Question

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

विकल्प

• $$\frac{1}{3}$$

   

• $$-\frac{1}{3}$$

   

• $$\frac{2}{3}$$

   

• $$-\frac{2}{3}$$

   

Answer 1

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

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

${\displaystyle x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0}$

So,

${\displaystyle k(3k-1)+3k(1-2k)+3(2k-3k)=0}$

So,

-3k² - k = 0

Take out the common terms,

-k (3k + 1) = 0

Therefore,

${\displaystyle k=-\frac{1}{3}}$