By definition you know that the parallel lines have the same slope and that the slopes of the perpendicular lines satisfy the equation
[tex]\begin{gathered} m_2=\frac{-1}{m_1} \\ \text{ Where }m_1\text{ and }m_2\text{ are the slopes of lines 1 and 2, respectively} \end{gathered}[/tex]So, you can take the equation of the second line to the form
[tex]\begin{gathered} y=mx+b \\ \text{ Where} \\ m\text{ is slope of the line} \\ b\text{ is y-intercept} \end{gathered}[/tex]Then you have
[tex]\begin{gathered} 10x-5y=15 \\ \text{ Subtract 10x from both sides of the equation} \\ 10x-5y-10x=15-10x \\ -5y=15-10x \\ \text{ Divide by -5 on both sides of the equation} \\ \frac{-5y}{-5}=\frac{15}{-5}-\frac{10x}{-5} \\ y=-3+2x \\ y=2x-3 \end{gathered}[/tex]Now you know that the slope of the first line is -1/2 and the slope of the second line is 2, that is
[tex]\begin{gathered} m_1=-\frac{1}{2} \\ m_2=2 \end{gathered}[/tex]Let is see if the lines are perpendicular
[tex]\begin{gathered} m_2=\frac{-1}{m_1} \\ 2=\frac{\frac{-1}{1}}{\frac{-1}{2}} \\ 2=\frac{-1\cdot2}{1\cdot-1} \\ 2=\frac{2}{1} \\ 2=2 \end{gathered}[/tex]Since we arrive at a true statement, then the lines are perpendicular.
Therefore, the correct answer is B. Perpendicular.