Question

If the vertices of DABC are A (3,4), B(0,0), and C(6,0); then the length of median AD is ??​

Answer 1

Corrected question:-

If the vertices of triangle ABC are $A(3,4),\ B(0,0),\ C(6,0)$, what is the length of the median $\overline{AD}$?

Before solving:-

In method 1, since $\overline{AD}$ is a median, it divides $\overline{BC}$ equally. The point $D$ lies on the middle of

In method 2, we're going to use a theorem called the "Apollonius theorem" and prove it by coordinate geometry. Let's do it.

Solution:-

Method 1.

A point $D$ divides $\overline{BC}$ in equally. Here we can use the midpoint formula, which  is used when the lengths are divided in a $1:1$ ratio.

Point D

$\equiv(\dfrac{0+6}{2} ,\dfrac{0+0}{2} )$

$\equiv\boxed{(3,0)}$

So the length of $\overline{AD}$ can be found by the distance formula, which is used when we know two endpoints of the segment.

Length of $\overline{AD}$

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

$=\sqrt{0^2+4^2}$

$=\sqrt{4^2}$

$=\boxed{4\ \mathrm{units}}$.

Method 2.

Using Apollonius' theorem,

$\overline{AB}^2+\overline{AC}^2=2(\overline{AD}^2+\overline{BD}^2)$

$\rightarrow 5^2+5^2=2(\overline{AD}^2+3^2)$

$\rightarrow 50=2(\overline{AD}^2+9)$

$\rightarrow 25=\overline{AD}^2+9$

$\rightarrow \overline{AD}^2=16$

$\rightarrow \overline{AD}=\boxed{\mathrm{4\ units}}$

Learn more:-

Apollonius' Theorem

→ The relation between the two sides, median, halved length of a triangle is as follows.

$\overline{AB}^2+\overline{AC}^2=2(\overline{AD}^2+\overline{BD}^2)$

• $\overline{AB},\ \overline{AC}$ are the two sides
• $\overline{AD}$ is the median
• $\overline{BD}$ is the halved length

Proof

Consider drawing $\triangle {ABC}$ on a plane, and $\overline{AD}$ as a median.

And consider the parallel movement of $D(0,0)$. Then $B(a,b)$ and $C(-a,-b)$ because they are symmetric against the origin.

And let the remaining vertice be $A(c,d)$.

[Eqn. 1] $\overline{AB}^2=a^2-2ac+c^2+b^2-2bd+d^2$

[Eqn. 2] $\overline{AC}^2=a^2+2ac+c^2+b^2+2bd+d^2$

[Eqn. 3] $\overline{AD}^2=c^2+d^2$

[Eqn. 4] $\overline{BD}=a^2+b^2$

According to [Eqn. 1,2],

$\rightarrow \overline{AB}^2+\overline{AC}^2=2(a^2+b^2+c^2+d^2)$

Then according to [Eqn. 3,4],

$\rightarrow 2(\overline{AD}^2+\overline{BD}^2)=2(a^2+b^2+c^2+d^2)$

Therefore

$\overline{AB}^2+\overline{AC}^2=2(\overline{AD}^2+\overline{BD}^2)$.