Correct answer is (c) 1 : 2
Let AB be divided by the x - axis in the ratio k :1 at the point P.
Then, by section formula, the coordinates of P are
\(P(\frac{5k\,+\,2}{k\,+\,1},\frac{6k\,-\,3}{k\,+\,1})\)
But P lies on the x - axis so, its ordinate is 0.
\(\frac{6k\,-\,3}{k\,+\,1}=0\)
\(\Rightarrow\) 6k - 3 = 0
\(\Rightarrow\) 6k = 3
\(\Rightarrow\) k = \(\frac{1}{2}\)
Hence, the required ratio is \(\frac{1}{2}:1\) which is same as 1 : 2.