Question: The angle between two lines is 45°. If the slope of one of them is $\dfrac{1}{2}$, find the slope of the other.

Solution:
Given,

Angle between two lines ($\theta$) = 45°

Slope of one ($m_1$) = $\dfrac{1}{2}$

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 45° = \left ( \pm \dfrac{m_1 - m_2}{1 + m_1×m_2} \right ) $

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

$or, 1 + m_1 × m_2 = \pm ( m_1 - m_2)$

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


Taking positive sign;

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

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

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

$or, 2 + m_2 = 1 - 2m_2$

$or, 2m_2 + m_2 = 1-2$

$or, 3m_2 = -1$

$\therefore m_2 = - \dfrac{1}{3}$


Taking negative sign,

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

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

$or, 2 + m_2 = - ( 1 - 2m_2)$

$or, 2 + m_2 = -1 + 2 m_2$

$or, 2m_2 - m_2 = 2 + 1$

$\therefore m_2 = 3$

Hence, the slope of the other line is either $3$ or $- \dfrac{1}{3}$.



Related Notes and Solutions:

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

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