# Graphing lines... perpendicular and whatnot

• Oct 28th 2007, 11:50 AM
Melancholy
Graphing 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 AM
Jhevon
Quote:

Originally Posted by Melancholy
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

if two lines are perpendicular, then their slopes are the negative inverses of each other. that is, if one's slope is $\displaystyle m$ then the other's is $\displaystyle - \frac 1m$

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:

$\displaystyle y - y_1 = m(x - x_1)$

where $\displaystyle m$ is the slope of your line, and $\displaystyle (x_1,y_1)$ is a point it passes through
• Oct 28th 2007, 12:10 PM
Melancholy
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 PM
Jhevon
Quote:

Originally Posted by Melancholy
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).

actually no. the slope of our line is -2. but if it was 2 (as you thought it was) it would be correct

Quote:

You should write a book... about geometry... lol!
i'm not that good at geometry
• Oct 28th 2007, 12:30 PM
Melancholy
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 PM
Jhevon
Quote:

Originally Posted by Melancholy
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.

yes, the slope is 1. good job catching yourself there. now what is the equation of the line you want?
• Oct 28th 2007, 01:23 PM
Melancholy
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 PM
Jhevon
Quote:

Originally Posted by Melancholy
It would be y = x - 2 wouldn't it?

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!
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, 01:47 PM
Melancholy
Quote:

the y-intercept is 2. find the x-intercept. then just draw a straight line through those two points, and viola! you're done
And... how do I find the x-intercept?! ... :o
• Oct 28th 2007, 10:32 PM
earboth
Quote:

Originally Posted by Melancholy
...
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!

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.

$\displaystyle y = \frac{\overbrace{-4}^{y-direction} }{\underbrace{3}_{x-direction} } \cdot x +2$