Find the angles between the pairs of straight lines:
$(m^{2} - m n) y = (m n + n^{2}) x + n^{3}$ and $(m n + m^{2}) y = (m n - n^{2}) x + m^{3}$

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Solution

From the given equations
$(m^{2} - m n) y = (m n + n^{2}) x + n^{3}$ and $(m n + m^{2}) y = (m n - n^{2}) x + m^{3}$
The slopes are :
$m_{1} = \frac{m n + n^{2}}{m^{2} - m n}$ and $m_{2} = \frac{m n - n^{2}}{m n + m^{2}}$
Angle between straight lines with slopes $m_{1}$ and $m_{2}$ is given by
$tan θ = ∣ ∣ ∣ \frac{m_{2} - m_{1}}{1 + m_{2} m 1} ∣ ∣ ∣$
$\Rightarrow tan θ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ \frac{\frac{m n - n^{2}}{m n + m^{2}} - \frac{m n + n^{2}}{m^{2} - m n}}{1 + \frac{(m n)^{2} - n^{4}}{m^{4} - (m n)^{2}}} ∣ ∣ ∣ ∣ ∣ ∣ ∣$
On simplification we get,
$\Rightarrow tan θ = \frac{4 m^{2} n^{2}}{m^{4} - n^{4}}$

$\Rightarrow θ = {tan}^{- 1} (\frac{4 m^{2} n^{2}}{m^{4} - n^{4}})$

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Similar questions

(m^{2} - m n) y = (m n + n^{2}) x + n^{3}

(m n + m^{2}) y = (m n - n^{2}) x + m^{3}

m > n

Find the angle between each of the following pairs of straight lines:

(i) $3 x + y + 12 = 0$ and $x + 2 y - 1 = 0$

(ii) $3 x - y + 5 = 0$ and $x - 3 y + 1 = 0$

(iii) $3 x + 4 y - 7 = 0$ and $4 x - 3 y + 5 = 0$

(iv) $x - 4 y = 3$ and $6 x - y = 11$

(v) $(m^{2} - m n) y = (m n + n^{2}) x + n^{2}$ and $(m n + m^{2}) y = (m n + n^{2}) x + m^{3}$ .

\frac{m^{2} - n^{2}}{{(m + n)}^{2}} \times \frac{m^{2} + m n + n^{2}}{m^{3} - n^{3}}

\frac{m^{2} - n^{2}}{{(m + n)}^{2}} \times \frac{m^{2} + m n + n^{2}}{m^{3} - n^{3}}

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