A) \[(4-\sqrt{5})x-(3-2\sqrt{5})y+(2-4\sqrt{5})=0\]
B) \[(4+\sqrt{5})x-(3+2\sqrt{5})y+(2+4\sqrt{5})=0\]
C) \[(4+\sqrt{5})x+(3+2\sqrt{5})y+(2+4\sqrt{5})=0\]
D) None of these
Correct Answer: A
Solution :
The equations of the bisectors of the angles between the lines are \[\frac{x-2y+4}{\sqrt{1+4}}=\pm \frac{4x-3y+2}{\sqrt{16+9}}\] Taking positive sign, then \[(4-\sqrt{5})x-(3-2\sqrt{5})\text{ }y-(4\sqrt{5}-2)=0\] .....(i) and negative sign gives \[(4+\sqrt{5})x-(2\sqrt{5}+3)y+(4\sqrt{5}+2)=0\] .....(ii) Let \[\theta \]be the angle between the line (i) and one of the given line, then \[\tan \theta =\left| \frac{\frac{1}{2}-\frac{4-\sqrt{5}}{3-2\sqrt{5}}}{1+\frac{1}{2}.\frac{4-\sqrt{5}}{3-2\sqrt{5}}} \right|=\sqrt{5}+2>1\] Hence the line (i) bisects the obtuse angle between the given lines.
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