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If the lines \(3 x+y-2=0, p x+2 y-3=0\) and \(2 x-y-3=0\) are concurrent, then \(p=\)
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1728 Upvotes
Verified Answer
The correct answer is:
5
It is given that lines
\(3 x+y-2=0, p x+2 y-3=0\) and \(2 x-y-3=0\)
are concurrent, so
\(\begin{array}{lll}
\left|\begin{array}{rrr}
3 & 1 & -2 \\
p & 2 & -3 \\
2 & -1 & -3
\end{array}\right|=0 \\
\Rightarrow 3(-6-3)-1(-3 p+6)-2(-p-4)=0 \\
\Rightarrow -27+3 p-6+2 p+8=0 \\
\Rightarrow 5 p-25=0 \Rightarrow p=5
\end{array}\)
Hence, option (b) is correct.
\(3 x+y-2=0, p x+2 y-3=0\) and \(2 x-y-3=0\)
are concurrent, so
\(\begin{array}{lll}
\left|\begin{array}{rrr}
3 & 1 & -2 \\
p & 2 & -3 \\
2 & -1 & -3
\end{array}\right|=0 \\
\Rightarrow 3(-6-3)-1(-3 p+6)-2(-p-4)=0 \\
\Rightarrow -27+3 p-6+2 p+8=0 \\
\Rightarrow 5 p-25=0 \Rightarrow p=5
\end{array}\)
Hence, option (b) is correct.
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