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

*print*Print*list*Cite

### 3 Answers

Let ABC be the triangle such that:

mid point AB = ((xa+xb)/2, (yb+yb)/2) = (2,2)

==> xa+xb/2 = 2==> xa+xb = 4......(1)

==> (ya+yb)/2 = 2 ==> ya+yb = 4.......(2)

midpoint AC= (xa+xc)/2, (ya+yc,2) = (2,3)

==> (xa+xc)/2 =2 ==> xa+xc = 4 .......(3)

==> (ya+yc)/2 = 3==> ya+yc = 6........(4)

midpoint BC = ((xb+xc)/2, (yb+yc)/2) = (4,6)

==> (xb+xc)/2 = 4 ==> xb+xc = 8..........(5)

==> (yb+yc)/2 = 6==> yb+yc = 12..........(6)

Now to solve:

Let us solve the system with equation (1), (3) , and (5)

xa+xb = 4 .........(1)

xa +xc = 4..........(2)

xb+xc = 8 .........(3)

Subtract (1) from (2)

==> xc - xb = 0 ==> xc = xb

==> xc + xc = 8

==> **xc=4, xb =4, xa= 0**

Now let us solve the system with equations (2), (4), and (6)

ya+yb = 4.......(2)

ya + yc = 6 .......(4)

yb + yc = 12.......(6)

Subtract (2) from (4)

==> yc-yb = 2 .....(7)

now add (7) and (6)

==> 2yc = 14

==> **yc = 7 , yb = 5, ya = -1**

**Then A(0, -1) , B(4,5) , and c(4,7)**

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).**

The mid Points of the triangle are :

(2,2),(2,3) and (4,6)

Let ABC Be the triangle and D(2,2), E(2,3) and F (4,6) be the mid points of AB,BC, and CA respectively.

We know that the line joining mid points of any two sides in a triangle is || to the 3rd side.

Therefore the slope of BC = the slope of DF = (yF - yD)/(xF-xD) = (6-2)/(4-2) = 2. So the equation of BC which passes through E(2,3) is:

y-3 = 2(x-3) . Or 2x-y -3 = 0

CA ||DE . So slope CA = (yE-YD)/(xE-xD) = (3-2)/(2-2) = 1/0 = infinite. So CA has the equation x = k which passes through F(4,6). So the equation of CA is x = 4. Or x-4 = 0.

AB|| EF. So the slope of AB = slope of EF = (yF-yE)/(xF-xE) = (6-3)/4-2) = 3/2. So the equation of EF is y = (3/2)x+const.But this line passes through (2,2). So y-2 = (3/2)(x-2) is the equation of AB. Or 2y-4 = 3x-6. Or 3x-2y -2 = 0 .