The lines `(p-q)x+(q-r)y+(r-p)=0`
`(q-r)x+(r-p)y+(p-q)=0`
`(r-p)x+(p-q)y+(q-r)=0`
are concurrent

Answer

To determine whether the given lines are concurrent, we can use the condition that the determinant formed by the coefficients of the lines must be equal to zero. The lines given are:

1. \((p-q)x + (q-r)y + (r-p) = 0\)
2. \((q-r)x + (r-p)y + (p-q) = 0\)
3. \((r-p)x + (p-q)y + (q-r) = 0\)

### Step 1: Write the coefficients in a matrix form

The coefficients of the lines can be arranged in a determinant as follows:

\[ \Delta = \begin{vmatrix} p-q & q-r & r-p \\ q-r & r-p & p-q \\ r-p & p-q & q-r \end{vmatrix} \]

### Step 2: Calculate the determinant

To check for concurrency, we need to calculate the determinant \(\Delta\) and set it to zero. We can perform row operations to simplify the determinant.

Let's perform the operation \(R_1 \to R_1 + R_2 + R_3\):

\[ R_1 = (p-q) + (q-r) + (r-p) = 0 \]
So, the first row becomes \(0\).

Now the determinant becomes:

\[ \Delta = \begin{vmatrix} 0 & 0 & 0 \\ q-r & r-p & p-q \\ r-p & p-q & q-r \end{vmatrix} \]

### Step 3: Evaluate the determinant

Since the first row is all zeros, the value of the determinant \(\Delta\) is zero:

\[ \Delta = 0 \]

### Conclusion

Since the determinant is zero, the lines are concurrent.