Number of triangles formed by the lines  $\mathrm{a}\left(\mathrm{x}+\mathrm{y}-2\right)+\mathrm{b}\left(3\mathrm{x}+4\mathrm{y}+7\right)=0$ and $5\mathrm{x}-4\mathrm{y}-1=0$

Any two lines among three lines are parallel or if three line are passing through the same point then the number of triangles formed by those lines is zero, otherwise number of triangles is 1. 
The equation $\mathrm{a}\left(\mathrm{x}+\mathrm{y}-2\right)+\mathrm{b}\left(3\mathrm{x}+4\mathrm{y}-7\right)=0$ represents the set of lines passing through the point of intersection of $\mathrm{x}+\mathrm{y}-2=0$ and  $3\mathrm{x}+4\mathrm{y}+7=0$
Eliminate  y
$\begin{array}{c}4\left(\mathrm{x}+\mathrm{y}-2\right)-\left(3\mathrm{x}+4\mathrm{y}-7\right)=0\\ 4\mathrm{x}+4\mathrm{y}-8-3\mathrm{x}-4\mathrm{y}+7=0\\ \mathrm{x}-1=0\end{array}$ 
Plug in x=1 in the equation  $\mathrm{x}+\mathrm{y}-2=0$
$\begin{array}{c}1+\mathrm{y}-2=0\\ \mathrm{y}-1=0\\ \mathrm{y}=1\end{array}$ 
Hence the point of intersection of $\mathrm{x}+\mathrm{y}-2=0$ and $3\mathrm{x}+4\mathrm{y}+7=0$  is $\left(1,1\right)$ , but this point lies on the third line  $5\mathrm{x}-4\mathrm{y}-1=0$
It implies that all three lines are concurrent, hence the number of triangles formed by the given three lines is zero