Question

Derive the formula of angle between two straight lines.
Don't spam....​

Answer 1

Answer:

Answer for the given problem is given

Step-by-step explanation:

Answer 2

Proof

Let l₁ and l₂ be two non Perpendicular lines , neither of which is parallel to Y - axis

Let m₁ and m₂ be the slopes of two Given lines l₁ and l₂

respectively . Let θ₁ and θ₂ the inclination of these lines

$\sf \therefore \: \: \: m_1 = \tan \theta_1 \: \: \: \: and \: \: m_2 = \tan \theta_2$

$\sf \to \: let \: \theta \: \: and \: \: \pi - \theta \: \: be \: the \: \: angles \: \: between \: \: the \: lines \: \bigg( \theta \not = \dfrac{\pi}{2} \bigg)$

Then

$\sf\to \: \theta_2 = \theta + \theta_1 \: \: \: \: or \: \: \: \theta = \theta _2+ \theta_1$

$\sf \to \: tan \theta \: = tan( \theta_2 - \theta_1)$

$\sf \to \: \tan \theta \: = \bigg( \dfrac{tan \theta_2 - tan \theta_1}{1 + tan \theta_2 \tan \theta_1} \bigg) = \bigg( \dfrac{m_2 - m_1}{1 + m_1m_2} \bigg) \: \: \: \: \: \: \: \: .....(i)$

$\sf \to \: tan( \pi - \theta) = - tan = - \bigg(\dfrac{m_2 - m_1}{1 + m_1m_2} \bigg) \: \: \: \: \: \: \: .....(ii)$

From eqs (i) and (ii) the angle between two lines of slopes

m₁ and m₂ is given by

$\sf \to \: tan \theta \: = \pm \bigg( \dfrac{m_2 - m_1}{1 + m_1m_2} \bigg)$

$\sf \to \theta \: = \tan ^{ - 1} \bigg \{ \pm \bigg(\dfrac{m_2 - m_1}{1 + m_1m_2} \bigg) \bigg\}$

Hence the acute angle between the line l₁ and l₂ is given by

$\sf \to \: \theta = \tan^{ - 1} \bigg|\dfrac{m_2 - m_1}{1 + m_1m_2} \bigg|$