<s>[INST] Slopes of perpendicular diagonals of a quadrilateral and its sides Let ABCD be a quadrilateral, let E be the intersection of lines AB and CD, and let F be the intersection of lines BC and AD. If the lines AC and BD of this quadrilateral are perpendicular and the slopes of the equations of its sides AB, BC, CD, DA are, respectively, $-1, 2, 3, 4$, which is the slope of equation of line EF?
Besides using the obvious relation between the slopes of lines AC e BD, what do you do?
 [/INST] The quadrilateral looks concave. Without loss of generality, assume that $D$ lies on the origin of the $xy$-coordinate system, hence, $D=(0,0)$.
Since $m_{AD}=4$ and $m_{CD}=3$, then lines $AD$ and $CD$ should have the following equations:
$$AD\implies f(x):=4x\\
CD\implies g(x):=3x$$
Thus, we know that $A=(a,f(a))$ and $C=(c,g(c))$ for $a,c \in \mathbb R$. Since the diagonals of $ABCD$ are perpendicular, then we know that: 
$$m_{AC}=-\frac1{m_{DB}}$$
Since we already know points $A,C$, then $m_{AC}$ is:
$$m_{AC}=\frac{g(c)-f(a)}{c-a}\implies m_{DB}=-\frac{c-a}{g(c)-f(a)}\\
DB\implies h(x):=-\frac{c-a}{g(c)-f(a)}x\\
\implies B=(b,h(b))\,\,\,\,\forall b \in \mathbb R$$
Now, we have to solve the following equations for $a,c$ in terms of $b$ in order to satisfy the slope requirements:
$$m_{AB}=-1=\frac{h(b)-f(a)}{b-a}\\
m_{BC}=2=\frac{h(b)-g(c)}{b-c}$$
And we get one of the solutions as:
$$\left\{a\to \frac{1}{55} \left(\sqrt{346} b+26 b\right),c\to \frac{1}{11} \left(\sqrt{346} b-7 b\right)\right\}\tag{1}$$

Now, we can write the following equations of the sides of the quadrilateral:
$$AB\implies y=\frac{f\left(a\right)-h\left(b\right)}{a-b}\left(x-b\right)+h\left(b\right)\\
BC\implies y=\frac{g\left(c\right)-h\left(b\right)}{c-b}\left(x-b\right)+h\left(b\right)\\
CD\implies y=3x\\
AD\implies y=4x$$
Solving for the intersection of $AB$ and $CD$ to get $E$, and $BC$ and $AD$ to get $F$, we find that $F$ and $E$ are:
$$E=\left(\frac{a b (17 a-13 c)}{4 a^2+13 a b-3 a c-10 b c},\frac{3 a b (17 a-13 c)}{4 a^2+13 a b-3 a c-10 b c}\right)\\
F=\left(\frac{b c (13 a-10 c)}{17 a b-4 a c-13 b c+3 c^2},\frac{4 b c (13 a-10 c)}{17 a b-4 a c-13 b c+3 c^2}\right)\tag{2}$$
Now, using $(1)$ above and substituting it in $(2)$, we get simpler definitions for $E,F$, to wit (again, this should be for all $b\in\mathbb R$):
$$E=\left(\frac{1}{44} \left(\sqrt{346}+26\right) b,\frac{3}{44} \left(\sqrt{346}+26\right) b\right)\\
F=\left(\frac{1}{22} \left(\sqrt{346}-7\right) b,\frac{2}{11} \left(\sqrt{346}-7\right) b\right)$$
And thus, we get the slope of $EF$, which should be invariant to $b$, as:
$$m_{EF}=\frac{\frac{3}{44} \left(\left(\sqrt{346}+26\right) b\right)-\frac{4}{22} \left(\left(\sqrt{346}-7\right) b\right)}{\frac{1}{44} \left(\sqrt{346}+26\right) b-\frac{1}{22} \left(\sqrt{346}-7\right) b}$$
Which simplifies to:
$$\bbox[10px, border:2px solid red]{\therefore m_{EF}=\frac{1}{19} \left(55-\sqrt{346}\right)}\tag3$$ 

You can check this Desmos implementation. You can check that all the slope requirements were met, and $(3)$ should give you the correct answer.
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