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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Reduce the following equation into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive $x$ - axis. $ x-y=4$

$\begin{array}{1 1}(A)\;p=\sqrt 2 \: units ; \alpha = 315^{\circ} \\(B)\; p= 2 \: units ; \alpha = 45^{\circ} \\(C)\; p=2\sqrt 2 \: units ; \alpha = 315^{\circ} \\(D)\;p=2\sqrt 2 \: units ; \alpha = 45^{\circ} \end{array} $

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  • Equation of a line in the normal form is $ x \cos \alpha + y \sin \alpha=p$ where p is the perpendicular distance from the origin and $ \alpha$ is the angle between the perpendicular and the positive $x$ - axis.
Given equation is $ x-y=4$
This can be written as
$1.x+(-1).y=4$
$ \sqrt{1^2+(-1)^2}=\sqrt 2$
dividing on both sides by $\sqrt 2$ we get,
$ \large\frac{1}{\sqrt 2}$$x+ \bigg( - \large\frac{1}{\sqrt 2} \bigg)$$y=\large\frac{4}{\sqrt 2}$
$ \Rightarrow x \cos \bigg( 2\pi - \large\frac{\pi}{4} \bigg)$$ + y \sin \bigg( 2\pi - \large\frac{\pi}{4} \bigg)$$= 2 \sqrt 2$
i.e., $x \cos 315^{\circ}+y \sin 315^{\circ}= 2\sqrt 2$----------(1)
This is of the form
$ x \cos \alpha + y \sin \alpha=p$----------(2)
Comparing the two equations we get,
$ p = 2\sqrt 2 \: and \: \alpha = 315^{\circ}$
Hence the perpendicular distance is $2\sqrt 2$ units and angle = $315^{\circ}$
answered May 7, 2014 by thanvigandhi_1
 

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