Problems on Slope

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September 15th 2008, 08:10 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. :)
September 15th 2008, 08:35 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 :)
September 15th 2008, 08:41 PM
PandaPanda

Alright, thanks for clarifying.
However, in $Y = b(x-x)+y$ , the x's cancel out. How do I change that?
September 15th 2008, 08:44 PM
Jhevon

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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
September 15th 2008, 08:46 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.
September 15th 2008, 08:50 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
September 15th 2008, 08:55 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
September 15th 2008, 08:57 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?