Question

The equation of the lines which pass through the origin and are inclined at an angle ${{\tan }^{-1}}m$ to the line y = mx + c, are: -

(a) $y=0,2mx+\left( 1-{{m}^{2}} \right)y=0$

(b) $y=0,2mx+\left( {{m}^{2}}-1 \right)y=0$

(c) $x=0,2mx+\left( {{m}^{2}}-1 \right)y=0$

(d) None of these

Answer 1

Assume ‘$\theta$’ as the angle between the lines y = mx + c and the required equation which we have to determine. Now, assume this required equation of line passing through the origin as $y={{m}_{1}}x$ where ${{m}_{1}}$ is its slope. Apply the formula: - $\tan \theta =\left| \dfrac{{{m}_{1}}-m}{1+{{m}_{1}}.m} \right|$ and substitute the value of $\tan \theta =m$. Remove the modulus sign and consider two cases, one with a positive sign and the other with a negative sign. Find the value of ${{m}_{1}}$ in each case and substitute in the equation: - $y={{m}_{1}}x$ to get the answer.

Complete step by step answer:

Here, we have been given that the lines passing through the origin are inclined at an angle ${{\tan }^{-1}}m$ to the line y = mx + c and we have to determine these equations of the lines.

Now, let us assume the equation of lines passing through the origin is $y={{m}_{1}}x$, where ${{m}_{1}}$ is its slope. We have been provided with the angle of inclination of $y={{m}_{1}}x$ with respect to y = mx + c, that means we have been provided with the angle between their slopes. Here, we are assuming ‘$\theta$’ as this angle of inclination. So, we have,

$\Rightarrow \theta ={{\tan }^{-1}}m$

$\Rightarrow \tan \theta =m$ - (1)

Now, we know that if we have ‘a’ and ‘b’ as the slopes of two lines then their angle of intersection ‘$\theta$’ is given as: -

$\Rightarrow \tan \theta =\left| \dfrac{a-b}{1+ab} \right|$

So, in the given question, we have,

$\Rightarrow \tan \theta =\left| \dfrac{{{m}_{1}}-m}{1+{{m}_{1}}.m} \right|$

Using equation (1), we get,

$\Rightarrow m=\left| \dfrac{{{m}_{1}}-m}{1+{{m}_{1}}.m} \right|$

Removing modulus sign, we get,

$\Rightarrow \dfrac{{{m}_{1}}-m}{1+{{m}_{1}}m}=\pm m$

Cancelling each sign one – by – one, we get,

1. Case 1: - Considering positive sign (+),

$\Rightarrow \dfrac{{{m}_{1}}-m}{1+{{m}_{1}}m}=m$

By cross – multiplication we get,

$\Rightarrow {{m}_{1}}-m=m+{{m}_{1}}{{m}^{2}}$

$\Rightarrow {{m}_{1}}\left( 1-{{m}^{2}} \right)=2m$

$\Rightarrow {{m}_{1}}=\dfrac{2m}{\left( 1-{{m}^{2}} \right)}$

So, substituting the value of ${{m}_{1}}$ in this case in the equation of required line, we get,

$\Rightarrow y=\left( \dfrac{2m}{1-{{m}^{2}}} \right)x$

$\Rightarrow \left( 1-{{m}^{2}} \right)y=2mx$

$\Rightarrow 2mx-\left( 1-{{m}^{2}} \right)y=0$

$\Rightarrow 2mx+\left( {{m}^{2}}-1 \right)y=0$

2. Case 2: - Considering negative sign (-)

$\Rightarrow \dfrac{{{m}_{1}}-m}{1+{{m}_{1}}m}=-m$

$\Rightarrow {{m}_{1}}-m=-m+\left( -{{m}_{1}}{{m}^{2}} \right)$

$\Rightarrow {{m}_{1}}=-{{m}_{1}}{{m}^{2}}$

$\Rightarrow {{m}_{1}}+{{m}_{1}}{{m}^{2}}=0$

$\Rightarrow {{m}_{1}}\left( 1+{{m}^{2}} \right)=0$

Here, $\left( 1+{{m}^{2}} \right)$ can never be 0, so we have,

$\Rightarrow {{m}_{1}}=0$

So, substituting the value of ${{m}_{1}}$ in this case in the equation of required line, we get,

$\Rightarrow y=0.x$

$\Rightarrow y=0$

So, from the above two cases the equation of required lines passing through the origin can be: -

$\Rightarrow y=0,2mx+\left( {{m}^{2}}-1 \right)y=0$

So, the correct answer is “Option (b)”.

Note: One may note that after removing the modulus sign we have considered two cases, one with (+) sign and the other with (-) sign. This is because we don’t know if ‘$\theta$’ is acute or obtuse in nature. So, if ‘$\theta$’ is acute then the (+) sign was considered and if ‘$\theta$’ is obtuse then the (-) sign was considered.