How do you write an equation of a line passing through (6, 1), perpendicular to #3x + y = 7#?

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How do you write an equation of a line passing through (6, 1), perpendicular to #3x + y = 7#?

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Answer 1 · 2016-07-18T00:48:36

#x-3y=3#

Explanation:

Any line perpendicular to #3 x+y = 7# must be of the form

#x-3y=c#

where #c# is a constant (this is easy to see from the fact that the slopes #m_1# and #m_2# of two mutually perpendicular lines obey #m_1 m_2 = -1#). Since the point #(6,1)# lies on this line, the constant #c# is given by

#c = 6-3 xx 1 = 3#

The straight line that we seek is

#x-3y=3#

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How do you write an equation of a line passing through (6, 1), perpendicular to #3x + y = 7#?

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