How do you write an equation of a line passing through (6, 1), perpendicular to 3x + y = 73x+y=7?

1 Answer
Ananda Dasgupta
Jul 18, 2016

x-3y=3x−3y=3

Explanation:

Any line perpendicular to 3 x+y = 73x+y=7 must be of the form

x-3y=cx−3y=c

where cc is a constant (this is easy to see from the fact that the slopes m_1m1 and m_2m2 of two mutually perpendicular lines obey m_1 m_2 = -1m1m2=−1). Since the point (6,1)(6,1) lies on this line, the constant cc is given by

c = 6-3 xx 1 = 3c=6−3×1=3

The straight line that we seek is

x-3y=3x−3y=3

Related questions
See all questions in Equations of Perpendicular Lines
Impact of this question
1668 views around the world
You can reuse this answer
Creative Commons License