# Find the lengths of medians of the triangle $ABC$ where $A(0,0,6),\:B(0,4,0)\:and\:C(6,0,0)$

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• Median is the segment joining the mid point of any one side of the triangle with the third vertex.
• Mid point formula:The coordinates of the point $C$ that divides the segment joining the points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ in the ratio $1:1$ (mid point) is given by $C\big(\large\frac{x_2+x_1}{2},\frac{y_2+y_1}{2},\frac{z_2+z_1}{2}\big)$
• Distance between two points $(x_1,y_1,z_1)\:\;and \:\:(x_2,y_2,z_2)$ is given by $\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Step 1
Given three vertices of the triangle are
$A(0,0,6),\:B(0,4,0)\:and\:C(6,0,0)$
Mid point of $AB$ is given by $D\big(\large\frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2}\big)$
$=D(0,2,3)$
Step 2
Mid point of $BC$ is given by $E\big(\large\frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2}\big)$
$=E(3,2,0)$
Step 3
Mid point of $AC$ is given by $F\big(\large\frac{0+6}{2},\frac{0+0}{2},\frac{6+0}{2}\big)$
$=F(3,0,3)$
Step 4
The three medians are $CD,\:AE\:and\:BF$
median $CD=\sqrt {(6-0)^2+(0-2)^2+(0-3)^2}=\sqrt {49}=7$
median $AE=\sqrt {(0-3)^2+(0-2)^2+(6-0)^2}=\sqrt {49}=7$
median $BF=\sqrt {(0-3)^2+(4-0)^2+(0-3)^2}=\sqrt {34}$
edited May 11, 2014