Question

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

Answer 1

`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`