Question

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

Answer 1

`"y=1/6x-4`

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

Explanation:

`color(blue)("General introduction")`

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

`y=mx+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 `[-1xx 1/m]` so its equation is:
`color(white)(.)`

`y=[(-1)xx1/m]x+k" "->" "y=-1/mx+k`

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

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the given line equation")`

Given `" "color(green)(6x+y=1)`

Subtract `color(red)(6x)` from both sides

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

But `6x-6x=0`

`0+y=-6x+1`

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

So `m=-6`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Determine perpendicular line equation")`

`y=-1/m x+k" "->" "y=-(1/(-6))x+k`

`y=+1/6 x+k" " larr" Perpendicular line"`

We are told that this passes through the known point

`(x,y)->(6,-3)`

Substitute these values in the equation to find `k`

`y=1/6 x+k" "->" "-3=1/(cancel(6))(cancel(6))+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" "|)))`