Question

In each of the following find the value of 'k' for which the point are collinear. (-1,2) (3,5) (6,k)​

Answer 1

Solution :-

Let :-

The points be A ( -1 , 2 ), B ( 3 , 5 ) and C ( 6 , k ).

Given that :-

These points are collinear.

That is,

Area of triangle ABC is equal to zero.

$\boxed{\bf Area \ of \ triangle = \dfrac{1}{2} \times [ \ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}$

Here :-

$\bullet \sf \ x_1=-1 \ , \ y_1=2 \\\\\bullet \ x_2=3 \ , \ y_2=5 \\\\\bullet \ x_3= 6 \ , \ y_3= k$

$\longrightarrow \sf \dfrac{1}{2} \times [ \ -1(5-k)+3(k-2)+6(2-5) \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ -5+k+3k-6+(6 \times -3 ) \ ] =0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ -5+k+3k-6-18 \ ] =0 \\\\\longrightarrow \dfrac{1}{2}\times [ \ 4k-11-18 \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ 4k-29 \ ] = 0 \\\\\longrightarrow 4k-29 = 0 \\\\\longrightarrow 4k = 29 \\\\\longrightarrow \bf k = \dfrac{29}{4}$