How do you find the slope perpendicular to #7x+2y=0#?

3 Answers
Jun 20, 2018

Answer:

By expressing this in the form #y=ax+b# and then calculating #-1/a#

Explanation:

Firstly, let's transform #7x+2y=0#
#7x+2y=0iff2y+7x-7x=0-7x iff (2y)/2=(7x)/2 iff y=7/2x#
You might know, that to find a perpendicular line to any line you just have to find #-1/(slope)#.
If you don't know why, tell me in the comments and I'll do my best explaining it.

Jun 20, 2018

Answer:

The perpedicular is #2x-7y=0#

Explanation:

The given line has a slope equal to #-7/2# (#y_1=-7/2x#)

Or if you will: For each 7 units x increases, y decreases 2 units.

The perpendicular, therefore, will have that for each 2 units x increases, y increases 7 units.

Therefore:
#y_2=2/7x# or #2x-7y=0#

If you look at the graphs, I think you can see how this must be so.
I, therefore, leave the actual proof to you. (Look at the angles of the two graphs relative to the two axes.)
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Jun 20, 2018

Answer:

The slope perpendicular to #7x+2y=0# is #2/7#.

Explanation:

First, I would rewrite the equation #7x+2y=0# into slope-intercept form, which is #y=mx+b#. Remember, m stands for slope and b is the y-intercept. Bring #7x# to the other side of the equation, and you will get #2y=-7x+0#. To get #y# by itself, you divide by 2 on both sides. This will result in #y=-7/2x+0#. The slope for this equation is #-7/2# but since you're looking for the perpendicular slope, which is the opposite reciprocal, the perpendicular slope for this equation is #2/7#.