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 ( _ ) so no need going over that....you know, if it has anything to do with understanding this.
Thanks-a-Bunch!
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.
Hello, amai!
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!