A) 1
B) 2
C) 1/2
D) 3
Correct Answer: B
Solution :
[b]: If the pair of lines is \[a{{x}^{2}}+2hxy+b{{y}^{2}}=0\]and it has slopes \[{{m}_{1}}\]and \[{{m}_{2}}\], then\[{{m}_{1}}+{{m}_{2}}=-\frac{2h}{b}\]and\[{{m}_{1}}{{m}_{2}}=\frac{a}{b}\] \[\therefore \]\[{{({{m}_{1}}-{{m}_{2}})}^{2}}={{({{m}_{1}}+{{m}_{2}})}^{2}}-4{{m}_{1}}{{m}_{2}}=\frac{4({{h}^{2}}-ab)}{{{b}^{2}}}\] Here \[a={{\tan }^{2}}\theta +{{\cos }^{2}}\theta ,h=-\tan \theta \]and\[b={{\sin }^{2}}\theta \] \[\therefore \]\[{{({{m}_{1}}-{{m}_{2}})}^{2}}=\frac{4}{{{\sin }^{4}}\theta }\left[ {{\tan }^{2}}\theta -(ta{{n}^{2}}\theta +co{{s}^{2}}\theta )si{{n}^{2}}\theta \right]\]\[=\frac{4}{{{\sin }^{4}}\theta }\left[ \frac{1}{{{\cos }^{2}}\theta }-{{\tan }^{2}}\theta -{{\cos }^{2}}\theta \right]\] \[=\frac{4}{{{\sin }^{2}}\theta }(1-{{\cos }^{2}}\theta )=4\] \[\therefore \]\[|{{m}_{1}}-{{m}_{2}}|=2\].
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