Question

Find what straight lines are represented by the following equation and determine the angles between them.
$y^3-x{y}^2-14x^{2}y+24x^3=0.$

Answer 1

For such type of problems we find the first root by hit and trial method.

$y^3-x{y}^2-14x^{2}y+24x^3=\mathrm{0.......}\left(i\right)$

let us put $y=x$ first

$\Rightarrow x^3-x.x^2-14x^2.x+24x^3=10x^3\ne 0$

Put $y=2x$

$\Rightarrow {8x}^3-x.4x^2-14x^2.2x+24x^3=0$

$\Rightarrow$ $y=2x$ is a root of the equation or $y-2x$ is the factor of the equation $\left(i\right)$

Dividing $\left(i\right)$ by $y-2x$

$\frac{y^3-x{y}^2-14x^{2}y+24x^3}{y-2x}=y^2+xy-12x^2......\left(ii\right)$

Factorising equation $\left(ii\right)$

$y^2+xy-12x^2=y^2+4xy-3xy-12x^2=y(y+4x)-3x(y+4x)=(y-3x)(y+4x)$

$\Rightarrow y^3-x{y}^2-14x^{2}y+24x^3=(y-2x)(y-3x)(y+4x)=0$

So the equation of lines are $y-2x=0,y+4x=0$ and $y-3x=0$

Angle between two lines that is $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\left(a\right)$ angle between $y=2x$ and $y=3x$

$\tan \theta =\left|\frac{2-3}{1+2\times 3}\right|=\frac{1}{7}\Rightarrow \theta ={\tan }^{-1}\left(\frac{1}{7}\right)$

$\left(b\right)$ Angle between $y=3x$ and $y=-4x$

$\tan \theta =\left|\frac{3-(-4)}{1+3\times -4}\right|=\frac{7}{11}\Rightarrow \theta ={\tan }^{-1}\left(\frac{7}{11}\right)$

$\left(c\right)$ Angle between $y=-4x$ and $y=2x$

$\tan \theta =\left|\frac{-4-\left(2\right)}{1+(-4)\times 2}\right|=\frac{6}{7}\Rightarrow \theta ={\tan }^{-1}\left(\frac{6}{7}\right)$