Question

Find the value(s) of $p$ in the following pair of equations: $-x+py=1$ and $px-y=1$, if the pair of equations has no solution.

Answer 1

Step 1: Compute the ratios of coefficients.

Given: pair of linear equations is,

$$\begin{gathered} -x+py=1 \\ px-y=1 \end{gathered}$$

On comparing with $ax+by+c=0$, we get,

$$\begin{gathered} a_1=-1,b_1=p,c_1=-1 \\ {a}_2=p,b_2=-1,c_2=-1 \end{gathered}$$

$$\begin{gathered} \frac{a_1}{a_2}=\frac{-1}{p}, \\ \frac{b_1}{b_2}=\frac{p}{-1}=-p, \\ \frac{c_1}{c_2}=1 \end{gathered}$$

Step 2: Compute the required values.

Since the equations of the lines have no solution i.e., both lines are parallel to each other.

The condition that satisfies parallel lines are:

$$\begin{gathered} \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2} \\ \Rightarrow \frac{-1}{p}=-p\ne 1 \end{gathered}$$

On considering the last two parts, we get

$p\ne -1$

On considering the first two parts, we get

$$\begin{gathered} p^2=1 \\ \Rightarrow p=\pm 1 \end{gathered}$$

Since, $p$can not be equal to $-1$

Therefore, $p=1$.

Hence, the given pair of linear equations are parallel for all real values of $p$ except $-1$.