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Question
Using determinants prove that the points (a, b), (a', b') and (a − a', b − b') are collinear if ab' = a'b.
Solution
\[\begin{vmatrix}a & b & 1 \\ a' & b' & 1 \\ a - a' & b - b' & 1\end{vmatrix}\]
\[ \Rightarrow ∆ = \begin{vmatrix}a & b & 1 \\ a' - a & b' - b & 0 \\ a - a' & b - b' & 1\end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1 \right]\]
\[ \Rightarrow ∆ = \begin{vmatrix}a & b & 1 \\ a' - a & b' - b & 0 \\ - a' & - b' & 0\end{vmatrix} \left[\text{ Applying }R_3 \to R_3 - R_1 \right]\]
\[ \Rightarrow ∆ = \begin{vmatrix}a' - a & b' - b \\ - a' & - b'\end{vmatrix}\]
\[ \Rightarrow ∆ = - b'\left( a' - a \right) + a'\left( b' - b \right)\]
\[ = - b'a' + b'a + a'b' - a'b\]
\[ = b'a - a'b\]
If the points are collinear, then ∆ = 0. So,
ab' − a'b = 0
Thus, ab' = a'b
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