# (I) Find the a. midpoint and b. distance of each line segment, AB with the given endpoint 1. A(2,-5), B(6,-7)  2. A(5,5), B(-9,3)  3. A(-2,3), B(-10,10) (II) The midpoint of line segment AB is M. The coordinates of M are (3,-2) and the coordinates of A are (-1.0).  What are the coordinates of B? (I) To solve for the midpoint of two points `(x_1, y_1)` and `(x_2, y_2)` , use the formula:

`x_(mid)=(x_2+x_1)/2`

`y_(mid)=(y_2+y_1)/2`

And to solve for the distance between two point, apply the formula:

`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

(1) A(2,-5), B(6, -7)

`x_(mid)=(6+2)/2=8/2=4`

`y_(mid)=(-7+(-5))/2=(-12)/2=-6`

`d=sqrt((6-2)^2+(-7-(-5))^2)=sqrt(4^2+(-2)^2)=sqrt(20) = 2sqrt5`

Hence, the midpoint of these two points is...

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(I) To solve for the midpoint of two points `(x_1, y_1)` and `(x_2, y_2)` , use the formula:

`x_(mid)=(x_2+x_1)/2`

`y_(mid)=(y_2+y_1)/2`

And to solve for the distance between two point, apply the formula:

`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

(1) A(2,-5), B(6, -7)

`x_(mid)=(6+2)/2=8/2=4`

`y_(mid)=(-7+(-5))/2=(-12)/2=-6`

`d=sqrt((6-2)^2+(-7-(-5))^2)=sqrt(4^2+(-2)^2)=sqrt(20) = 2sqrt5`

Hence, the midpoint of these two points is (4,-6) and their distance is` 2sqrt5` units.

(2) A(5,5), B(-9,3)

`x_(mid)=(-9+5)/2=(-4)/2=-2`

`y_(mid)=(3+5)/2=8/2=4`

`d=sqrt((-9-5)^2+(3-5)^2)=sqrt((-14)^2+(-2)^2)=sqrt200=10sqrt2`

Thus, the midpoint of these two points is (-2,4) and the distance between them is `10sqrt2` units.

3. A(-2,3), B(-10,10)

`x_(mid)=(-10+(-2))/2=(-12)/2=-6`

`y_(mid)=(10+3)/2=13/2`

`d=sqrt((-10-(-2))^2+(10-3)^2)=sqrt((-8)^2+7^2)=sqrt113`

Hence, their midpoint is `(-6,13/2)`  and their distance is `sqrt113` units.

(II) The midpoint of line segment AB is M. The coordinates of M are (3,-2) and the coordinates of A are (-1.0). What are the coordinates of B?

To solve, apply the midpoint formula. Let's consider the x_mid first.

`x_(mid)=(x_2+x_1)/2`

Plug-in `x_(mid)=3` and `x_1=-1` .

`3=(x_2+(-1))/2`

`3=(x_2-1)/2`

`6=x_2-1`

`7=x_2`

Next, consider the y_mid.

`y_(mid)=(y_2+y_1)/2`

Plug-in `y_(mid)=-2` and `y_1= 0` .

`-2=(y_2+0)/2`

`-2=y_2/2`

`-4=y_2`

Therefore, point B is (7,-4).

Approved by eNotes Editorial Team