Question

The equations of \[{L_1}\] and \[{L_2}\] are \[y = mx\] and \[y = nx\], respectively. Suppose \[{L_1}\] makes twice as large of an angle with the horizontal (measured counterclockwise from the positive –axis) as does \[{L_2}\] and that \[{L_1}\] has 4 times the slope of \[{L_2}\]. If \[{L_1}\] is not horizontal, then the value of the product\[{\text{(mn)}}\] equals.
A. \[\dfrac{{\sqrt 2 }}{2}\]
B. \[ - \dfrac{{\sqrt 2 }}{2}\]
C. \[2\]
D. \[ - 2\]

Answer 1

Hint:
First, we will take \[m = \tan 2\theta \] and \[n = \tan \theta \] and then use the property \[\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\] to simplify the values of \[m\] and \[n\]. Then substitute the obtained values in the product \[mn\].

Complete step by step solution:
Given that \[{L_1}\] is \[y = mx\] and \[{L_2}\] is \[y = nx\].

We know that the slope of \[{L_1}\] is \[m\] and \[{L_2}\] is from the above equations.

Since it is given that the \[{L_1}\] has 4 times the slope of \[{L_2}\], \[m = 4n\].

Take \[m = \tan 2\theta \] and \[n = \tan \theta \], we get

\[m = 4n\]

We will use the property of tangential function, that is, \[\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\], where \[\theta \] is the angle.

Using the above property in the above equation \[m = \tan 2\theta \], we get

\[m = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\]

Taking \[n = \tan \theta \] in the above equation, we get

\[m = \dfrac{{2n}}{{1 - {n^2}}}\]

Substituting this value of \[m\] in the given equation \[m = 4n\], we get
\[4n = \dfrac{{2n}}{{1 - {n^2}}}\]

Dividing the above equation by \[2n\] on each of the sides, we get

\[ \Rightarrow 2 = \dfrac{1}{{1 - {n^2}}}\]

Cross-multiplying the above equation, we get

\[
   \Rightarrow 2\left( {1 - {n^2}} \right) = 1 \\
   \Rightarrow 2 - 2{n^2} = 1 \\
\]

Subtracting the above equation by 2 on both sides, we get

\[
   \Rightarrow 2 - 2{n^2} - 2 = 1 - 2 \\
   \Rightarrow \dfrac{{ - 2{n^2}}}{{ - 2}} = \dfrac{{ - 1}}{{ - 2}} \\
   \Rightarrow {n^2} = \dfrac{1}{2} \\
\]
\[ \Rightarrow - 2{n^2} = - 1 \\
\]

Dividing the above equation by \[ - 2\] on each of the sides, we get



Taking square root in the above equation on both sides, we get

\[ \Rightarrow n = \pm \dfrac{1}{{\sqrt 2 }}\]

Since we know that the slope of a line can never be negative, so the value \[ - \dfrac{1}{{\sqrt 2 }}\] is discarded.

Substituting the positive value of \[n\] in the given equation \[m = 4n\], we get

\[m = \dfrac{4}{{\sqrt 2 }}\]

Now we will find the product \[mn\] from these values of \[m\] and \[n\].

\[
  mn = \dfrac{4}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }} \\
   \Rightarrow mn = \dfrac{4}{2} \\
   \Rightarrow mn = 2 \\
\]

Thus, the product \[mn\] equals 2.

Hence, the correct option is C.

Note:
In this question, we can also solve it by taking the angle made by \[y = mx\] with positive direction of –axis is \[{\tan ^{ - 1}}m\] and the angle made by line by \[y = nx\] is \[{\tan ^{ - 1}}n\].
Now, we will take \[{\tan ^{ - 1}}m = 2{\tan ^{ - 1}}n\].

We will use the property of tangential function, that is, \[2{\tan ^{ - 1}}a = {\tan ^{ - 1}}\dfrac{{2a}}{{1 - {a^2}}}\].

Using the above property in our assumed equation \[{\tan ^{ - 1}}m = 2{\tan ^{ - 1}}n\], we get

\[
   \Rightarrow {\tan ^{ - 1}}m = {\tan ^{ - 1}}\dfrac{{2n}}{{1 - {n^2}}} \\
   \Rightarrow m = \dfrac{{2n}}{{1 - {n^2}}} \\
\]

Substituting this value of \[m\] in the equation \[m = 4n\], we get

\[ \Rightarrow \dfrac{{2n}}{{1 - {n^2}}} = 4n\]

Dividing the above equation by \[2n\] on each of the sides, we get

\[ \Rightarrow \dfrac{1}{{1 - {n^2}}} = 2\]

Cross-multiplying the above equation, we get

\[
   \Rightarrow 1 = 2\left( {1 - {n^2}} \right) \\
   \Rightarrow 1 = 2 - 2{n^2} \\
\]

Subtracting the above equation by 2 on both sides, we get

\[
   \Rightarrow 1 - 2 = 2 - 2{n^2} - 2 \\
   \Rightarrow - 1 = - 2{n^2} \\
\]

Dividing the above equation by \[ - 2\] on each of the sides, we get

\[
   \Rightarrow \dfrac{{ - 1}}{{ - 2}} = \dfrac{{ - 2{n^2}}}{{ - 2}} \\
   \Rightarrow \dfrac{1}{2} = {n^2} \\
   \Rightarrow {n^2} = \dfrac{1}{2} \\
\]

Taking square root in the above equation on both sides, we get

\[ \Rightarrow n = \pm \dfrac{1}{{\sqrt 2 }}\]

Since we know that the slope of a line can never be negative, so the value \[ - \dfrac{1}{{\sqrt 2 }}\] is discarded.

Substituting the positive value of \[n\] in the given equation \[m = 4n\], we get

\[m = \dfrac{4}{{\sqrt 2 }}\]

Now we will find the product \[mn\] from these values of \[m\] and \[n\].

\[
  mn = \dfrac{4}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }} \\
   \Rightarrow mn = \dfrac{4}{2} \\
   \Rightarrow mn = 2 \\
\]

Thus, the product \[mn\] equals 2.