Question

Prove that the straight lines (a + b) x + (a − b ) y = 2ab, (a − b) x + (a + b) y = 2ab and x + y = 0 form an isosceles triangle whose vertical angle is 2 tan⁻¹ \[\left( \frac{a}{b} \right)\].

Answer 1

The given lines are
(a + b) x + (a − b ) y = 2ab      ... (1)
(a − b) x + (a + b) y = 2ab       ... (2)
x + y = 0                                    ... (3)
Let m_1, m₂ and m₃ be the slopes of the lines (1), (2) and (3), respectively.
Now,

Slope of the first line = m₁ = \[- \left( \frac{a + b}{a - b} \right)\]

Slope of the second line = m₂ = \[- \left( \frac{a - b}{a + b} \right)\]

Slope of the third line = m₃ = \[- 1\] 

Let \[\theta_1\]  be the angle between lines (1) and (2),  \[\theta_2\] be the angle between lines (2) and (3) and \[\theta_3\] be the angle between lines (1) and (3).

\[\therefore \tan \theta_1 = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|\]

\[ = \left| \frac{- \left( \frac{a + b}{a - b} \right) + \frac{a - b}{a + b}}{1 + \frac{a + b}{a - b} \times \frac{a - b}{a + b}} \right|\]

\[ \Rightarrow \tan \theta_1 = \left| \frac{2ab}{a^2 - b^2} \right|\]

\[ \Rightarrow \theta_1 = \tan^{- 1} \left| \frac{2ab}{a^2 - b^2} \right|\]

\[ = 2 \tan^{- 1} \left( \frac{a}{b} \right)\]

\[\therefore \tan \theta_2 = \left| \frac{m_2 - m_3}{1 + m_2 m_3} \right|\]

\[ = \left| \frac{- \left( \frac{a - b}{a + b} \right) + 1}{1 + \frac{a - b}{a + b}} \right|\]

\[ \Rightarrow \tan \theta_2 = \left| \frac{b}{a} \right|\]

\[ \Rightarrow \theta_2 = \tan^{- 1} \left( \frac{b}{a} \right)\]

\[\therefore \tan \theta_3 = \left| \frac{m_1 - m_3}{1 + m_1 m_3} \right|\]

\[ = \left| \frac{- \left( \frac{a + b}{a - b} \right) + 1}{1 + \frac{a + b}{a - b}} \right|\]

\[ \Rightarrow \tan \theta_3 = \left| \frac{b}{a} \right|\]

\[ \Rightarrow \theta_3 = \tan^{- 1} \left( \frac{b}{a} \right)\]

Here,

\[\theta_2 = \theta_3 \text { and } \theta = 2 \tan^{- 1} \left( \frac{a}{b} \right)\]

Hence, the given lines form an isosceles triangle whose vertical angle is \[2 \tan^{- 1} \left( \frac{a}{b} \right)\].