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If the straight lines $2 x-y+1=0$, $4 x+y+2=0$ and $x+y-k=0$ are concurrent, then $k$ equals
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Verified Answer
The correct answer is:
$\frac{-1}{2}$
Given lines,
$$
\begin{array}{r}
2 x-y+1=0 \\
4 x+y+2=0 \\
x+y-k=0
\end{array}
$$
are concurrent.
$$
\begin{aligned}
& \therefore & & \left|\begin{array}{rrc}
2 & -1 & 1 \\
4 & 1 & 2 \\
1 & 1 & -k
\end{array}\right|=0 \\
& & 2(-k-2)+1(-4 k-2)+1(4-1) & =0 \\
\Rightarrow & & -2 k-4-4 k-2+3 & =0 \\
\Rightarrow & & -6 k & =3 \\
\Rightarrow & & k & =-1 / 2
\end{aligned}
$$
$$
\begin{array}{r}
2 x-y+1=0 \\
4 x+y+2=0 \\
x+y-k=0
\end{array}
$$
are concurrent.
$$
\begin{aligned}
& \therefore & & \left|\begin{array}{rrc}
2 & -1 & 1 \\
4 & 1 & 2 \\
1 & 1 & -k
\end{array}\right|=0 \\
& & 2(-k-2)+1(-4 k-2)+1(4-1) & =0 \\
\Rightarrow & & -2 k-4-4 k-2+3 & =0 \\
\Rightarrow & & -6 k & =3 \\
\Rightarrow & & k & =-1 / 2
\end{aligned}
$$
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