Find the angles between the pairs of straight lines:
$y = 3 x + 7$ and $3 y - x = 8$

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Solution

The given straight lines are $y = 3 x + 7$ and $3 y - x = 8$
As $m_{1} = 3$ & $m_{2} = \frac{1}{3}$ .................. $(1)$
$tan θ = ∣ ∣ ∣ \frac{m_{2} - m_{1}}{1 + m_{2} m 1} ∣ ∣ ∣$
Substitute one in above equation, we get

$tan θ = ∣ ∣ ∣ ∣ ∣ ∣ \frac{\frac{1}{3} - 3}{1 + 1} ∣ ∣ ∣ ∣ ∣ ∣$
$\Rightarrow tan θ = \frac{4}{3}$
$\Rightarrow θ = {tan}^{- 1} (\frac{4}{3})$

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Similar questions

x - y \sqrt{3} = 5

\sqrt{3} x + y = 7

y = (2 - \sqrt{3}) x + 5

y = (2 + \sqrt{3}) x - 7

\sqrt{3} x + y = 1

x + \sqrt{3} y = 1

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

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