Question

The equations 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) \[x = 0,2mx + \left( {{m^2} - 1} \right)y = 0\]
B) \[y = 0,2mx + \left( {{m^2} - 1} \right)y = 0\]
C) \[y = 0,2mx + \left( {1 - {m^2}} \right)y = 0\]
D) None of these

Answer 1

Hint: First using the given line equation find the slope of that line then using that inclined angle find the slope of the required line after that using line equation we get the required answer.

Complete step by step solution:
Given, equation of line is \[y = mx + c \ldots \left[ 1 \right]\]
As we know the slope of the line of any line is the coefficient of x
Here, in equation [1] m is the coefficient of x
So, slope of \[y = mx + c\] is m.
Also slope can be written as m = gradient =\[\tan \alpha \]
Since, the required lines inclined at an angle \[{\tan ^{ - 1}}m\] to the line [1]
\[\therefore \] Slope of required line \[ = \tan \left( {{{\tan }^{ - 1}}m} \right)\]
               \[ = m\]
Equation of lines
\[\left( {y - 0} \right) = \left( {\dfrac{{m \pm m}}{{1 \mp {m^2}}}} \right)\left( {x - 0} \right)\]
\[ \Rightarrow y = \left( {\dfrac{{2m}}{{1 - {m^2}}}} \right)\left( {x - 0} \right)\]and \[y = 0\]
\[ \Rightarrow y\left( {1 - {m^2}} \right) = 2mx\] and \[y = 0\]
\[ \Rightarrow 2mx - y\left( {1 - {m^2}} \right) = 0\] and \[y = 0\]
\[ \Rightarrow 2mx + y\left( {{m^2} - 1} \right) = 0\] and \[{y = 0}\]
Hence, option B, \[y = 0,2mx + \left( {{m^2} - 1} \right)y = 0\] is the correct answer.

Note: when the line equation is given coefficient of x is the slope of the given line. The standard line equation is \[y = mx + c\] where m is slope and c is the intercept point where the line crosses the y-axis.
Example: \[y = 4x + 6\]\[ \Rightarrow m = 4\& 6\] is the intercept point.