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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 passing through the origin.

$\begin {array} {1 1} (A)\;k=-6 \: or \: k = 1 & \quad (B)\;k=6 \: or \: k =-1 \\ (C)\;k= -6 \: or \: k =-1 & \quad (D)\;k = 6 \: or \: k =1 \end {array}$

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  • If a line passes through the origin, then the point (0,0) satisfies the given equation of the line.
Given equation of the line is
$(k-3)x-(4-k^2)y+k^2-7k+6=0$
Since it passes through (0,0)
$(k-3)(0)-(4-k^2)0+k^2-7k+6=0$
(i.e.,) $k ^2-7k+6=0$
On factorizing we get,
$(k-6)(k-1)=0$
Hence k = 6 or k = 1
answered May 18, 2014 by thanvigandhi_1
 

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