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 $(c,a+b).$

Answer 1

For points $(x_1,y_1)$ $(x_2,y_2)$ $(x_3,y_3)$

Area of $△$ABC is given by

$Area=\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$

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

$Area=\frac{1}{2}|a\times (c+a-a-b)+b\times (a+b-b-c)+c\times (b+c-c-a\left)\right|$

$\Rightarrow Area=\frac{1}{2}|ac-ab+ab-bc+bc-ac|$

$\Rightarrow Area=0$

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