The value of ' \(k\) ' for which the equation \(x^2-4 x y-y^2+6 x+2 y+k=0\) represents a pair of straight…
The value of ' \(k\) ' for which the equation \(x^2-4 x y-y^2+6 x+2 y+k=0\) represents a pair of straight lines is equal to ........
- \(\frac{4}{5}\)
- \(\frac{-3}{5}\)
- \(\frac{-4}{5}\)
- \(\frac{3}{5}\)
Solution
\(\begin{aligned}
& x^2-4 x y-y^2+6 x+2 y+k=0 \quad \ldots (i) \\
& a=1,2 h=-4, b=-1,2 g=6,2 f=2, c=k
\end{aligned}\)
Eq. (i) Represents a pair of straight line
\(\begin{aligned}
& \Delta=0 \\
& \Rightarrow a b c+2 f g h-a f^2-b g^2-c h^2=0 \\
&(\mathrm{l})(-1)(k)+2(1)(3)(-2)-1(1)^2+1(3)^2-k(-2)^2=0 \\
&-k-12-1+9-4 k=0 \\
&-5 k-4=0 \\
& k=\frac{-4}{5}
\end{aligned}\)
Hence, option (c) is correct.
Asked in: AP EAMCET 2020 (18 Sep Shift 2)
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