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 $y=m_{1}x,y=m_{2}x$ and $y=m_{3}x$
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
$\Rightarrow d{\left(\frac{3}{d}\right)}^3-3{\left(\frac{3}{d}\right)}^2+3\left(\frac{3}{d}\right)+3=0\to d=-3$
and the slope of one of the lines is $m_3=-1$
The given equation then becomes
$x^2+x^{2}y-x{y}^2-y^3=0or(x+y)(x^2-y^2)=0$
Showing that the slope of the other two are 1 and -1