Two lines are perpendicular if the product of their slopes is -1.
When the variable y is isolated and the expressions are reduced to its lowest terms, the coefficient of the variable x corresponds to the slope of the line.
Isolate y from both equations to find if the lines are perpendicular or not.
First expression:[tex]\begin{gathered} 3-\frac{5y-x}{2}=2x+2 \\ \\ \Rightarrow-\frac{5y-x}{2}=2x-1 \\ \\ \Rightarrow5y-x=-2(2x-1) \\ \\ \Rightarrow5y-x=-4x+2 \\ \\ \Rightarrow5y=-3x+2 \\ \\ \Rightarrow y=-\frac{3}{5}x+\frac{2}{5} \end{gathered}[/tex]Second expression:[tex]\begin{gathered} 3x-5y=13 \\ \\ \Rightarrow5y=3x-13 \\ \\ \Rightarrow y=\frac{3}{5}x-\frac{13}{5} \end{gathered}[/tex]
The slope of the first line is -3/5 and the slope of the second line is 3/5. The product of the slopes is:
[tex]\left(-\frac{3}{5}\right)\left(\frac{3}{5}\right)=-\frac{9}{25}[/tex]
Which is not equal to -1.
Therefore, the lines are not perpendicular.
