Question

Find the value of k for which (1,2), (3,0), (2,k) are collinear.

(a) 0

(b) -1

(c) 2

(d) 1

This question was addressed to me in an international level competition.

My question is taken from Area of a Triangle in division Determinants of Mathematics – Class 12

Answer 1

Correct choice is (d) 1

Explanation: The area of triangle formed by collinear points is zero.

Δ=\(\frac{1}{2}\) \(\begin{Vmatrix}1&2&1\\3&0&1\\2&k&1\end{Vmatrix}\)=0

Expanding along C2, we get

\(\frac{1}{2}\){-2(3-2)+0-k(1-3)}=0

\(\frac{1}{2}\) {-2+2k}=0

∴k=1