Question

The points (K, 3), (2, -4) and (-K + 1,
-2) are collinear. Find K.​

Answer 1

Solution :-

Let the points be A ( K , 3 ), B ( 2 , -4 ) and ( -K , 1 ).

Given that, the points are collinear.

That is, Area of ΔABC = 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 :-

$\bullet \sf \ x_1 = K \ , \ y_1 = 3 \\\\\bullet \ x_2 = 2 \ , \ y_2 = -4 \\\\\bullet x_3 = -K+1 \ , \ y_3 = -2$

$\longrightarrow \sf \dfrac{1}{2} \times [ \ K(-4-(-2)) +2(-2-3) +(-K+1)(3-(-4)) \ ]= 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ K ( -4+2) +2(-2-3)+(-K+1)(3+4) \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ (K \times -2 ) +(2 \times -5 )+( -K+1)\times 7 \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ -2K -10-7K+7 \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ -9K-3 \ ] = 0 \\\\\longrightarrow -9K-3 = 0 \\\\\longrightarrow -9K = 3$

$\longrightarrow\sf K = \dfrac{-3}{9}\\\\\longrightarrow \bf K = \dfrac{-1}{3}$