Prove that 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.

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Solution

The lines will be concurrent if
$∣ ∣ ∣ ∣ \begin{matrix} p - q & q - r & r - p q - r & r - p & p - q r - p & p - q & q - r \end{matrix} ∣ ∣ ∣ ∣ = 0$
Adding column no, $2$ and $3$ to column no. $1$ , we get
$∣ ∣ ∣ ∣ \begin{matrix} 0 & q - r & r - p 0 & r - p & p - q 0 & p - q & q - r \end{matrix} ∣ ∣ ∣ ∣ = 0$

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