Vertices of a triangle through mid points of its sides

[Guest]

The mid point of the sides of a triangle are at (-1,4);(5,2);(2,-1). Find its vertices.

[Guest]

The perpendicular bisector of the sides of the triangle passing through these points will each be a diameter of the circumscribed circle around the vertices of the triangle. Each side of the triangle will be a chord of the circumscribing circle. The intersection of the perpendicular bisectors will be the centre of the circumscribed circle

[Guest]

Let the coordinates of the vertices of the triangle be [tex](x_1,y_1), (x_2,y_2), (x_3,y_3)[/tex].

The coordinates of the midpoint of any pair of vertices is simply the average of the values of the [tex]x[/tex] and [tex]y[/tex] values.

So the coordinates of the three midpoints are
[tex]\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)[/tex], [tex]\left(\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2}\right)[/tex], [tex]\left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right)[/tex],
this must match (-1,4), (5,2), (2,-1) (the order doesn't matter it will ultimately just permute the order of the vertices).

This gives us 3 sets of equations involving [tex]x_1,x_2,x_3[/tex] and 3 sets of equations involving [tex]y_1,y_2,y_3[/tex]:
[tex]\frac{x_1+x_2}{2} = -1[/tex]
[tex]\frac{x_1+x_3}{2} = 5[/tex]
[tex]\frac{x_2+x_3}{2} = 2[/tex]

[tex]\frac{y_1+y_2}{2} = 4[/tex]
[tex]\frac{y_1+y_3}{2} = 2[/tex]
[tex]\frac{y_2+y_3}{2} = -1[/tex]

Solving the equations involving [tex]x[/tex] can be done as follows:
Take the first equation, rearrange it to get
[tex]x_1=-2-x_2[/tex]
Substitute it into the second equation to get
[tex]\frac{-2-x_2+x_3}{2} = 5[/tex]
rearrange it to get
[tex]x_2 = x_3-12[/tex]
Substitute it into the third equation to get
[tex]\frac{x_3-12+x_3}{2} = 2[/tex]
which simplifies to [tex]x_3=8[/tex],
which means [tex]x_2=x_3-12=8-12=-4[/tex]
and [tex]x_1=-2-x_2=-2--4=2[/tex].

Similarly we can solve the other 3 equations to get [tex]y_1=7[/tex], [tex]y_2=1[/tex], [tex]y_3=-3[/tex].
So the coordinates of the vertices of the triangle are [tex](2,7), (-4,1), (8,-3)[/tex].

Hope this helped,

R. Baber.

[Guest]

Just another variation for the solution......

Let the vertices be points A, B & C.
Let side AB have midpoint D(-1,4)
Let side BC have midpoint E(5,2)
Let side AC have midpoint F(2,-1)

A line joining the midpoints of any 2 sides is parallel to the 3rd (other) side of the triangle.
That means the slope is the same as in y = mx + c where m is the slope.

So line EF has same slope as AB.

m(EF) = m(AB) = (2 -(-1)) / (5-2) = 3 / 3 = 1
Eqn. AB ...> y-4 = 1*(X + 1) .....y = X + 5 .............EqnAB
................

m(DF) = m(BC) = (-1-4) / (2-(-1)) = -5 / 3
Eqn. BC ...> y-2 = (-5/3)*(X-5) ....y = (-5/3)X + 25/2 + 2 ..... y = (-5/3)X + 31/3
3y = -5X + 31 ....................................EqnBC
..................

Point B(x) ...> 3x+15 = -5X + 31 ....8X = 16 ....X = 2.
Point B(y) ...> 2 + 5 = 7 .............................B(x,y) = (2,7). .....Answer1
.....................

m(DE) = m(AC) = (2-4) / (5-(-1) = -2/6 = -1/3
Eqn. AC ...> y+1 = -1/3*(X-2) ...y = -1/3X -1/3 .... 3y = -X-1 ......EqnAC
....................

C(x) ...> 5X+31 = -X -1 ..... 32 = 4X ... 8 = X
C(y) ...> 3y = -8-1 ..... 3y = -9 ....y = -3 ...........C(x,y) = (8,-3). ....Answer2
...............

A(x) ...> 3x+15 = -X-1 ...4X = -16 .... X = -4
A(y) ...> 3y = 4-1 ....y = 3/3 = 1 .....................A(x.y) = (-4,1). ....Answer3