What is the slope of the line perpendicular to $y = - \frac{1}{3} x + 6$ $y=-1/3x+6$ ?

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What is the slope of the line perpendicular to $y = - \frac{1}{3} x + 6$ $y=-1/3x+6$ ?

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Answer 1 · 2018-04-08T10:51:30

$Slope of perpendicular line m_{1} = - \frac{1}{m} = 3$ $color(red)("Slope of perpendicular line " m_1 = - 1/m = 3$

Explanation:

Standard form of Slope-Intercept equation is $y = m x + c$ $y = m x + c$

Given equation is $y = - (\frac{1}{3}) x + 6$ $y = -(1/3) x + 6$

Hence $S l o p e m = - (\frac{1}{3})$ $Slope " " m = -(1/3)$

$Slope of perpendicular line m_{1} = - \frac{1}{m} = - (\frac{1}{- \frac{1}{3}}) = 3$ $color(red)("Slope of perpendicular line " m_1 = - 1/m = -(1/(-1/3)) = 3$

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What is the slope of the line perpendicular to $y = - \frac{1}{3} x + 6$ $y=-1/3x+6$ ?

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Explanation:

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