Interesting linear equations problem involving slopes of perpendicular lines.?

First put each line into standard form:

1) y = (n/2)x + 4

2) y = -(3/n)x - 6/n

Now we see the two slopes are n/2 and -3/n. In order for them to be perpendicular, they must be negative recipricals, so we must have

n/2 = n/3 ==> n/6 = 0 ==> n = 0.

With n = 0, line 1 is horizontal and line 2 is vertical so they are perpendicular.

Remember : the slope is tg of the angle a between x-axis and the functions line.

Then the angle between the perpendicular is tg(a+ 90 deg)= -cot a

-2y=-nx-8

y=n/2+4 is the first line (1)

ny= - 3x-6

y=-3/n- 6/n

a perpendicular to (1)

has then slope -1/(n/2)=-2/n. I cannot see that the second line can be perpendicular to (1) for any value but 0 and infinity.

Only because people are rude to you does not mean you have the right to insult anyone. If I am in second grade or whatever, that is not your problem. IT'S MINE. If you're not going to help then do not even comment.

By the way, I'm in college honeyyyyyyy :P

The same reason a 45degree slope has 12inches of rise/fall per 12inches of run, yet 6inches of fall per ft. is not 22.5 degrees

Just solve each for y.

y = (nx + 8)/2

y = -(3x + 6) / n

Haha. Idiot. get a life before you start getting but hurt about a question.

You've NEVER heard of math before? Go back in your cave, you'll be safe there...

hey, I wanted to reach u about my last question, the concentration is 0.5M and I would REALLY appreciate if u could help me out a bit as it seems like ur great at chemistry and I am struggling greatly.. ;/

thanks

Located in butthole of all atoms. Have f*cking change.

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