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July 25, 2014

July 25, 2014

Posted by **Celina** on Wednesday, December 10, 2008 at 4:03pm.

Here's one of the questions I must answer:

Use slope to verify that your point from part a (which was line segment PQ, with P being (2, 4) and Q being (-7, -2)) is on the line through P and Q, and use the distance formula to show that your point is just as far from P as it is from Q.

Thank you so much! This homework will be collected tomorrow, so I really need to be able to understand this and apply what you tell me to other problems. Thanks!

- Correction to what I wrote! Please read this instead! -
**Celina**, Wednesday, December 10, 2008 at 4:08pmSorry, what I wrote might not have made sense. Here's a re-written version.

I'm very confused with what I'm learning about right now. We are kind of talking about slopes, functions, graphs, etc. Today, my teacher taught about midpoint and distance, and did not explain very well at all. Would someone please explain to me how to find midpoint and distance? I have formulas, but I have no idea how to use them, and when I try them, I get the wrong answer.

Here's one of the questions I must answer:

Use slope to verify that your midpoint from part a (which was the midpoint of line segment PQ, with P being (2, 4) and Q being (-7, -2)) is on the line through P and Q, and use the distance formula to show that your midpoint is just as far from P as it is from Q.

Thank you so much! This homework will be collected tomorrow, so I really need to be able to understand this and apply what you tell me to other problems. Thanks!

- Algebra please help! -
**Anonymous**, Wednesday, December 10, 2008 at 4:22pmWell, first of all let's find a point halfway between P and Q.

That point will have an x value halfway between 2 and -7

That is (2-7)/2 = -5/2 = -2.5

The vy value of that halfway point will be halfway between 4 and -2

That is (4-2)/2 = 2/2 = 1

so our halfway point is

( -2.5 , 1 )

now they ask us to verify that

well what is the distance squared from

(2 , 4) to (-2.5 , 1) ?

x difference = -2.5 -2 = -4.5

y difference = 1 - 4 = -3

square of hypotenuse = 4.5^2+3^2 = 29.25

Now do the same thing from

(-2.5 , 1) to (-7 , -2)

x difference = -7 + 2.5 = -4.5

y difference = -2 -1 = -3

square of hypotenuse = same

done

- Algebra please help! -
**Celina**, Wednesday, December 10, 2008 at 4:32pmI kind of get what you're saying... but what is a hypotenuse?

Thanks for the help!

- Algebra please help! -
**Ms. Sue**, Wednesday, December 10, 2008 at 4:41pmhttp://www.mathwarehouse.com/geometry/triangles/right-triangle.html

- Algebra please help! -
**Celina**, Wednesday, December 10, 2008 at 4:45pmThank you!

- Algebra please help! -

- Algebra please help! -

- Algebra please help! -
- Algebra please help! -
**Damon**, Wednesday, December 10, 2008 at 4:28pmTo make sure you have the correct numerical answers, take the square root of 29.25 for your actual half distance.

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