Question

If two of the lines given by $3x^3+3x^{2}y-3x{y}^2+d{y}^3=0$ are at right angles then the slope of one of them is

• A. $-1$
• B. $1$
• C. $3$
• D. $-3$

Answer 1

Answer: (A), (B)

The correct options are
A $-1$
B $1$

Let the lines represented by the given equations be $y=m_{1}x,y=m_{2}xandy=m_{3}x$

Then $m_1{m}_2{m}_3=\frac{-3}{d}$

If two of the lines are perpendicular,

let $m_1{m}_2=-1\Rightarrow m_3=\frac{-3}{d}$

$\Rightarrow y=\left(\frac{3}{d}\right)x$ satisfies the given equation

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

Therefore, $d=-3$

Thus, the slope of one of the lines is $m_3=-1$

The given equation for $d=-3$ becomes

$x^3+x^{2}y-x{y}^2-y^3=0or(x+y)(x^2-y^2)=0$

The slope of the other two are $1$ and $-1$.