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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 ?

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  • $\begingroup$ Looks like the "someone else" ws computing AM, and you computed BM. $\endgroup$ Apr 11, 2016 at 13:51
  • $\begingroup$ @ThomasAndrews, sorry, typo, fixed now. $\endgroup$ Apr 11, 2016 at 14:14
  • $\begingroup$ 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? $\endgroup$ Apr 11, 2016 at 14:25
  • $\begingroup$ @ThomasAndrews, yeah another typo, really sorry. $\endgroup$ Apr 11, 2016 at 14:59
  • $\begingroup$ It is better style to define M in or before a question concerning it, not on the 8th line. $\endgroup$ Apr 11, 2016 at 15:15

1 Answer 1

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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 $AC$.

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  • $\begingroup$ I'm almost sure that my interpretation (and your) is the right. so M is the mid point of AC and B is the opposite side. $\endgroup$ Apr 11, 2016 at 15:02
  • $\begingroup$ My only concern is that you had both version (AM and BM) originally in your question, and you haven't given the complete answer from the other person. Nor does it appear that you've given us the exact question, as written. $\endgroup$ Apr 11, 2016 at 15:04

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