# Problems on Slope

• Sep 15th 2008, 09:10 PM
PandaPanda
Problems on Slope
Here's the problem:

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Problem: Recall that you can use two points on a line to determine an equation for the line. Suppose two points in a coordinate plane have coordinates (x1, y1) and (x2, y2).
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A. Write a formula that gives the slope b of the line containing the two points.
Alright, I know this is $b = \frac{y1-y2}{x1-x2}$. No problem there.
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B. Solve the equation for in Part a for y1 in terms of the other variables.
I don't quite remember how to do this. In the end, I got $Y1 = b(x1-x2)+y2$ Please correct me on this.
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C. In your equation from Part b, replace the point with coordinates (x1, y1), with a general point (x, y) on the line. Explain how this equation is now that of a line with slope b through the point (x2, y2).
This is the part I thought was a brain teaser/tongue twister. What do you do here?
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D. Use your equation from Part c to write the equation of the line that contains the points (8, 12) and has slope -2. What is the y-intercept of this line?
I don't have the equation for C. yet, so I'll wait for help on that first. :)
• Sep 15th 2008, 09:35 PM
Jhevon
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Originally Posted by PandaPanda
Here's the problem:

Alright, I know this is $b = \frac{y1-y2}{x1-x2}$. No problem there.

yup

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I don't quite remember how to do this. In the end, I got $Y1 = b(x1-x2)+y2$ Please correct me on this.
that's fine

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This is the part I thought was a brain teaser/tongue twister. What do you do here?
do as they say. change (x1, y1) to (x,y). simplify to get the equation in the slope-intercept form (y = mx + b, here, m is the slope and b is the y-intercept). you will be able to give them the info they require from this form

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I don't have the equation for C. yet, so I'll wait for help on that first. :)
well, i gave you the hint on how to do C, so now you can try again :)
• Sep 15th 2008, 09:41 PM
PandaPanda
Alright, thanks for clarifying.
However, in $Y = b(x-x)+y$, the x's cancel out. How do I change that?
• Sep 15th 2008, 09:44 PM
Jhevon
Quote:

Originally Posted by PandaPanda
Alright, thanks for clarifying.
However, in $Y = b(x-x)+y$, the x's cancel out. How do I change that?

remember, one of the x's is x1 and the other is x2. they are not the same, you don't change both of them, just x1
• Sep 15th 2008, 09:46 PM
Jameson
C) When you do the sub. you get $y=b(x-x_{2})+y_{2}$

This is obviously still a linear equation, so nothing weird here. You can confirm quickly that this still has a slope of b as when you solve for b you get the slope between the two points (x,y) and (x2,y2). If you recall the Point-Slope form of a linear equation, the form of this last equation fits it exactly. Look Linear equation - Wikipedia, the free encyclopedia there to brush up.
• Sep 15th 2008, 09:50 PM
PandaPanda
Ah, okay. Just one last question, because I'm slow :P

After expanding, the result is:
$Y = bx-bx2+y2$
How do you deal with -bx2? Pretty sure it's not supposed to be there.

Thanks very much. ~Panda
• Sep 15th 2008, 09:55 PM
Jhevon
Quote:

Originally Posted by PandaPanda
Ah, okay. Just one last question, because I'm slow :P

After expanding, the result is:
$Y = bx-bx2+y2$
How do you deal with -bx2? Pretty sure it's not supposed to be there.

Thanks very much. ~Panda

write as $y = bx + (y_2 - bx_2)$

that is in the form y = mx + b, your $(y_2 - bx_2)$ is just a constant
• Sep 15th 2008, 09:57 PM
Jameson
Ok so use the equation $y=b(x-x_{2})+y_{2}$. You have the point (8,12) and a slope of -2. Plug in (8,12) for (x2,y2) and -2 for b.

$y=-2(x-8)+12$ Simplifying this and moving towards a form of y=mx+b you get $y=-2x+16+12=-2x+28$. Now do you know how to get the y-intercept from here?