Question: The angle between two lines is 30°. If the slope of one of them is 2, find the slope of the other.

Solution:
Given,

Angle between two lines (\theta) = 30°

Slope of one (m_1) = 2

To find: slope of another (m_2) = ?

We know,

tan \theta = \left ( \pm \dfrac{m_1 - m_2}{1 + m_1×m_2} \right )

or, tan 30° = \left ( \pm \dfrac{m_1 - m_2}{1 + m_1×m_2} \right )

or, \dfrac{1}{\sqrt{3}} =  \left ( \pm \dfrac{m_1 - m_2}{1 + m_1×m_2} \right )

or, \dfrac{1 + m_1 × m_2}{\sqrt{3}} = \pm ( m_1 - m_2)

or, \dfrac{1 + 2 × m_2}{\sqrt{3}} = \pm \left ( 2 - m_2 \right )


Taking positive sign;

or, \dfrac{1 + 2×m_2}{\sqrt{3}} = 2 - m_2

or, 1 + 2m_2 = \sqrt{3} ( 2 - m_2)

or, 1 + 2m_2 = 2\sqrt{3} - \sqrt{3}m_2

or, 2m_2 + \sqrt{3}m_2 = 2\sqrt{3} - 1

or, m_2 ( 2 + \sqrt{3}) = 2\sqrt{3} - 1

or, m_2 = \dfrac{2\sqrt{3} - 1}{2 + \sqrt{3}}

or, m_2 = \dfrac{2\sqrt{3} - 1}{2 + \sqrt{3}} × \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}}

or, m_2 = \dfrac{(2\sqrt{3} - 1)(2- \sqrt{3})}{(2)^2 - (\sqrt{3})^2}

or, m_2 = \dfrac{2\sqrt{3}(2 - \sqrt{3}) - 1(2 - \sqrt{3})}{4 - 3}

or, m_2 = \dfrac{4\sqrt{3} - 2×3 - 2 +\sqrt{3}}{1}

\therefore m_2 = 5\sqrt{3} - 8


Taking negative sign,

or, \dfrac{1 + 2 × m_2}{\sqrt{3}} = - \left ( 2 - m_2 \right )

or, \dfrac{1 + 2 × m_2}{\sqrt{3}} = m_2 -2

or, 1 + 2m_2 = \sqrt{3} ( m_2 -2)

or, 1 + 2m_2 = \sqrt{3}m_2 - 2\sqrt{3}

or, 1 + 2\sqrt{3} = \sqrt{3} - 2m_2

or, \sqrt{3} - 2m_2 = 1 + 2\sqrt{3}

or, ( \sqrt{3} - 2) m_2) = 1 + 2\sqrt{3}

or, m_2 = \dfrac{1 + 2\sqrt{3}}{\sqrt{3} - 2}

or, m_2 = \dfrac{1 + 2\sqrt{3}}{\sqrt{3} - 2} × \dfrac{\sqrt{3} + 2}{\sqrt{3} + 2}

or, m_2 = \dfrac{(1 + 2\sqrt{3})(\sqrt{3} +2)}{(\sqrt{3})^2 - 2^2}

or, m_2 = \dfrac{1(\sqrt{3} +2) + 2\sqrt{3}(\sqrt{3} + 2)}{3 -4}

or, m_2 = \dfrac{\sqrt{3} +2 + 2×3 + 4\sqrt{3}}{-1}

or, m_2 = - \left ( 5 \sqrt{3} + 8 \right )

\therefore m_2 = - 5\sqrt{3} - 8

Hence, the slope of the other line is \pm \left ( 5\sqrt{3} + 8} \right )



Related Notes and Solutions:

Here is the website link to all the important formulae of Coordinate Geometry of Class 10.

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