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What is the equation of the line that passes through the point (6, 1) and is perpendicular to the line y = –3x + 2
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if the lines perpendicular each other then \[m _{1}m _{2}=-1\] we know second line`s slop is -3 ( because y=ax+b then m=a) \[m _{1}(-3)=-1\] \[m _{1}=1/3\] if we know slop and one point we can find the equation \[m=\frac{y-y _{1}}{x-x _{1}}\] \[\frac{1}{3}=\frac{y -1}{x -6}\] \[x-6=y-1\] \[y=x-5\]
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