We know that in geometry, a median of a triangle is a line segment joining a vertex to the midpoint of the opposing side. So, in a triangle there are three medians. We will find them.
As shown in the figure below, we have the medians:
P1C
P2A
P3B
We need to find P1, P2 and P3. The midpoint of the segment (x1, y1) to (x2, y2) is:
[tex](\frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2})[/tex]
Therefore:
For the segment AB:
[tex]P_{1} = ( \frac{4-6}{2}, \frac{7-1}{2})[/tex]
[tex]P_{1} = (-1,3)[/tex]
For the segment BC:
[tex]P_{2} = ( \frac{4-2}{2}, \frac{-1-9}{2})[/tex]
[tex]P_{2} = (1,-5)[/tex]
For the segment CA:
[tex]P_{3} = ( \frac{-6-2}{2}, \frac{7-9}{2})[/tex]
[tex]P_{3} = (-4,-1)[/tex]
We know that the distance d between two points P1(x1,y1) and P2(x2,y2) is given by the formula:
[tex]d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}- y_{1})^{2} }[/tex]
Then of each median is:
Median P1C:
[tex]d_{1} = \sqrt{(-9-3)^{2}+(-2-(-1))^{2}} = \sqrt{145} [/tex]
Median P2A:
[tex]d_{2} = \sqrt{(7-(-5))^{2}+(-6-1)^{2}} = \sqrt{193} [/tex]
Median P3B:
[tex]d_{3} = \sqrt{(-1-(-1))^{2}+(4-(-4))^{2} } = 8[/tex]
