The midpoints of the sides of a triangle are (2,2), (2,3), (4,6). Find the equations of the sides.

First, we'll write the relations for calculating the
coordinates of the mid-point of a segment.

The coordinates
of M are:

xM=(x1+x2)/2

yM=
(y1+y2)/2

Now, we'll calculate the mid-point of the side
AB.

xM = (xA+xB)/2

2 =
(xA+xB)/2 => xA+xB = 4 (1)

yM=
(yA+yB)/2

2 = (yA+yB)/2 => yA+yB = 4
(2)

We'll calculate the mid-point of
AC:

xN = (xA+xC)/2

2 =
(xA+xC)/2 => xA+xC = 4 (3)

yN=
(yA+yC)/2

3 = (yA+yC)/2 => yA+yC = 6
(4)

We'll calculate the mid-point of
BC:

xP = (xB+xC)/2

4 =
(xB+xC)/2 => xB+xC = 8 (5)

yP=
(yB+yC)/2

6 = (yB+yC)/2 => yB+yC = 12
(6)

We'll subtract (3) from
(1):

 xA+xB-xA-xC = 0

xB - xC
= 0 (7)

We'll add (7) to
(5):

xB - xC + xB+xC = 8

We'll
eliminate like terms:

2xB =
8

xB =
4

But, from (7), xB - xC = 0 and xB = 4,
so:

4 - xC =
0

xC =
4

From  (3), xA+xC = 4 and xC = 4,
so:

xA+4 = 4

xA=
0

Now, we'll calculate the
coordinates yA, yB, yC.

We'll subtract (2)
from (4):

yA+yC-yA-yB =
2

We'll eliminate like
terms:

yC-yB = 2 (8)

We'll add
(8) and (6):

yC-yB+yB+yC =
12+2

We'll eliminate like
terms:

2yC =
14

yC =
7

But yB+yC = 12 and yC = 7,
so:

yB+7 =
12

yB =
5

From (2),  yA+yB = 4 and yB
= 5, so:

yA+5 =
4

yA =
-1

The coordinates of the
vertices of the triangle
are:

A (0,-1) , B(4,5) and
C(4,7).