ABC$ be a triangle of sides $AB$, $AC$ and $BC$ which has the following equations- $$AB \equiv a_1x+b_1y+c_1=0 \\ AC \equiv a_2x+b_2y+c_2=0 \\ BC \equiv px+qy+r=0$$ Then an angle $\angle A$ is acute if $$ \left | \begin{matrix} a_1 & b_1 \\ p & q \\ \end{matrix} \right | \left | \begin{matrix} p & q \\ a_2 & b_2 \\ \end{matrix} \right | (a_1a_2+b_1b_2)<0 $$ And the angle $\angle A$ is obtuse if $$ \left | \begin{matrix} a_1 & b_1 \\ p & q \\ \end{matrix} \right | \left | \begin{matrix} p & q \\ a_2 & b_2 \\ \end{matrix} \right | (a_1a_2+b_1b_2)>0 $$
I found this formula in my textbook without any proof to immediately learn if an internal angle of a triangle is acute or obtuse if we are given the equations of the sides of the triangle. But I absolutely have no idea how to prove this theorem. I have tried many approaches using coordinate geometry to proof it but to no avail. Note that I only know coordinate geometry and only I can proof through this as this is included in my "Two dimensional Coordinate Geometry" chapter. I know this site requires some effort should've been shown but I have nothing of value to show. This is a common formula as I understand it and some good coordinate geometry book may have it proven.
I need help to prove this formula. Your help is appreciated.
For this proof I will assume, naturally, that none of the lines are parallel and they do not all intersect in one point. I will also assume for the moment that $r\neq0$ so this line does not pass through the origin.
Consider two lines parallel to $AB$ and $AC$, call them $OP$ and $OQ$, whose equations, respectively are
$$a_1x+b_1y=0$$ and $$a_2x+b_2y=0$$
These lines meet $BC$ at points $P$ and $Q$. Since triangles $ABC$ and $OPQ$ are similar with $\angle A$ the same as $\angle O$, it is sufficient to derive the result for $\angle O=\theta$ instead.
The angle $\theta$ will be acute if $\cos\theta>0$ and obtuse if $\cos\theta<0$.
Since $$\cos\theta=\frac{\overrightarrow{OP}\cdot\overrightarrow{OQ}}{|OP||OQ|},$$ so for the acute angle, we just require $$\overrightarrow{OP}\cdot\overrightarrow{OQ}>0$$
Solving simultaneously, we get the coordinates of $P$ to be $$\left(\frac{rb_1}{a_1q-b_1p}, \frac{-ra_1}{a_1q-b_1p}\right)$$
Correspondingly, the coordinates of $Q$ are the same but will the suffix changed from $1$ to $2$.
Therefore the required dot product is $$\frac{rb_1}{a_1q-b_1p}\cdot\frac{rb_2}{a_2q-b_2p}+\frac{-ra_1}{a_1q-b_1p}\cdot\frac{-ra_2}{a_2q-b_2p}$$ $$=\frac{r^2(a_1a_2+b_1b_2)}{\left|\begin{matrix}a_1&b_1\\p&q\end{matrix}\right|\left|\begin{matrix}a_2&b_2\\p&q\end{matrix}\right|}$$
Noting that $r\neq0$, for the angle to be acute, therefore, we require $$\left|\begin{matrix}a_1&b_1\\p&q\end{matrix}\right|\left|\begin{matrix}a_2&b_2\\p&q\end{matrix}\right|(a_1a_2+b_1b_2)>0, $$ with the inequality reversed for the obtuse angle.
Note that this is the same formula as that provided by the OP, but expressed, in my opinion, in a better way.
In the case that $r=0$, then we would translate all three lines so that the intersection point of the first two lines became the origin, and the same argument would apply.