# Algebra

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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 - ,

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
= √[(a-c)^2 + (b-d)^2]

3. slope = (b-d)/(a-c)

compare (a,b) and (c,d) with your given points and evaluate.
Let me know what you get

• Algebra - ,

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 = [(Y2-Y1)^2 + (X2-X1)^2 ]
= [ 3^2 + 6^2 ]
= [ 9 + 36 }
= 45
so
d = sqrt 45 = 3 sqrt 5

slope = (Y2-Y1) / (X2-X1)
I think you can carry that division out

• Algebra - ,

(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 - ,

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 - ,

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)

1/2 is .5 not .05

• Algebra - ,

oops I misplaced the decimal. So yea (6,0.5) is the solution I came up with.

• Algebra - ,

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