What is the slope of a line perpendicular to \(3 x+4 y=13\)?



Solve the given equation for y to find the slope of the given line:

\(3 x+4 y=13\)

\(3 x+4 y-3 x=13-3 x\)

\(4y = -3x + 13\)

\(\frac{4 y}{4}=\frac{-3 x+13}{4}\)

\(y=\frac{-3}{4} x+\frac{13}{4}\)


A perpendicular line has a slope that is the negative reciprocal, so its slope is 43

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