Posted by Kate on .
Consider the points (3, –1) and
(9, 2).
(a) State the midpoint of the line segment with the given endpoints.
(b) Find the distance between the points. Show some work. Give the exact answer (involving a radical), and also an approximation to three decimal places.
(c) Find the slope of the line passing through the two given points. Show some work.

Algebra 
Reiny,
I will give you the formulas for the problems,
for any two points (a,b) and (c,d)
1. Midpoint = ((a+c)/2,(b+d)/2)
2. Distance between the two points
= √[(ac)^2 + (bd)^2]
3. slope = (bd)/(ac)
compare (a,b) and (c,d) with your given points and evaluate.
Let me know what you get 
Algebra 
Anonymous,
This has got to be in your text book!
The midpoint is the average x and the average y
Xav = (3+9)/2 = 6
Yav = (1+2)/2 = 1/2
so
(6 , 1/2)
d^2 = [(Y2Y1)^2 + (X2X1)^2 ]
= [ 3^2 + 6^2 ]
= [ 9 + 36 }
= 45
so
d = sqrt 45 = 3 sqrt 5
slope = (Y2Y1) / (X2X1)
I think you can carry that division out 
Algebra 
Kate,
(x1 + x2)/2, y1 + y2)/2 =
3+9/2 = 12/2 = 6
1+2/2 = 1/2 = .05
This is what I got for the mid point. is this correct? I got 6 for the first set of numbers, and .05 for the second. How do I determine the mid point? 
Algebra 
Kate,
I think I figured it out. The answer can contain a fraction along with whole numbers. So i guess the mid point would be (6,.05) correct?

Algebra 
Reiny,
"Anonymous" already gave you the answer.
why don't you just replace the values ?
you started with ...
((x1 + x2)/2, (y1 + y2)/2) , notice I corrected your "bracket errors"
= ((3+9)/2,(1+2)/2)
= (6,1/2)
you had .05
1/2 is .5 not .05 
Algebra 
Kate,
oops I misplaced the decimal. So yea (6,0.5) is the solution I came up with.

Algebra 
Kate,
Now I will work on the distance and slope parts. Hopefully they will be correct to!