Question

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. Find the lengths of the median of the triangle with vertices A(1,2,3), B(-2,0,5), and C(4,1,5)

Answer 1

Answer:

$5/2units$.

$\sqrt{47/2}units$.

$\sqrt{41/2}units$.

Step-by-step explanation:

The given points are A(1,2,3), B(-2,0,5) and C(4,1,5). The triangle is represented in the attach file where the three possible median are length AE, BF, and CD. We determine the coordinate of point D,E and F using the midpoint equation which is for any point A(x,y,z) and point B(a,b,c), the midpoint D is determine by

$D=(\frac{x+a}{2},\frac{y+b}{2},\frac{z+c}{2})$.

Hence going by the above formula we determine the coordinate of point D,E and F

$D=(\frac{-2+1}{2},\frac{0+2}{2},\frac{5+3}{2})$.

$D=(\frac{-1}{2},1,4)$.

point E

$E=(\frac{4-2}{2},\frac{1+0}{2},\frac{5+5}{2})$.

$E=(1,\frac{1}{2},5)$.

Point F

$F=(\frac{4+1}{2},\frac{1+2}{2},\frac{5+3}{2})$.

$F=(\frac{5}{2},\frac{3}{2},4)$.

To determine the length of each median line we use the formula for distance between two points which is express as

$AB=\sqrt{(y_{2}-y_{1} )^{2} +(x_{2}-x_{1} )^{2}+(z_{2}-z_{1} )^{2}}$.

Using the above formula we determine the length of line AE,BF and CD.

$AE=\sqrt{(1-1 )^{2} +(1/2-2)^{2}+(5-3 )^{2}}$.

$AE=\sqrt{0 +9/4+4}$.

$AE=\sqrt{25/4}$.

$AE=5/2units$.

For point BF

$BF=\sqrt{(5/2+2 )^{2} +(3/2-0)^{2}+(4-5 )^{2}}$.

$BF=\sqrt{81/4 +9/4+1}$.

$BF=\sqrt{47/2}$.

$BF=\sqrt{47/2}units$.

For point CD

$CD=\sqrt{(-1/2-4 )^{2} +(1-1)^{2}+(4-5 )^{2}}$.

$BF=\sqrt{81/4 +0+1}$.

$BF=\sqrt{41/2}$.

$BF=\sqrt{41/2}units$.