Question

If $a{x}^3+b{y}^3+c{x}^{2}y+dx{y}^2=0$ so that each line is angular bisector of other two, then

• A. $b+3a=0$
• B. $a+3c=0$
• C. $c+3b=0$
• D. $3a+d=0$

Answer 1

Answer: (C)

$c+3b=0$

$a{x}^3+b{y}^3+c{x}^{2}y+dx{y}^2=0$

Divide by $x^3$

$a+b{\left(\frac{y}{x}\right)}^3+c\left(\frac{y}{x}\right)+d{\left(\frac{y}{x}\right)}^2=0$

Let $m=\frac{y}{x}$

$b{m}^3+d{m}^2+cm+a=0$

As each line is angular bisector of other $2$, implies angle between any

$2$ pair of lines will be same and Let it be $\theta$

$\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}\Rightarrow m_1-m_2=\tan \theta (1+m_1{m}_2)$

Similarly for the remaining lines we have.

$m_2-m_3=\tan \theta (1+m_2{m}_3)$....(ii)

$m_3-m_1=\tan \theta (1+m_1{m}_3)$....(iii)

Adding (i),(ii) and (iii)

$(m_1-m_2)+(m_2-m_3)+(m_3-m_1)=\tan \theta \left(3+m_1{m}_2+m_2{m}_3+m_3{m}_1\right)$

$\tan \theta (3+\sum m_1{m}_2)=0$

$\tan \ne 0$

$\therefore 3+\sum m_1{m}_2=0$

From cubic equation $\sum m_1{m}_2=\frac{c}{b}$

$3+\frac{c}{b}=0$

$3b+c-0$