Question

If a#b#0,prove that the points (a,a^),(b,b^),(0,0) will not be collinear

(^ is square)

Answer 1

$\boxed{\textsf{Solution by finding line's equation :}}$

The three given points are (a, a²), (b, b²) and (0, 0)

So, the equation of the line passing through the points (a, a²) and (0, 0) is

  $\dfrac{y-a^{2}}{a^{2}-0} = \dfrac{x-a}{a-0}$

  ⇒ $\dfrac{y-a^{2}}{a^{2}} = \dfrac{x-a}{a}$

  ⇒ a (y - a²) = a² (x - a)

  ⇒ y - a² = a (x - a) , dividing both sides by a (≠ 0)

  ⇒ y - a² = ax - a²

  ⇒ y = ax , cancelling a² from both sides

  ⇒ y - ax = 0 .....(i)

The line passing through the points (a, a²) and (0, 0) is y - ax = 0

Now, we satisfy the LHS of (i) no. equation with the point (b, b²). We get

b - ab² ≠ 0 , since none of a and b is zero

Hence, the point (b, b²) doesn't lie on the line (i)

Therefore, we conclude that the three points (a, a²), (b, b²) and (0, 0) aren't collinear.

$\boxed{\textsf{Solution by finding triangle's area :}}$

  If we can show that the area of the triangle formed by the points (a, a²), (b, b²) and (0, 0) is non - zero, we can conclude that the points aren't collinear.

Now, the area of the considered triangle be

  = ½ [ a (b² - 0) - a² (b - 0) + 1 (0 - 0) ] units

  = ½ [ ab² - a²b ]

  ≠ 0 , since both a and b are non - zero.

Therefore, the given three points (a, a²), (b, b²) and (0, 0) aren't collinear.