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

Hint: 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.