Prove (by showing that the area of the triangle formed by them is zero) that the following set of three points are in a straight line
$(a, b + c), (b, c + a)$ and $(c, a + b) .$

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Solution

For points $(x_{1}, y_{1})$ $(x_{2}, y_{2})$ $(x_{3}, y_{3})$

Area of $△$ ABC is given by

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

Now, for the given points $(a, b + c), (b, c + a)$ and $(c, a + b) .$

$A r e a = \frac{1}{2} | a \times (c + a - a - b) + b \times (a + b - b - c) + c \times (b + c - c - a) |$

$\Rightarrow A r e a = \frac{1}{2} | a c - a b + a b - b c + b c - a c |$

$\Rightarrow A r e a = 0$

$\Rightarrow$ All 3 points are in a straight line.

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Using the concept of slope, prove that the points (a, b +c), (b,c+a) and (c, a+b) are in a straight line.

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(c - a) x + (a - b) y = b - c

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