Calculus with Applications (10th Edition)

$5x-8y=-40$
Solving for y, we find the slope of $8\mathrm{x}+5\mathrm{y}=3:$ $8x+5y=3$ $5y=-8x+3$ $y=\displaystyle \frac{-8}{5}x+\frac{3}{5}$ $m_{1}=-\displaystyle \frac{8}{5}$ The line perpendicular to it has slope $m= -\displaystyle \frac{1}{m_{1}}=\frac{5}{8}.$ Use $y-y_{1}=m(x-x_{1}) \ \ \$ (the point-slope form) and rearrange : $y-5=\displaystyle \frac{5}{8}(x-0)\qquad/\times 8$ $8(y-5)=5x$ $8y-40=5x$ $-40=5x-8y$ $5x-8y=-40$