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

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

1 Answer

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Answer 1 · 2016-07-18T00:48:36

$x - 3 y = 3$ $x-3y=3$

Explanation:

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

$x - 3 y = c$ $x-3y=c$

where $c$ $c$ is a constant (this is easy to see from the fact that the slopes $m_{1}$ $m_1$ and $m_{2}$ $m_2$ of two mutually perpendicular lines obey $m_{1} m_{2} = - 1$ $m_1 m_2 = -1$ ). Since the point $(6, 1)$ $(6,1)$ lies on this line, the constant $c$ $c$ is given by

$c = 6 - 3 \times 1 = 3$ $c = 6-3 xx 1 = 3$

The straight line that we seek is

$x - 3 y = 3$ $x-3y=3$

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

1 Answer

Explanation:

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

1 Answer

Explanation:

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