How to find the coordinates of the vertices of a triangle given the midpoints..help!

[sarah3]

the midpoints of the triangle are: L(4,2) M(2,3) N(5,4)
how to i find the coordinates of the vertices of the triangle? I've tried to do this by using the midpoint formula and then finding x for LM, LN, and MN and then finding y for the same ones but i ended up with the same coordinates that were given. Please help!

 

[pka]

the midpoints of the triangle are: L(4,2) M(2,3) N(5,4) how to i find the coordinates of the vertices of the triangle

Say the vertices are \(\displaystyle Aa,b),{\kern 1pt} {\kern 1pt} Ee,f)\;\& \,Hh,g)\).

Now $\displaystyle \dfrac{{a + e}}{2} = 4,\quad \dfrac{{a + h}}{2} = 2,\;\& \,\dfrac{{e + h}}{2} = 5$

AND $\displaystyle \dfrac{{b + f}}{2} = 2,\quad \dfrac{{b + g}}{2} = 3,\;\& \,\dfrac{{f + g}}{2} = 4$

The solution of the first set gives the x-coordinates.

The solution of the second set gives the y-coordinates.

 

[Mrspi]

the midpoints of the triangle are: L(4,2) M(2,3) N(5,4)
how to i find the coordinates of the vertices of the triangle? I've tried to do this by using the midpoint formula and then finding x for LM, LN, and MN and then finding y for the same ones but i ended up with the same coordinates that were given. Please help!

It might help to remember that a segment joining the midpoint of two sides of a triangle is parallel to the third side. So, for example, you know that one side of your triangle must be parallel to segment ML. Could we write an equation for the line containing that side of the triangle?

Let's see...parallel lines have the same slope. Segment ML has a slope of (2 - 3) / (4 - 2), or -1/2. So any line parallel to ML must have a slope of -1/2. We also know that the third side of the original triangle must have point N as its midpoint. So, we need a line with a slope of -1/2 which contains point N(5, 4):

y = mx + b
We know that the slope (m) is -1/2, and that a point (x, y) on the line is (5, 4). Substitute those values:

4 = (-1/2)*5 + b
4 = (-5/2) + b
4 + (5/2) = b
13/2 = b

And the equation of the third side of the original triangle, the side with N as its midpoint, can be written as
y = (-1/2)x + (13/2)

Now...you can use the same sort of procedure to write the equation of the lines containing each of the other two sides (the one with M as its midpoint, and the one with L as its midpoint).

When you have the equations of the lines containing the three sides of the triangle, you can take them two at a time and solve for the point of intersection of the sides, pair by pair, to find the three vertices of the original triangle.

There is probably a way to solve this problem which does not involve as much work as this method does....perhaps someone else will have a different suggestion.

 

[sarah3]

To pka: Thank you. Are those all the steps I need to solve this? There's a lot of websites that show doing this in different ways..
To Mrspi: Thank you, i will try to do it the way you showed it as well!

 

[pka]

To pka: Are those all the steps I need to solve this?

Both of those systems of three equations in three unknowns are quite easy to solve.

 

[soroban]

Hello, sarah3!

The midpoints of the triangle are: L(4,2), M(2,3), N(5,4)
How to i find the coordinates of the vertices of the triangle?

Mrspi has the best approach.
We can almost "eyeball" the solution.

      |
      |                     N
      |                   ◊(5,4)
      |               *   : 
      |       M   *       :1
      |  (2,3)◊ - - - - - +       A
      |             3             ♥(7,3)
      |                       *   :
      |        3      L   *       :1
      |   + - - - - - ◊ - - - - - +
      |  1:       * (4,2) -3
      |   :   *
      | B ♥
      | (1,1)
      |
  - - + - - - - - - - - - - - - - - - - -
      |

We have the three midpoints $\displaystyle L,M,N.$


The segment joining two midpoints of the sides of a triangle
. . is one-half the length and parallel to the third side.

Hence, side $\displaystyle AB\text{ (through }L)$ is twice the length of $\displaystyle MN$
. . and parallel to $\displaystyle MN.$

To go from $\displaystyle M$ to $\displaystyle N$, we move 3 units right and 1 unit up.

So, to go from $\displaystyle L$ to $\displaystyle A$, we move 3 units right and 1 unit up . . . $\displaystyle A(7,3)$

To go from $\displaystyle L$ to $\displaystyle B$, we move 3 units left and 1 unit down . . . $\displaystyle B(1,1)$

In a similar fashion, we can locate vertex $\displaystyle C.$