A) 2bc − 3ad = 0
B) 2bc + 3ad = 0
C) 3bc − 2ad = 0
D) 3bc + 2ad = 0
Correct Answer: C
Solution :
Solving the given lines we get the point of intersection P\[P\left( \frac{bc-ad}{ab},\frac{4ad-5bc}{2ab} \right)\] As point is equidistant from the axes. \[\Rightarrow \]\[\frac{bc-ad}{ab}=\left| \frac{4ad-5bc}{2ab} \right|\]which gives 3bc − 2ad = 0
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