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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Find the values of $k$ for which the line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ is Parallel to the $x$ axis.

$\begin {array} {1 1} (A)\;k=3 & \quad (B)\;k=-3 \\ (C)\;k=\large\frac{1}{3} & \quad (D)\;k=-\large\frac{1}{3} \end {array}$

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  • If two lines are parallel then their slopes are equal.
  • If the given line is parallel to $x$ - axis then its slope is zero.
The given equation of line is
If the given line is parallel to the $x$ - axis , the slope of the given line = slope of the $x$ - axis.
The given line can be written in the form of $y=mx+c$
(i.e.,) $y=\large\frac{(k-3)}{(4-k^2)}$$x+\large\frac{k^2-7k+6}{(4-k^2)}$
Hence the slope of the given line $ = \large\frac{(k-3)}{(4-k^2)}$
Slope of the $x$ - axis = 0
$ \therefore \large\frac{k-3}{4-k^2}=0$
$ \Rightarrow k-3=0$
$ \therefore k=3$
Hence the value of $ k = 3$.
answered May 18, 2014 by thanvigandhi_1

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