Example 2. Form the equation of the line passing through the intersection point of the lines $x+y-2=0$ and $3x+2y-5=0$ and perpendicular to the line $3x+4y-12=0$.
Solution. 1) The coordinates of the point $M_{0}$ of intersection of the lines are found by solving the system

$$
\left\{\begin{array}{l}
x+y=2 \\
3 x+2 y=5
\end{array} \quad \Rightarrow \quad M(1,1)\right.
$$

2) The slopes of perpendicular lines are related (point $3^{\circ}$) as follows: $k_{2}=-1 / k_{1}$. The slope of the given line is $k_{1}=-\frac{A}{B}=-\frac{3}{4}$ (point $1^{\circ}$). Therefore, $k_{2}=-\frac{1}{-3 / 4}=\frac{4}{3}$.
3) The equation of the desired line passing through the point $M_{0}(1,1)$ and having the slope $k_{2}$ (point $2^{\circ}$) can be written as $y-1=$ $=\frac{4}{3}(x-1)$. We convert it to the general form: $4 x-3 y-1=0$.