Question

If two of the straight lines represented by $a{x}^3+b{x}^{2}y+cx{y}^2+d{y}^3=0$ are at right angles, then

• A. $a^2+ac+bd-d^2=0$
• B. $a^2+ac-bd+d^2=0$
• C. $a^2-ac+bd+d^2=0$
• D. $a^2+ac+bd+d^2=0$

Answer 1

The correct option is D $a^2+ac+bd+d^2=0$

$a{x}^3+b{x}^{2}y+cx{y}^2+d{y}^3=0$ ...(1)

is a homogeneous equation of third degree is $x$ and $y$.

It, therefore, represents three straight lines through the origin.

Let the slopes of the lines be $m_1,m_2,m_3$.

Then, $m_1,m_2$ and $m_3$ are the roots of

$d{m}^3+{cm}^2+bm+a=0$ ...(2)

Product of the roots $=m_1{m}_2{m}_3=-\frac{a}{d}$ ...(3)

Since the two lines represented by (1) are at right angles, let the lines with slopes $m_1,m_2$ be perpendicular.

Then, $m_1{m}_2=-1$

$\therefore$ from (3), $(-1)m_3=-\frac{a}{d}$ or $m_3=\frac{a}{d}$ ...(4)

But $m_3$ is a root of (2)

$\therefore {{dm}_3}^3+{{cm}_3}^2+{bm}_3+a=0$

Substituting the value of $m_3$ from (4),

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

$\Rightarrow a^3+a^{2}c+abd+{ad}^2=0$ $\Rightarrow a^2+ac+bd+d^2=0$,

Which is the required condition.