For lines $kx + 3y = k - 3$ and $12x + ky = k$ to be coincident, $k$ equals

For lines $kx + 3y = k - 3$ and $12x + ky = k$ to be coincident, $k$ equals
  1. $6$
  2. $3$
  3. $2$
  4. $12$

Solution

$\dfrac{k}{12} = \dfrac{3}{k} \Rightarrow k^{2} = 36 \Rightarrow k = 6$ (positive).

Asked in: MH-SSC-9

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