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  • If two of the lines given by the equation <img alt="\mathrm{a x^3+b x^2 y+c x y^2+d y

If two of the lines given by the equation \mathrm{a x^3+b x^2 y+c x y^2+d y^3=0(a, d \neq 0)} make complementary angles with x-axis in anticlockwise sense then 

 

Option: 1

\mathrm{a(a-c)-d(b-d)=0}

 


Option: 2

\mathrm{d(a-c)+a(b-d)=0}


Option: 3

\mathrm{a(a-c)+d(b-d)=0}


Option: 4

none of these


Answers (1)

best_answer

Let y = mx be one of the lines represented by the given equation, then

\mathrm{a x^3+b m x^3+c m^2 x^3+d m^3 x^3=0 \text { or } d m^3+c m^2+b m+a=0}

Let its roots be \mathrm{m_1, m_2, m_3}

\therefore\mathrm{ \quad m_1 m_2 m_3=-\frac{a}{d} }\\

\text { If } m_1=\tan \alpha \text {, then } m_2=\tan \left(90^{\circ}-\alpha\right) \text { (given) } \\

\therefore \quad m_2=\cot \alpha

From (i) and (ii) \mathrm{m_3=-\frac{a}{d}} ,

Since \mathrm{m_3} is the root of the above cubic equation 

we have, 

\mathrm{d\left(-\frac{a^3}{d^3}\right)+c\left(\frac{a^2}{d^2}\right)+b\left(-\frac{a}{d}\right)+a=0 \Rightarrow-\frac{a^3}{d^2}+\frac{c a^2}{d^2}-\frac{a b}{d}+a=0}

\Rightarrow \quad\mathrm{-a^3+c a^2-a b d+a d^2=0 \Rightarrow-a^2+c a-b d+d^2=0}

Or      \mathrm {a(a-c)+d(b-d)=0}

 

Posted by

Ajit Kumar Dubey

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