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

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 -

\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.