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If the lines \(3 x+4 y-5=0,2 x+3 y-4=0\) and \(p x+4 y-6=0\) all meet at the same point, then \(p\) is equal to
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The correct answer is:
2
\(\begin{aligned}
3 x+4 y-5 & =0 \quad \ldots (i) \\
2 x+3 y-4 & =0 \quad \ldots (ii) \\
\text {and } \quad p x+4 y-6 & =0 \quad \ldots (iii)
\end{aligned}\)
From Eqs. (i) and (ii), we get
\(x=-1, y=2\)
So, \(\quad-p+8-6=0 \Rightarrow p=2\)
3 x+4 y-5 & =0 \quad \ldots (i) \\
2 x+3 y-4 & =0 \quad \ldots (ii) \\
\text {and } \quad p x+4 y-6 & =0 \quad \ldots (iii)
\end{aligned}\)
From Eqs. (i) and (ii), we get
\(x=-1, y=2\)
So, \(\quad-p+8-6=0 \Rightarrow p=2\)
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