Find the value(s) of $ p $ for the following pair of equations:
$ -x+p y=1 $ and $ p x-y=1 $,
if the pair of equations has no solution.
Given:
Given pair of linear equations is:
\( -x+p y=1 \) and \( p x-y=1 \).
To do:
We have to find the value of $p$ if the given system of equations has no solution.
Solution:
Comparing the given pair of linear equations with the standard form of linear equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$, we get,
$a_1=-1, b_1=p$ and $c_1=-1$
$a_2=p, b_2=-1$ and $c_2=-1$
A system of equations has no solution if the lines are parallel to each other.
Here,
$\frac{a_1}{a_2}=\frac{-1}{p}$
$\frac{b_1}{b_2}=\frac{p}{-1}$
$\frac{c_1}{c_2}=\frac{-1}{-1}=1$
Therefore,
$\frac{a_1}{a_2}=\frac{b_1}{b_2}$
$\frac{-1}{p}=-p$
$-p^2=-1$
$p^2=1$
$p=\sqrt1$
$p=\pm 1$
$\frac{a_1}{a_2}≠\frac{c_1}{c_2}$
$\frac{-1}{p}≠1$
$p≠-1$
This implies,
$p=1$
Therefore, the value of $p$ is 1.
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