Question

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

Answer 1

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.

Answer 2

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.)

Answer 3

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`.