Question

In Q. No. 33, what is the value of \[\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}\]?

Answer 1

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 `(x_1, y_1), (x_2, y_2), (x_3, y_3)` is-

`((x_1 + x_2 + x_3 )/3, (y_1 + y_2 + y_3)/3)`

So,

`(0, 0) = ((a + b + c)/3, (b + c + a)/3)`

Compare individual terms on both the sides-

`(a + b + c)/3 = 0`

Therefore,

a + b + c = 0

We have to find the value of -

`= (a^2)/(bc) + (b^2)/(ca) + (c^2)/(ab) ` 

Multiply and divide it by (abc) to get,

`= (1/(abc)) (a^3 + b^3 + c^3)`

Now as we know that if,

a  + b + c = 0

Then,

`a^3 + b^3 + c^3 = 3abc`

So,

`(a^2 ) /( bc) +(b^2)/(ca) + (c^2)/(ab) = (1/(abc)) (a^3 + b^3 +c^3)`

`= (1/(abc))(3abc)`

= 3