A) 2
B) 3
C) 1
D) None of these
Correct Answer: C
Solution :
Given, pair of equations is |
\[-x+py-1=0\] ...(i) |
and \[px-y-1=0\] ...(ii) |
On comparing the given equations with standard form i.e. \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\] we get |
\[{{a}_{1}}=-1,\,{{b}_{1}}=p,\,{{c}_{1}}=-1\] |
and \[{{a}_{2}}=p,\,{{b}_{2}}=-1,\,{{c}_{2}}=-1\] |
For parallel lines, |
\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\] |
(iii) |
On taking I and II terms, we get |
\[\frac{-1}{\,p}=\frac{p}{-1}\] |
\[\Rightarrow \,\,\,{{p}^{2}}=1\,\,\Rightarrow \,p=\pm 1\] |
Since, \[p=-1\]does not satisfy the last two terms of Eq. (iii). |
\[\therefore \] p = 1 is the required value. |
Hence, for p = 1, the given system of equations will represent parallel lines. |
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