What is the slope of any line perpendicular to 6x+3y-1=0?
6x+3y-1=0
3y = -6x + 1
y = -2x + 1/3
The slope is -2, so the perpendicular slope line is the negative reciprocal, 1/2.
=^._.^=
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\(6x+3y-1=0\\ y=-\frac{6}{3}x+\frac{1}{3}\\ y=-2x+\frac{1}{3}\)
m n
\(m_{perpendicular}=-\frac{1}{m}=-\frac{1}{-2}=\color{blue}\frac{1}{2}\)
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