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 which of the following options is correct?
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

Hint: A second-degree homogeneous equation of the form $a{x^2} + 2hxy + b{y^2} = 0$ represents a pair of straight lines passing through the origin. Similarly, we are provided a third-degree homogeneous equation of the form $a{x^3} + b{x^2}y + cx{y^2} + d{y^3} = 0$ in the question. This equation represents three straight lines passing through the origin. Then using product of roots we can get the correct answer.

Formula Used: The general form of a cubic equation, that is, a three-degree equation is $a{x^3} + b{x^2} + cx + d = 0$ . The product of roots of this equation is given by $ - \dfrac{d}{a}$ .

Complete step by step solution:
Given equation:
$a{x^3} + b{x^2}y + cx{y^2} + d{y^3} = 0$ … (1)
Clearly, the given equation is a third-degree homogeneous equation in $x$ and $y$ .
This means that it is representing three straight lines passing through the origin.
Let their slopes be ${m_1},{m_2},{m_3}$ respectively.
For a general form of a line passing through the origin, let $y = mx$ be the solution of (1).
Then, substituting $y = mx$ in (1), we get:
$a{x^3} + bm{x^3} + c{m^2}{x^3} + d{m^3}{x^3} = 0$
Simplifying further, we get:
$a + bm + c{m^2} + d{m^3} = 0$ … (2)
Now, this is a three-degree equation.
Hence,
Product of the roots, ${m_1}{m_2}{m_3} = - \dfrac{a}{d}$ … (3)
It is given that two of the lines are perpendicular.
Let them be the one with slopes ${m_1}$ and ${m_2}$ .
Then, ${m_1}{m_2} = - 1$ .
Substituting this in equation (3), we get:
${m_3} = \dfrac{a}{d}$ … (4)
Now, ${m_3}$ is one of the solutions of (2), therefore,
$a + b{m_3} + c{m_3}^2 + d{m_3}^3 = 0$
Substituting the value of ${m_3}$ from (4),
$a + b\left( {\dfrac{a}{d}} \right) + c{\left( {\dfrac{a}{d}} \right)^2} + d{\left( {\dfrac{a}{d}} \right)^3} = 0$
On further simplifying, we get:
$a{d^2} + abd + {a^2}c + {a^3} = 0$
As $a \ne 0$ , therefore,
\[{a^2} + ac + bd + {d^2} = 0\]
Thus, the correct option is D.

Note: In the above solution, avoid making any mistakes while calculating the product of the slopes. While calculating the product of slopes for a third-degree homogeneous equation, proceed in the same way as you’d do for any second-degree equation.