Question

If two of the lines $a{x}^3+b{x}^{2}y+cx{y}^2+d{y}^3=0(a\ne 0)$ make complementary angles with x-axis in anti-clockwise sense then

• A. $a(a-c)-d(b-d)=0$
• B. $d(a-c)+a(b-d)=0$
• C. $a(a-c)+d(b-d)=0$
• D. None of these

Answer 1

The correct option is C $a(a-c)+d(b-d)=0$

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

${ax}^3+bm{x}^3+{cm}^2{x}^3+{dm}^2{x}^3=0$

or, ${dm}^3+{cm}^2+bm+a=0$

Let its roots be $m_1,m_2,m_3$

$\therefore m_1{m}_2{m}_3=-\frac{a}{d}$ ...(1)

If $m_1=\tan \alpha$, then $m_2=\tan ({90}^0-\alpha )$ $($Given$)$

$\therefore m_2=\cot \alpha$

$\therefore m_1{m}_2=1$

From(1), $m_3=-\frac{a}{d}$

Since $m_3$ is the root of the above cubic equation, we have

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

$\Rightarrow -\frac{a^3}{d^2}-\frac{c{a}^2}{d^2}-\frac{ab}{d}=0$

$\Rightarrow a(a-c)+d(b-d)=0$