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

Answer:

#"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