# medians....hypotenuse.....yeah

• Mar 19th 2007, 01:04 AM
amai
medians....hypotenuse.....yeah
Next topic is geometry....in class. I'm trying to get a head start before I'm left in the dust........again.

Given the points A(-1,1), B(-2,-2) and C(4,0) as the vertices of a triangle ABC. Determine the length of the median AM:

OR

Given A(-3,0), B(3,-2) and C(1,-4)as the vertices of a triangle ABC. Show that the midpoint M of the hypotenuse is equidistant from the vertices of the triangle.

BTW, I've learned how to find the midpoint of a line and length of a line........on my own (:D _:D ) so no need going over that....you know, if it has anything to do with understanding this.

Thanks-a-Bunch!
• Mar 19th 2007, 05:38 AM
CaptainBlack
Quote:

Originally Posted by amai
Next topic is geometry....in class. I'm trying to get a head start before I'm left in the dust........again.

Given the points A(-1,1), B(-2,-2) and C(4,0) as the vertices of a triangle ABC. Determine the length of the median AM:

The mid point M of BC is ((-2+4)/2, (-2)/2) = (1, -1)

The length of AM is: sqrt[(A(1)-M(1))^2 + (A(2)+M(2))^2]

..................................... = sqrt[(-1-1)^2 + (1+1)^2] = 2 sqrt(2)

RonL
• Mar 19th 2007, 06:40 AM
ThePerfectHacker
Here is another way to do it (the nice thing about it is that you do not need coordinates to use it).

Let ABC be a triangle with sides a,b,c.
Let m_a represent the median drawn to side a, we have the following:

m_a=(1/2)*sqrt(2(b^2+c^2)-a^2)

Thus, you find all the three sides of this triagnle (through distance formula) and then you can apply that theorem.
• Mar 19th 2007, 07:06 AM
Soroban
Hello, amai!

Quote:

Given: A(-3,0), B(3,-2) and C(1,-4)as the vertices of right triangle ABC.
Show that the midpoint M of the hypotenuse is equidistant from the vertices of the triangle.

BTW, I've learned how to find the midpoint of a line and length of a line. . Good!

Now put all that knowledge together . . .

Make a sketch and AB seems to be the hypotenuse.

The midpoint of AB is: .x = ˝(-3 + 3) = 0, .y = ˝(0 -2) = -1
. . Hence, we have: .M(0,-1)

Now we're to show that: .MA = MB = MC.

. . MA˛ .= .(-3 - 0)˛ + (0 + 1)˛ . . MA = √10

. . MB˛ .= .(3 - 0)˛ + (-2 + 1)˛ . . MB = √10

. . MC˛ .= .(1 - 0)˛ + (-4 + 1)˛ . . MC = √10

. . . ta-DAA!