Question

Triangle ABC has vertices ofA(–6, 7), B(4, –1), and C(–2, –9).Find the length of the median from

Answer 1

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:

$(\frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2})$

Therefore:

For the segment AB:

$P_{1} = ( \frac{4-6}{2}, \frac{7-1}{2})$
$P_{1} = (-1,3)$

For the segment BC:
$P_{2} = ( \frac{4-2}{2}, \frac{-1-9}{2})$
$P_{2} = (1,-5)$


For the segment CA:
$P_{3} = ( \frac{-6-2}{2}, \frac{7-9}{2})$
$P_{3} = (-4,-1)$

We know that the distance d between two points P1(x1,y1) and P2(x2,y2) is given by the formula:

$d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}- y_{1})^{2} }$

Then of each median is:

Median P1C:

$d_{1} = \sqrt{(-9-3)^{2}+(-2-(-1))^{2}} = \sqrt{145}$

Median P2A:

$d_{2} = \sqrt{(7-(-5))^{2}+(-6-1)^{2}} = \sqrt{193}$

Median P3B:

$d_{3} = \sqrt{(-1-(-1))^{2}+(4-(-4))^{2} } = 8$