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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Choose the correct answer from the given four options. The distance between the lines $ y = mx + c_1$ and $ y = mx + c_2$ is

$\begin {array} {1 1} (A)\;\large\frac{c_1-c_2}{\sqrt{m^2+1}} & \quad (B)\;\large\frac{|c_1-c_2|}{\sqrt{1+m^2}} \\ (C)\;\large\frac{c_2-c_1}{\sqrt {1+m^2}} & \quad (D)\;0 \end {array}$

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  • Distance between parallel lines is $ d = \bigg| \large\frac{c_1-c_2}{\sqrt{A^2+B^2}} \bigg|$
Step 1 :
The equation of the given lines are
$ y = mx+c_1$
$ \Rightarrow mx-y+c_1=0$ $ \therefore $ slope of this line is m and $y=mx+c_2$
$ \Rightarrow mx-y+c_2=0 \therefore $ slope of this line is m.
Since they have same slopes, the lines are parallel.
$ \therefore d = \bigg| \large\frac{c_1-c_2}{\sqrt{m^2+(-1)^2}} \bigg|$
$ = \bigg| \large\frac{ c_1-c_2}{\sqrt{1+m^2}} \bigg|$
Hence 'B' is the correct answer.
answered Jul 3, 2014 by thanvigandhi_1

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