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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. $ y-2=0$

$\begin{array}{1 1}(A)\;p = 1\: units\: and \: \alpha = 90^{\circ} \\(B)\; p = 2\: units\: and \: \alpha = 60^{\circ} \\(C)\; p = 2\: units\: and \: \alpha = 120^{\circ} \\(D)\;p = 2\: units\: and \: \alpha = 90^{\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.
The given equation is $y-2=0$
This can be written as
$0.x+1.y=2$
$ \sqrt{0^2+1^2}=1$
Hence dividing by 1 on both sides we get
$0.x+1.y=2$
i.e., $ x \cos 90^{\circ}+ y \sin 90^{\circ} = 2$
The above equation can be compared with the normal form of the equation.
$x \cos \alpha + y \sin \alpha = p$
Here $p=2$ and the angle $ \alpha = 90^{\circ}$.
Hence the perpendicular distance is 2 units and the angle $ \alpha = 90^{\circ}$
answered May 7, 2014 by thanvigandhi_1
 

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