Prove that the straight line $(a + b) x + (a - b) y = 2 a b, (a - b) x + (a + b) y = 2 a b$ and $x + y = 0$ form an isosceles triangle whose vertical angle is $2 t a n^{- 1} (\frac{a}{b})$ .

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Solution

Given lines
$(a + b) x + (a - b) y - 2 a b = 0 - - - - (1), (a - b) x + (a + b) y - 2 a b = 0 - - - - (2)$ and $x + y = 0 - - - - - (3)$
On comparing above eq (1) (2) and (3) with $y = m x + c$ , we get
$m_{1} = - \frac{a + b}{a - b}$
$m_{2} = - \frac{a - b}{a + b}$ and $m_{3} = - 1$
Angle between (1) and (3) by formula
$tan α = ∣ ∣ ∣ \frac{m_{1} - m_{3}}{1 + m_{1} m_{3}} ∣ ∣ ∣$

$tan α = ∣ ∣ ∣ ∣ ∣ ∣ \frac{- \frac{a + b}{a - b} + 1}{1 + \frac{a + b}{a - b}} ∣ ∣ ∣ ∣ ∣ ∣$

$tan α = ∣ ∣ ∣ \frac{- a - b + a - b}{a - b + a + b} ∣ ∣ ∣$

$tan α = ∣ ∣ ∣ \frac{- 2 b}{2 a} ∣ ∣ ∣$

$tan α = \frac{b}{a}$

Angle between (2) and (3) by formula
$tan β = ∣ ∣ ∣ \frac{m_{2} - m_{3}}{1 + m_{2} m_{3}} ∣ ∣ ∣$

$tan β = ∣ ∣ ∣ ∣ ∣ ∣ \frac{- \frac{a - b}{a + b} + 1}{1 + \frac{a - b}{a + b}} ∣ ∣ ∣ ∣ ∣ ∣$

$tan β = ∣ ∣ ∣ \frac{- a + b + a + b}{a + b + a - b} ∣ ∣ ∣$

$tan β = ∣ ∣ ∣ \frac{2 b}{2 a} ∣ ∣ ∣$

$tan β = \frac{b}{a}$

Here $tan β = tan α$
Hence traingle is isosceles triangle and the vertical angle is $π - 2 {tan}^{- 1} \frac{b}{a} = 2 {tan}^{- 1} \frac{a}{b}$

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Similar questions

Prove that the straight line $(a + b) x + (a - b) y = 2 a b, (a - b) x + (a + b) y = 2 a b$ and $x + y = 0$ form an isosceles triangle whose vertical angle is $2 t a n^{- 1} (\frac{a}{b})$ .

Q. Prove that the lines:

(a + b) x + (a - b) y - 2 a b = 0....... (1)

(a - b) x + (a + b) y - 2 a b = 0....... (2)

and

x + y = 0...... (3)

form an isosceles triangle whose vertical angle is

2 {tan}^{- 1} (a / b)

Q. 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})

Q. The lines

(a + b) x + (a - b) y - 2 a b = 0, (a - b) x + (a + b) y - 2 a b = 0

and

x + y = 0

forms an isosceles triangle whose vertical angle is

Q. Find the acute angle between the lines,

(a + b) x + (a - b) y = 2 a b, (a - b) x + (a + b) y = 2 a b

where a > b.

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Prove that the straight line (a+b)x+(a−b)y=2 ab,(a−b)x+(a+b)y=2ab and x+y=0 form an isosceles triangle whose vertical angle is 2tan−1(ab).

Prove that the straight line $(a + b) x + (a - b) y = 2 a b, (a - b) x + (a + b) y = 2 a b$ and $x + y = 0$ form an isosceles triangle whose vertical angle is $2 t a n^{- 1} (\frac{a}{b})$ .