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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Find the distance between parallel lines $ l(x+y)+p=0$ and $l(x+y)-r=0$.

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  • Distance between parallel lines is $d=\large\frac{|c_1-c_2|}{\sqrt{A^2+B^2}}$
The given parallel lines are $l(x+y)+p=0\: and \: l(x+y)-r=0$
Here $A=l, B = l, C_1=p \:and \: C_2=r$.
Hence substituting the values in $d = \large\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$ we get,
$ d = \large\frac{|p-(-r)|}{\sqrt{l^2+l^2}}$$ = \large\frac{|p+r|}{l\sqrt 2}$
Hence the required distance is $ \large\frac{1}{\sqrt 2}$$ \bigg| \large\frac{p+r}{l} \bigg|$ units.
answered May 9, 2014 by thanvigandhi_1
 

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