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