What is the slope of a line perpendicular to \(3 x+4 y=13\)?
Explanation
Solve the given equation for y to find the slope of the given line:
\(3 x+4 y=13\)
→ \(3 x+4 y-3 x=13-3 x\)
→ \(4y = -3x + 13\)
→ \(\frac{4 y}{4}=\frac{-3 x+13}{4}\)
→ \(y=\frac{-3}{4} x+\frac{13}{4}\)
→ \(m=\frac{-3}{4}\)
A perpendicular line has a slope that is the negative reciprocal, so its slope is 43