# Solve the following

Question:

In Q. No. 9, what is the value of $\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}$ ?

Solution:

The co-ordinates of the vertices are (a, b); (b, c) and (c, a)

The co-ordinate of the centroid is (0, 0)

We know that the co-ordinates of the centroid of a triangle whose vertices are $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)$ is-

$\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$

So,

$(0,0)=\left(\frac{a+b+c}{3}, \frac{b+c+a}{3}\right)$

Compare individual terms on both the sides-

$\frac{a+b+c}{3}=0$

Therefore,

$a+b+c=0$

We have to find the value of -

$=\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}$

Multiply and divide it by $(a b c)$ to get,

$=\left(\frac{1}{a b c}\right)\left(a^{3}+b^{3}+c^{3}\right)$

Now as we know that if,

$a+b+c=0$

Then,

$a^{3}+b^{3}+c^{3}=3 a b c$

So,

$\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}=\left(\frac{1}{a b c}\right)\left(a^{3}+b^{3}+c^{3}\right)$

$=\left(\frac{1}{a b c}\right)(3 a b c)$

$=3$