Question

Find the acute angle between the lines \[(a + b)x + (a - b)y = 2ab\], \[(a - b)x + (a + b)y = 2ab\] where \[{\mathbf{a}} > {\mathbf{b}}\].

Answer 1

Hint: To solve the question first we have to find out the slopes of respective lines which can be obtained from the coefficients of x and y is the line equations. Then we must apply the formula of the angle between two lines which is related to the slopes of the lines.

Complete step-by-step solution:
From the question, we get the equations of two lines are given by -
\[ {L_1}:(a + b)x + (a - b)y = 2ab \\
   \Rightarrow (a + b)x + (a - b)y - 2ab = 0 \\ \] ……………………………… (1)
And
\[ {L_2}:(a - b)x + (a + b)y = 2ab \\
   \Rightarrow (a - b)x + (a + b)y - 2ab = 0 \\ \] …………………………………… (2)
The slope of a line is defined as the tangent of the angle with which the line makes in the positive direction of the X-axis. From the equation of the line, we can obtain the slope which is as follows.
The slopes of a line of general equation \[Ax + By + C = 0\] is given by
\[m = - \dfrac{A}{B}\] ………………………. (3)
Now comparing the general equation of line with equation of $L_1$, we can get
\[A = a + b\] and \[B = a - b\]
Hence the slope of the line $L_1$ is given by
\[{m_1} = - \left( {\dfrac{{a + b}}{{a - b}}} \right)\] …………………………. (4)
Similarly, comparing the general equation of line with equation of $L_1$, we can get the slope of the line \[{L_2}\] which is given by
\[{m_2} = - \left( {\dfrac{{a - b}}{{a + b}}} \right)\] ……………………………… (5)
As \[a > b\], from eq. (4) and eq. (5) we see that \[{m_2} > {m_1}\].
We know that the acute angle between two lines having slopes \[{m_1}\] and \[{m_2}\] for \[{m_2} > {m_1}\] is given by \[\theta = {\tan ^{ - 1}}\left( {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right)\] ………………………… (6)
Substituting the values of $m_1$ and $m_2$ from eq. (4) and eq. (5) in eq. (6) we get,
\[ \theta = {\tan ^{ - 1}}\left( {\dfrac{{ - \left( {\dfrac{{a - b}}{{a + b}}} \right) + \left( {\dfrac{{a + b}}{{a - b}}} \right)}}{{1 + \left( {\dfrac{{a + b}}{{a - b}}} \right)\left( {\dfrac{{a - b}}{{a + b}}} \right)}}} \right) \\
   = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{{{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2}}}{{\left( {a + b} \right)(a - b)}}}}{{1 + 1}}} \right) \\
   = {\tan ^{ - 1}}\left( {\dfrac{{4ab}}{{2({a^2} - {b^2})}}} \right) \\
   = {\tan ^{ - 1}}\left( {\dfrac{{2ab}}{{{a^2} - {b^2}}}} \right) \\ \]
Here we got the acute angle between the given lines is \[{\tan ^{ - 1}}\left( {\dfrac{{2ab}}{{{a^2} - {b^2}}}} \right)\].

Note: When two lines intersect with each other, then one pair each of acute angles and obtuse angles are formed. The obtuse angle between two lines having slopes \[{m_1}\] and \[{m_2}\] for \[{m_2} > {m_1}\] is given by \[\tan \theta = - \left( {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right)\]. While applying the formula it should be observed that for determination of acute angle \[\left( {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right)\] must be positive quantity and for obtuse angle \[\left( {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right)\] must be negative quantity.