Question

10. If the points (k, 2k), (3k, 3k) and (3, 1) are collinear, then k=​

Answer 1

$\huge\boxed{Answer:-}$

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Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \purple{Answer\::}$

Let A = ( k , 2k ) ; B = ( 3k , 3k ) ; C = ( 3,1)

$\red{We \: know \: that \: , \: these \: points \: are \: </p><p> collinear \: then \: : \: Area \: of \: triangle \: is \: 0.</p><p>\:}$

➩ x1(y2-y3) + x2(y3-y1) + x3(y1-y2) = 0

➩ k (3k-1) + 3k (1-2k) + 3 (2k-3k) = 0

➩ -3k^2 - k = 0

$\pink{ஃ \: k \: = - \frac{1}{3}\:}$