Question

If the points (k, 2k), (3k, 3k) and (3, 1) are collinear, then k

Options

• \[\frac{1}{3}\]

   

• \[- \frac{1}{3}\]

   

• \[\frac{2}{3}\]

   

• \[- \frac{2}{3}\]

   

Answer 1

We have three collinear points A(k, 2k) : B(3k, 3k) : C (3, 1) .

In general if `A(x_1 , y_1) ; B (x_2 ,y_2) ; C(x_3 ,y_3) `are collinear then, area of the triangle is 0.

`x_1 (y_2 - y_3 ) + x_2 (y_3 - y_1) + x_3 (y_1- y_2 ) = 0`

So,

`k (3k - 1 ) + 3k(1 - 2k) + 3 (2k - 3k) = 0`

So,

-3k² - k = 0

Take out the common terms,

-k (3k + 1) = 0

Therefore,

`k = -1/3`