Question

For what value of k are the points (1, 1), (k, 3) and (1, –4) collinear?​

Answer 1

Solution :-

Let :-

The points be A ( 1 , 1 ), B ( k , 3 ) and C ( 1 , -4 )

Given that, these points are collinear.

That is,

Area of triangle ABC = 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 :-

$\bullet \sf \ x_1 = 1 \ , y_1 = 1 \\\\\bullet \ x_2 = k \ , \ y_2 = 3 \\\\\bullet \ x_3 = 1 \ , y_3 = -4$

$\longrightarrow \sf \dfrac{1}{2} \times [ \ 1(3-(-4))+k(-4-1)+1(1-3) \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ 1(3+4)+k(-4-1)+1(1-3) \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ (1 \times 7 )+ ( k \times -5 )+( 1 \times -2 ) \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ 7-5k-2 \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ -5k+5 \ ] = 0 \\\\\longrightarrow -5k+5 = 0 \\\\\longrightarrow -5k = -5 \\\\\longrightarrow \bf k = 1$

Value of k = 1