## College Algebra 7th Edition

To determine if the lines are parallel or perpendicular, we compare their slopes. We start with the line $5x-8y=3$: $5x-8y=3$ $8y=5x-3$ $y=\frac{5}{8}x-\frac{3}{8}$ So this line has a slope of $\frac{5}{8}$. Next, we look at the line $10y+16x=1$: $10y+16x=1$: $10y=-16x+1$ $y=-\frac{8}{5}x+\frac{1}{10}$ This line has a slope of $-\frac{8}{5}=-\frac{1}{5/8}$ Since the two slopes are negative reciprocals of each other, the lines are perpendicular.