Question

Prove that the equation ${y^3} - {x^3} + 3xy(y - x) = 0$ represents three straight lines equally inclined to one another.

Answer 1

Solve the given equation to find three equations representing straight lines using the formula ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$ and find the angle they are inclined to each other using the formula $\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$ where ${m_1}$ and ${m_2}$ are the slopes of the lines. Then prove that the angles are equal.

Complete step-by-step answer:
The equation of straight lines involves linear power of x and y. The given equation contains cubic terms of x and y, representing the product of the equation of three lines. We solve the equation to find the equation of three lines.
Consider the equation ${y^3} - {x^3} + 3xy(y - x) = 0$ and we need to simplify the left-hand side of the equation.
${y^3} - {x^3} + 3xy(y - x) = 0..........(1)$
We know that ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$, using this formula in equation (1), we get:
$(y - x)({y^2} + xy + {x^2}) + 3xy(y - x) = 0$
Taking (y - x) as a common term, we get:
$(y - x)\left( {{y^2} + xy + {x^2} + 3xy} \right) = 0$
Simplifying, we get:
$(y - x)\left( {{y^2} + {x^2} + 4xy} \right) = 0$
Hence, we get two equations:
$y - x = 0...........(2)$
${y^2} + {x^2} + 4xy = 0.......(3)$
Simplifying equation (3) using formula for quadratic root, that is, the roots of the equation $a{x^2} + bx + c = 0$ is as follows:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}........(4)$
Using formula (4) in equation (3), we get:
$y = \dfrac{{ - 4x \pm \sqrt {{{(4x)}^2} - 4.1.{x^2}} }}{2}$
Simplifying this we get:
$y = \dfrac{{ - 4x \pm \sqrt {16{x^2} - 4{x^2}} }}{2}$
$y = \dfrac{{ - 4x \pm \sqrt {12{x^2}} }}{2}$
$y = \dfrac{{ - 4x \pm 2x\sqrt 3 }}{2}$
$y = - 2x \pm x\sqrt 3$
The two solutions are:
$y = - 2x + x\sqrt 3 ..........(5)$
$y = - 2x - x\sqrt 3 .........(6)$
Equations (2), (5) and (6) are the equations of three lines. Let the lines be A, B and C respectively.
Given a line equation as $y = mx + c$, m is called the slope of the line and c is called its y-intercept. Comparing this equation with (2), (5), and (6), we have:
Slope of line A is 1.
Slope of line B is $\sqrt 3 - 2$.
Slope of line C is $- \sqrt 3 - 2$.
Angle between two lines with slopes ${m_1}$ and ${m_2}$ is given as follows:
$\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$
Hence, the angle between lines A and B is as follows:
$\tan \theta = \left| {\dfrac{{1 - (\sqrt 3 - 2)}}{{1 + 1.(\sqrt 3 - 2)}}} \right|$
$\Rightarrow \tan \theta = \left| {\dfrac{{3 - \sqrt 3 }}{{1 + \sqrt 3 - 2}}} \right|$
$\Rightarrow \tan \theta = \left| {\dfrac{{\sqrt 3 (\sqrt 3 - 1)}}{{\sqrt 3 - 1}}} \right|$
$\Rightarrow \tan \theta = \sqrt 3$
$\theta = 60^\circ$
Then angle between lines A and C is as follows:
$\tan \theta = \left| {\dfrac{{1 - ( - \sqrt 3 - 2)}}{{1 + 1.( - \sqrt 3 - 2)}}} \right|$
$\Rightarrow \tan \theta = \left| {\dfrac{{3 + \sqrt 3 }}{{1 - \sqrt 3 - 2}}} \right|$
$\Rightarrow \tan \theta = \left| {\dfrac{{\sqrt 3 (\sqrt 3 + 1)}}{{ - \sqrt 3 - 1}}} \right|$
$\Rightarrow \tan \theta = \left| { - \sqrt 3 } \right|$
$\Rightarrow \tan \theta = \sqrt 3$
$\theta = 60^\circ$
The angle between lines B and C is as follows:
$\tan \theta = \left| {\dfrac{{\sqrt 3 - 2 - ( - \sqrt 3 - 2)}}{{1 + (\sqrt 3 - 2).( - \sqrt 3 - 2)}}} \right|$
$\Rightarrow \tan \theta = \left| {\dfrac{{\sqrt 3 - 2 + \sqrt 3 + 2}}{{1 - (3 - 4)}}} \right|$
$\Rightarrow \tan \theta = \left| {\dfrac{{2\sqrt 3 }}{2}} \right|$
$\Rightarrow \tan \theta = \sqrt 3$
$\theta = 60^\circ$
Hence, they are equally inclined to each other at an angle of $60^\circ$.

Note: Usually students think linear equations represent straight lines and all higher equations represent curves that are not straight lines but this notion is wrong. As we see in this question, even a third-degree equation can represent three straight lines and not a curve.