A) equally inclined
B) perpendicular
C) bisector of the angle
D) None of these
Correct Answer: A
Solution :
The equation of the bisectors of the angle between the lines given by\[a{{x}^{2}}+2xy+b{{y}^{2}}=0\]is \[\frac{{{x}^{2}}-{{y}^{2}}}{a-b}=\frac{xy}{h}\] ?(i) And the equation of the bisectors of the angle between the lines given by \[{{a}^{2}}{{x}^{2}}+2h(a+b)xy+{{b}^{2}}{{y}^{2}}=0\]is \[\frac{{{x}^{2}}-{{y}^{2}}}{{{a}^{2}}-{{b}^{2}}}=\frac{xy}{h\,(a+b)}\] \[\Rightarrow \] \[\frac{{{x}^{2}}-{{y}^{2}}}{a-b}=\frac{dy}{h\,}\] ?(ii) From Eqs. (i) and (ii), it is clear that both the pair of straight lines have the same bisector, hence, the given two given two pairs of straight lines are equally inclined.
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