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

Nov 22, 2016

"y=1/6x-4

Sorry the explanation is a bit long. Tried to give full explanation of what is going on.

#### Explanation:

$\textcolor{b l u e}{\text{General introduction}}$

consider the equation of a straight line in the standard form of:

$y = m x + c$

In this case $m$ is the slope (gradient) and $c$ is some constant value

A straight line that is perpendicular to this would have the gradient of $\left[- 1 \times \frac{1}{m}\right]$ so its equation is:
$\textcolor{w h i t e}{.}$

$y = \left[\left(- 1\right) \times \frac{1}{m}\right] x + k \text{ "->" } y = - \frac{1}{m} x + k$

Where $k$ is some constant value that is different to that of $c$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the given line equation}}$

Given $\text{ } \textcolor{g r e e n}{6 x + y = 1}$

Subtract $\textcolor{red}{6 x}$ from both sides

color(green)(6xcolor(red)(-6x)+y" "=" "1color(red)(-6x)

But $6 x - 6 x = 0$

$0 + y = - 6 x + 1$

color(blue)(y=-6x+1)" "->" "y=mx+c" "color(blue)(larr" Given line")

So $m = - 6$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Determine perpendicular line equation}}$

$y = - \frac{1}{m} x + k \text{ "->" } y = - \left(\frac{1}{- 6}\right) x + k$

$y = + \frac{1}{6} x + k \text{ " larr" Perpendicular line}$

We are told that this passes through the known point

$\left(x , y\right) \to \left(6 , - 3\right)$

Substitute these values in the equation to find $k$

$y = \frac{1}{6} x + k \text{ "->" } - 3 = \frac{1}{\cancel{6}} \left(\cancel{6}\right) + k$

$- 3 = 1 + k$

Subtract 1 from both sides

$- 4 = k$

So the equation is

y=-1/mx+k" "->" "color(blue)(ul(bar(|color(white)(2/2)y=1/6x-4" "|))) 