How do you write an equation of a line passing through (6, 4), perpendicular to #y=3x - 2#?

1 Answer
Jul 15, 2016

Answer:

#x+3y-18=0#

Explanation:

Slope intercept form of equation of a line is #y=mx+c#, where #m# is its slope and #c# is its intercept on #y#-axis. As such slope of #y=3x-2# is #3#.

As product of slopes of two lines perpendicular to each other is #-1#, hence slope of line perpendicular to #y=3x-2# is #-1/3#.

Equation of a line passing through #(x_1,y_1)# and having a slope of #m# is #(y-y_1)=m(x-x_1)#. Hence equation of line passing through #(6,4)# and slope of #-1/3# is

#(y-4)=-1/3(x-6)# or

#3y-12=-x+6# or

#x+3y-12-6=0# i.e.

#x+3y-18=0#