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What is an equation of the line that passes through the point (1,6)(1,6) and is perpendicular to the line x+3y=27x+3y=27?

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Answer:

y = 3x + 3

Explanation:

x + 3y = 27

3y = -x + 27

y = - 1/3 + 27/3

y = - 1/3x + 9

Slope: -1/3 (slope of the perpendicular line: 3)

(1, 6)

y - 6 = 3(x - 1)

y - 6 = 3x - 3

y = 3x + 3

User Frank Shearar
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