First slide

Pair of straight lines

Question

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

Moderate

A

all real values of d
B

two real values of d
C

three real values of d
D

no real value of d

Solution

Let $m_{1}, m_{2}, m_{3}$ be the slopes of the three lines represented 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 $y = m_{3} x$ i.e. $x = d y$ satisfies the given equation, we get
$\begin{array}{l} d^{3} - 6 d^{2} + 3 d + d = 0 \\ \Rightarrow d (d^{2} - 6 d + 4) = 0 \end{array}$
If $d = 0$ the given equation represents the lines x = 0 and $x^{2} - 6 x y + 3 y^{2} = 0$ which are not perpendicular.
$∴ d \neq 0$ and $d^{2} - 6 d + 4 = 0$
$\Rightarrow d = \frac{6 \pm \sqrt{36 - 16}}{2} = 3 \pm \sqrt{5}$
which gives two real values of d.

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