Question

# $$\displaystyle ax^{2}+2hxy+by^{2}=0$$ represents a pair of straight lines through origin & angle between them is given by$$\displaystyle \tan \theta=\frac{2\sqrt{h^{2}-ab}}{a+b}$$. If the lines are perpendicular then $$\displaystyle a+b=0$$ and the equation of bisectors is given by  $$\displaystyle \frac{x^{2}-y^{2}}{a-b}=\frac{xy}{h}$$The general equation of second degree given by$$\displaystyle ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$$ represent a pair of straight lines if $$\displaystyle \triangle =0$$ or $$\displaystyle \begin{vmatrix}a&h &g \\ h&b &f \\ g&f &c \end{vmatrix}=0$$ or $$\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$$On the basis of above information answer the following questionThe Joint equation of the bisectors of the angle between the lines represented by $$\displaystyle ax^{2}+2hxy+by^{2}=0$$ is

A
a pair of perpendicular lines
B
a pair of parallel lines
C
a pair of intersecting lines but not er
D
None of these

Solution

## The correct option is B a pair of perpendicular linesEquation of bisectors of angle between the pair of straight line $$\displaystyle ax^{2}+2hxy+by^{2}=0$$ is given by $$\displaystyle \frac{x^{2}-y^{2}}{a-b}=\frac{xy}{h}$$$$\displaystyle\Rightarrow hx^{2}-\left ( a-b \right )xy-hy^{2}=0$$Here coefficient of $$x^{2}+$$ coefficient of $$y^{2}=h+(-h)=0$$$$\displaystyle \therefore$$ above equation represent the pair of $$\perp$$er linesHence choice (a) is correct.Maths

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