Question

Two of the lines represented by $x^3-6x^{2}y+3x{y}^2+d{y}^3=0$ are perpendicular for

• A. all real values of $d$
• B. two real values of $d$
• C. three real values of $d$
• D. no real value of $d$

Answer 1

Answer: (B)

two real values of $d$

Let $m_1,m_2,m_3$ be the slopes of the three lines represnted by the given equation such that $m_1{m}_2=-1$

We have $m_1{m}_2{m}_3=-\frac{1}{d}$ so that

$m_3=\frac{1}{d}$

Since $y=m_{3}x\Rightarrow x=dy$ satisfies the given equation, we get

$d^3-6d^2+3d+d=0\Rightarrow d(d^2-6d+4)=0$

If $d=0,$ the given equation represents the line $x=0$ and $x^2-6xy+3y^2=0$ which are not perpendicular

$\therefore d\ne 0$ and $d^2-6d+4=0\Rightarrow d=\frac{6\pm \sqrt{36-16}}{2}=3\pm \sqrt{5}$

which gives two real values of $d$