Algebra please help!
posted by Celina on .
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 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!

Sorry, what I wrote might not have made sense. Here's a rewritten 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! 
Well, 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 (27)/2 = 5/2 = 2.5
The vy value of that halfway point will be halfway between 4 and 2
That is (42)/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 
I kind of get what you're saying... but what is a hypotenuse?
Thanks for the help! 
http://www.mathwarehouse.com/geometry/triangles/righttriangle.html

Thank you!

To make sure you have the correct numerical answers, take the square root of 29.25 for your actual half distance.