Given:
The line passes through point (7,-3) and perpendicular to line 2x-5y-8=0.
Explanation:
Simplify the equation 2x-5y-8=0 in slope intercept form.
[tex]\begin{gathered} 2x-5y-8=0 \\ 5y=2x-8 \\ y=\frac{2}{5}x-\frac{8}{5} \end{gathered}[/tex]So slope of line is 2/5.
Determine the slope of perpendicular line.
[tex]\begin{gathered} m\cdot\frac{2}{5}=-1 \\ m=-\frac{5}{2} \end{gathered}[/tex]The equation of line with slope m = -5/2 is,
[tex]y=-\frac{5}{2}x+c[/tex]Substitute 7 for x and -3 for y in the equation y = -5/2x + c to determine the value of c.
[tex]\begin{gathered} -3=-\frac{5}{2}\cdot(7)+c \\ c=-3+\frac{35}{2} \\ =\frac{-6+35}{2} \\ =\frac{29}{2} \end{gathered}[/tex]The value of c is 29/2.
The equation of line is,
[tex]\begin{gathered} y=-\frac{5}{2}x+\frac{29}{2} \\ y=\frac{-5x+29}{2} \\ 2y+5x-29=0 \end{gathered}[/tex]So answer is 2y + 5x -29 = 0