Question

If A(1,3) B(3,0) and C(0,K) are collinear, find K​

Answer 1

Answer:

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Answer 2

Solution :-

Given :-

The points A ( 1 , 3 ), B ( 3 , 0 ) and C ( 0 , k ) are collinear.

That is :-

Area of triangle ABC is 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 :-

$\bullet \sf \ x_1= 1 \ , \ y_1= 3 \\\\\bullet \ x_2 = 3 \ , \ y_2 = 0 \\\\\bullet \ x_3 = 0 \ , \ y_3 = k$

$\longrightarrow \sf \dfrac{1}{2} \times [ \ 1(0-k) +3(k-3) + 0(3-0) \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ -k+3k-9 \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ 2k - 9 \ ] = 0 \\\\\longrightarrow 2k - 9 = 0 \\\\\longrightarrow 2k = 9 \\\\\longrightarrow \bf k = \dfrac{9}{2}$

$\bf Value \ of \ k = \dfrac{9}{2}$