Graph the line that satisfies each condition.

contains Q(-1,3) and is perpendicular to line AB with A(-2,0) and B(4,3)

How do I do this? lol... :o

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- Oct 28th 2007, 11:50 AMMelancholyGraphing lines... perpendicular and whatnot
**Graph the line that satisfies each condition.**

contains Q(-1,3) and is perpendicular to line AB with A(-2,0) and B(4,3)

How do I do this? lol... :o - Oct 28th 2007, 11:57 AMJhevon
if two lines are perpendicular, then their slopes are the negative inverses of each other. that is, if one's slope is then the other's is

find the slope of the line connecting the last two points. then take it's negative inverse, that will give you the slope of the line you want to make.

then use the point-slope form:

where is the slope of your line, and is a point it passes through - Oct 28th 2007, 12:10 PMMelancholy
Dude. You rock my word.

y = 2x + 5, no? Which means the other point on the line containing Q... it's coordinates would be (1, -1).

You should write a book... about geometry... lol! - Oct 28th 2007, 12:16 PMJhevon
- Oct 28th 2007, 12:30 PMMelancholy
Alright. Then I must have worked the slope wrong... but I was close.

Another (parallel):

**contains M(1, -1) and is parallel to FG with F(3,5) and G(-3,-1)**

So, I found the slope of F and G which is 4... i think... but I have graphed all the points they've given me, and done the 'rise and run' thing with M, and they are soooooo not parallel. Aren't parallel lines supposed to have the same slope? I didn't find the right slope, did I?

Edit: no wait..... i did do the slope wrong... *fixing slope equation*

Edit 2: The slope is 1... making the two lines parallel. - Oct 28th 2007, 12:42 PMJhevon
- Oct 28th 2007, 01:23 PMMelancholy
It would be y = x - 2 wouldn't it?

Okay, on my next problem... they're all making me cry lol.

**slope = -4/3, contains K(3, -2)**

This is what I got:

m = 1

y + 2 = -4/3(x - 3)

y + 2 = -4/3x - 3

y + 2 = -4/3x + 4

___y = -4/3x + 2

BUT. It tells me to graph it... and how am I supposed to continue the line on from K if it is some whacky slope?! Oi! - Oct 28th 2007, 01:30 PMJhevon
yes (Clapping)

Quote:

Okay, on my next problem... they're all making me cry lol.

**slope = -4/3, contains K(3, -2)**

This is what I got:

m = 1

y + 2 = -4/3(x - 3)

y + 2 = -4/3x - 3

y + 2 = -4/3x + 4

___y = -4/3x + 2

BUT. It tells me to graph it... and how am I supposed to continue the line on from K if it is some whacky slope?! Oi!

- Oct 28th 2007, 01:47 PMMelancholyQuote:

the y-intercept is 2. find the x-intercept. then just draw a straight line through those two points, and viola! you're done

- Oct 28th 2007, 10:32 PMearboth
Hello,

you know (or should know :D ) that the slope of a line is**always**a fraction and the slope is determined by a right triangle. The legs of the triangle are parallel to the coordinate axes.

With your problem:

That means: Start at K, go 3 units to the right, go 4 units down: There is the a point which belongs to the line. Connect this brand new point with K and you are done.

If you don't like the point K then start at the y-intercept: Go 3 units to the right, 4 units down: There is a point which belongs to the line (and which is known to you: It's good ol' K)