Given the information, we have that the line passes through the points (-5,5) and (-2,0), therefore, we can find the equation like this:
[tex]\begin{gathered} \text{First we find the slope:} \\ m=\frac{y_2-y_1}{x_2-x_1}=\frac{0-5}{-2-(-5)}=-\frac{5}{3} \\ m=\frac{5}{3} \\ \text{Now we use the point-slope formula (using the point (-2,0)):} \\ y-y_0=m(x-x_0) \\ \Rightarrow y-0=-\frac{5}{3}(x-(-2)) \\ \Rightarrow y=-\frac{5}{3}x-\frac{10}{3} \\ \text{Multiplying both sides by 3 we get:} \\ 3y=-5x-10 \\ \Rightarrow3y+5x=-10 \end{gathered}[/tex]
Now that we have the equation of the line, we can easily find the parallel and perpendicular. For the parallel, we need that the line use the same slope (m=-5/3), therefore, from the options, the equation 5x+3y=13 is the only parallel to the original line. Now, for the perpendicular we have:
[tex]\begin{gathered} m_p=-\frac{1}{m}=-\frac{1}{-\frac{5}{3}}=\frac{3}{5} \\ \text{ Using the same point as before (-2,0) we get:} \\ y-0=\frac{3}{5}(x+2) \\ \Rightarrow y=\frac{3}{5}x+\frac{6}{5} \\ \text{Multiplying both sides by 5 we get:} \\ 5y=3x+6 \\ \Rightarrow-3x+5y=6 \end{gathered}[/tex]
Therefore, the equation of the perpendicular line is -3x+5y=6
