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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Show that the tangent of an angle between the lines $ \large\frac{x}{a}$$+\large\frac{y}{b}$$=1$ and $ \large\frac{x}{a}$$-\large\frac{y}{b}$$=1$ is $ \large\frac{2ab}{a^2-b^2}$

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  • Angle between two lines is $ \theta = \tan^{-1} \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$ where $m_1$ and $m_2$ are the slopes of the two lines.
Step 1:
the equation of the given lines are :
$ \large\frac{x}{a}$$+\large\frac{y}{b}$$=1$-----(1)
$ \large\frac{x}{a}$$-\large\frac{y}{b}$$=1$--------(2)
This can be written as
$ bx+ay=ab$
or $y=-\large\frac{bx}{a}$$+b$----------(3)
$bx-ay=ab$
or $ y = \large\frac{b}{a}$$x-b$--------(4)
Hence the slopes of the lines (1) and (2) are $ -\large\frac{b}{a}$ and $ \large\frac{b}{a}$ respectively.
Step 2 :
The angle between the lines is
$ \theta = \tan^{-1} \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
Substituting for $m_1$ and $m_2$ we get,
$ \theta = \tan^{-1} \bigg| \large\frac{ -\Large\frac{b}{a}-\Large\frac{b}{a}}{1+\Large\frac{b}{a} \times \large\frac{b}{a}} \bigg|$
$ = \tan^{-1} \bigg| \large\frac{-2\Large\frac{b}{a}}{\Large\frac{a^2+b^2}{a^2}} \bigg|$
$ \therefore \tan \theta = \large\frac{2ab}{a^2+b^2}$
Hence proved.
answered Jun 30, 2014 by thanvigandhi_1
 

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