Prove that the points $(a b,), (a_{1}, b_{1})$ and $(a - a_{1}, b - b_{2})$ are collinear if $a b_{1} = a_{1} b$

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Solution

Consider the following points $A (a, b), B (a 1, b 1), C (a - a 1, b - b 1)$

Since the given points are collinear, we have $a r e a (△ A B C) = 0$

First find the area of $a r e a (△ A B C)$ as follows:

$a r e a (△ A B C) = \frac{1}{2} | x_{1} (y_{1} - y_{3}) + x 1 (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$= | a (b_{1} - (b - b_{1})) + a_{1} ((b - b_{1}) - b) + (a - a_{1}) (b - b_{1}) |$

$= | a (b_{1} - b + b_{1}) + a_{1} (b - b_{1} - b) + a (b - b_{1}) - a 1 (b - b_{1}) |$

$= | - a b - a_{1} b_{1} + a b - a b_{1} + a_{1} b + a_{1} b_{1} |$

$= | - (a b_{1} - a_{1} b) |$

$= (a b_{1} - a_{1} b)$

This gives, $a b_{1} - a_{1} b = 0$

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