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If the lines $4 x+3 y-1=0, x-y+5=0$ and $k x+5 y-3=0$ are concurrent then $k$ is equal to
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Verified Answer
The correct answer is:
6
We have,
$$
\begin{array}{r}
4 x+3 y-1=0 \\
x-y+5=0 \\
k x+5 y-3=0
\end{array}
$$
are concurrent lines :
$$
\begin{array}{rrrr}
& & \left|\begin{array}{rrr}
4 & 3 & -1 \\
1 & -1 & 5 \\
k & 5 & -3
\end{array}\right|=0 \\
\Rightarrow & & 4(3-25)-3(-3-5 k)-1(5+k)=0 \\
\Rightarrow & & -88+9+15 k-5-k=0 \\
& 14 k=84 \Rightarrow k & =6
\end{array}
$$
$$
\begin{array}{r}
4 x+3 y-1=0 \\
x-y+5=0 \\
k x+5 y-3=0
\end{array}
$$
are concurrent lines :
$$
\begin{array}{rrrr}
& & \left|\begin{array}{rrr}
4 & 3 & -1 \\
1 & -1 & 5 \\
k & 5 & -3
\end{array}\right|=0 \\
\Rightarrow & & 4(3-25)-3(-3-5 k)-1(5+k)=0 \\
\Rightarrow & & -88+9+15 k-5-k=0 \\
& 14 k=84 \Rightarrow k & =6
\end{array}
$$
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