Write the equation of a line perpendicular to y=3x-2, that passes through the point (6, 2)?

1 Answer
Oct 5, 2017

#y'=-1/3x+4#

Explanation:

The slopes of perpendicular lines are negative reciprocals of each other. (So, negative, and flip the fraction.) Since the slope of #y=3#, the slope of any line perpendicular to it is #m=-1/3#, where #m=#slope.

Now, writing the second equation in slope-intercept form, with #y'# (read "y prime"), we have #y'=-1/3x+b#.

To find that second #b#, or #b'#, plug #(6,2)# in for #x and y# in the #y'# equation:
#2=-1/3(6)+b#, which simplifies to #2=-2+b#, and finally, #b=4#.

Now you've got the slope and the y-intercept for your perpendicular line!
#y'=-1/3x+4#