Consider that the equation of a line with slope 'm' and y-intercept 'c' is given by,
[tex]y=mx+c[/tex]
Convert the given equation,
[tex]\begin{gathered} 2x-5y=5 \\ 5y=2x-5 \\ y=\frac{2}{5}x-1 \end{gathered}[/tex]
So the slope of the given equation is,
[tex]m=\frac{2}{5}[/tex]
Let the equation of the perpendicular line be,
[tex]y=m^{\prime}x+c^{\prime}[/tex]
For the lines to be parallel, the product of slopes must be -1.
[tex]\begin{gathered} m^{\prime}\times m=-1 \\ m^{\prime}\times\frac{2}{5}=-1 \\ m^{\prime}=\frac{-5}{2} \end{gathered}[/tex]
So the equation becomes,
[tex]y=\frac{-5}{2}x+c^{\prime}[/tex]
Given that the perpendicular passes through the point (2,-9),
[tex]\begin{gathered} -9=\frac{-5}{2}(2)+c^{\prime} \\ -9=-5+c^{\prime} \\ c^{\prime}=-4 \end{gathered}[/tex]
Substitute the value in the equation,
[tex]\begin{gathered} y=\frac{-5}{2}x-4 \\ 2y=-5x-8 \\ 5x+2y+8=0 \end{gathered}[/tex]
Thus, the equation of the perpendicular line is 5x + 2y + 8 = 0 .
