Well, it's easier to check it out graphing it than saying it. Though, the perpendicular one's slope is MINUS 1/5, the sign is very important

I think the rest is right, though

Yeah, for perpendicular lines you need to take the opposite reciprocal (sometimes called the multiplicative inverse) of the slope and plug it in.

So with a standard equation in slope-intercept form: y=mx+b, a perpendicular line can always be found with the formula y=(-1/m)x+b (but you will have to take into account the special case of an undefined slope when m=0).

For parallel you just plug in the point the line must pass through and solve for b (as you've done above).

As for the linear algebra class, you should be doing something similar to this in that class with interpolating polynomials iirc.

You need to use the point-slope form y-y1 = m(x - x1) to find your new line. So you put the y coordinate in for y1 and the x coordinate in for x1 and solve! The end.

The equations have to be equal, so whatever you do on one side you must do to the other. If you add 5 to one side you add 5 to the other, if you divide by x on one side you divide by x on the other, and so forth.

i wish I was back in grade 8 math....Those were the days. Back when it didn't take 30 minutes and 3 pages to do one question.

What class are you in?

My highest mathematics class was Calculus IV and I never came across such a long problem.

Oui! Mais moi je suis à Québec :( lol

it would be negative 5 because a negative multiplied by a postive is a negeative.... i hated linear equasions

i always drew a picture....