EQUATION OF A STRAIGHT LINE IN NORMAL FORM

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Definition

A normal to a line is a line segment drawn from a point perpendicular to the given line.

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Let p be the length of perpendicular form from the origin to a line, which subtends an angle'\alpha 'with the positive direction of x-axis as shown below:

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Let AB be a straight line on coordinate axes. p be the perpendicular distance from the origin to the line at clip_image010 be the angle of perpendicular.

Draw PQ \bot OA

In \Delta POQ,

\begin{array}{l}
\sin \alpha  = \frac{{PQ}}{p},\,\,PQ = p\sin \alpha \\
\cos \alpha  = \frac{{OQ}}{p},\,OQ = p\cos \alpha 
\end{array}

\therefore The coordinates of point P are \left( {p\cos \alpha ,\,p\sin \alpha } \right) .

Slope of OP, {m_{op}} = \frac{{p\sin \alpha  - 0}}{{p\cos \alpha  - 0}}

\begin{array}{l}
 \Rightarrow {m_{op}}\, = \,\tan \alpha \\
\therefore \,\,\,OP\, \bot \,\,AB\\
\therefore \,\,{m_{AB}}\, =  - \frac{1}{{\tan \alpha }}
\end{array}

{m_{AB}} =  - \frac{{\cos \alpha }}{{\sin \alpha }}

clip_image018[1] The equation of the line AB:

 y - p\sin \alpha  =  - \frac{{\cos \alpha }}{{\sin \alpha }}\left( {x - p\cos \alpha } \right)

 \Rightarrow \,y\sin \alpha  - p{\sin ^2}\alpha  =  - x\cos \alpha  + p{\cos ^2}\alpha

 \Rightarrow \,x\cos \alpha  + y\sin \alpha  = p{\cos ^2}\alpha  + p{\sin ^2}\alpha

 \Rightarrow x\cos \alpha  + y\sin \alpha  = p\left( {{{\cos }^2}\alpha  + {{\sin }^2}\alpha } \right)

 \Rightarrow x\cos \alpha  + y\sin \alpha  = p

Therefore, the equation of straight line in normal form is:

x\cos \alpha  + y\sin \alpha  = p

 

Example:

Determine the equation of the line with \alpha  = 135^\circ and the perpendicular distancep = \sqrt 2 from the origin.

The standard equation of the straight line in the normal is:

x\cos \alpha  + y\sin \alpha  = p

\alpha  = 135^\circ ,p = \sqrt 2

Substitute the known values as follows:

 x\cos 135^\circ  + y\sin 135^\circ  = \sqrt 2

 \Rightarrow \,x\left( { - \frac{1}{{\sqrt 2 }}} \right) + y\left( {\frac{1}{{\sqrt 2 }}} \right) = \sqrt 2

 \Rightarrow \,\frac{{ - x + y}}{{\sqrt 2 }} = \sqrt 2

 \Rightarrow \, - x + y = 2

 

Question 1:

Determine the equation of the line whose perpendicular distance from the origin is 8 units and makes an angle of 90o in positive direction of x-axis.

Solution 1:

The standard equation of the straight line in the normal is:

x\cos \alpha  + y\sin \alpha  = p

\alpha  = 90^\circ ,p = 8

Substitute the known values as follows:

\begin{array}{c}
x\cos 90^\circ  + y\sin 90^\circ  = 8\\
x\left( 0 \right) + y\left( 1 \right) = 8\\
y - 8 = 0
\end{array}

The required equation of a straight line is{y - 8 = 0}.

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