# Median BM of triangle ABC two results

Given

• Calculate the measure of the median $\overline{BM}$ of ABC triangle, given A (-6.1); B (-5,7) and C (2,5)

I get this result:

$Xm = \frac{Xc - Xa}{2} + Xa$

$Xm = \frac{2-(-6)}{2} + (-6) = 4 - 6 = -2$

$Ym = \frac{(Yc - Ya)}{2} + Ya$

$Ym = \frac{5-1}{2} + 1 = 2 + 1 = 3$

$M(-2, 3)$

$d_{BM}^2 = (-5-(-2))^2 + (7-3)^2 = 9 + 16 = \sqrt{25} = 5$

and someone else get:

$Xm = \frac{-5+2}{2} = \frac{-3}{2}$

$Ym = \frac{7+5}{2} = \frac{12}{2} = 6$

$...$

So what's correct ?

• Looks like the "someone else" ws computing AM, and you computed BM. Apr 11, 2016 at 13:51
• @ThomasAndrews, sorry, typo, fixed now. Apr 11, 2016 at 14:14
• Your computation of $d_{BM}^2$ uses $B(-5,7)$ but your question says $B(5,7)$. The "someone else" also seems to think $B$ is $(-5,7)$, so another typo? Apr 11, 2016 at 14:25
• @ThomasAndrews, yeah another typo, really sorry. Apr 11, 2016 at 14:59
• It is better style to define M in or before a question concerning it, not on the 8th line. Apr 11, 2016 at 15:15

First, both you and someone else seem to be using $B(-5,7)$, not $B(5,7)$.
The "someone else" appears to be computing the midpoint between some other point $B(-5,7)$ and $C(2,5)$, and you are computing the midpoint between $A(-6,1)$ and $C(2,5)$.
If you are really trying to compute the length $BM$ where $M$ is the midpoint of $AC$, the side opposite to $B$, then your answer is correct.
We don't have the complete "someone else" answer, so it is unclear what distance that person is computing, but the $M$ that person is computed is the midpoint of $BC$, not the midpoint of $AC$.