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Question

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 (x1,y1) (x2,y2) (x3,y3)

Area of ABC is given by

Area=12|x1(y2y3)+x2(y3y1)+x3(y1y2)|

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

Area=12|a×(c+aab)+b×(a+bbc)+c×(b+cca)|

Area=12|acab+abbc+bcac|

Area=0

All 3 points are in a straight line.

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