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

1 Answer
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:

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" "|)))

Tony B