A) \[20\sqrt{2}\beta =11\alpha \]
B) \[20\sqrt{2}\alpha =11\beta \]
C) \[11\sqrt{2}\beta =20\alpha \]
D) None of these
Correct Answer: A
Solution :
Given \[\alpha \] be the distance between lines \[x-y+2=0\] and \[x-y-2=0.\] \[\therefore \] \[\alpha =\frac{|2+2|}{\sqrt{1+1}}=\frac{|4|}{\sqrt{2}}=2\sqrt{2}\] and \[\beta \] be the distance between the lines \[4x-3y-5=0\] and \[4x-3y+1/2=0\] \[\therefore \] \[\beta =\frac{\left| 5+\frac{1}{2} \right|}{\sqrt{{{(4)}^{2}}+{{(3)}^{2}}}}=\frac{|11|}{2\sqrt{25}}=\frac{11}{10}\] Now, \[\frac{\alpha }{\beta }=\frac{2\sqrt{2}}{11/10}\] \[\Rightarrow \] \[\frac{\alpha }{\beta }=\frac{20\sqrt{2}}{11}\] \[\Rightarrow \] \[20\sqrt{2}\,\,\beta =11\alpha \]
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