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 $(\frac{a}{b})$ .

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Solution

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 m₁_,m₂and m₃ be the slopes of the lines (1), (2) and (3), respectively.

Now,

Slope of the first line = m₁ = $- (\frac{a + b}{a - b})$

Slope of the second line = m₂ = $- (\frac{a - b}{a + b})$

Slope of the third line = m₃ = $- 1$

Let $θ_{1}$ be the angle between lines (1) and (2), $θ_{2}$ be the angle between lines (2) and (3) and $θ_{3}$ be the angle between lines (1) and (3).

$∴ \tan θ_{1} = |\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}| = |\frac{- (\frac{a + b}{a - b}) + \frac{a - b}{a + b}}{1 + \frac{a + b}{a - b} \times \frac{a - b}{a + b}}| \Rightarrow \tan θ_{1} = |\frac{2 a b}{a^{2} - b^{2}}| \Rightarrow θ_{1} = \tan^{- 1} |\frac{2 a b}{a^{2} - b^{2}}| = 2 \tan^{- 1} (\frac{a}{b})$

$∴ \tan θ_{2} = |\frac{m_{2} - m_{3}}{1 + m_{2} m_{3}}| = |\frac{- (\frac{a - b}{a + b}) + 1}{1 + \frac{a - b}{a + b}}| \Rightarrow \tan θ_{2} = |\frac{b}{a}| \Rightarrow θ_{2} = \tan^{- 1} (\frac{b}{a})$

$∴ \tan θ_{3} = |\frac{m_{1} - m_{3}}{1 + m_{1} m_{3}}| = |\frac{- (\frac{a + b}{a - b}) + 1}{1 + \frac{a + b}{a - b}}| \Rightarrow \tan θ_{3} = |\frac{b}{a}| \Rightarrow θ_{3} = \tan^{- 1} (\frac{b}{a})$

Here,
$θ_{2} = θ_{3} and θ = 2 \tan^{- 1} (\frac{a}{b})$

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

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