Question

Find the length of the medians of the triangle with vertices $A(0,0,6)B(0,4,0)$ and $C(6,0,0)$.

Answer 1

Here $A(0,0,6),B(0,4,0)$ and $C(6,0,0)$ are vertices of $\Delta ABC$

Now D is mid point of BC

$\therefore$ Coordinates of D is $(\frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2})$

= $(3,2,0)$

$\therefore AD=\sqrt{(0-3{)}^2+(0-2{)}^2+(6-0{)}^2}$

= $\sqrt{9+4+36}=7$ units.

Also E is mid point of AC

$\therefore$ Coordinates of E is $(\frac{0+6}{2},\frac{0+0}{2},\frac{6+0}{2})$

= $(3,0,3)$

$\therefore BE=\sqrt{(0-3{)}^2+(4-0{)}^2+(0-3{)}^3}$

= $\sqrt{9+16+9}=\sqrt{34}$ units.

Also F is mid point of AB

$\therefore$ Coordinates of F is $(\frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2})$

= $(0,2,3)$

$\therefore CF=\sqrt{(6-0{)}^2+(0-2{)}^2+(0-3{)}^2}$

= $\sqrt{36+4+9}=7$ units.