Question
Determine whether the sets of points are collinear?
(a, b + c), (b, c + a) and (c, a + b)
(a, b + c), (b, c + a) and (c, a + b)
✓
Answer
Let the points be $A(a, b+c), B(b, c+a)$ and $C(c, a+b)$
Area of the triangle $=\frac{1}{2}\left[\left(x_1 y_2+x_2 y_3+x_3 y_1\right)-\left(x_2 y_1+x_3 y_2+x_1 y_3\right)\right]$
$
\begin{aligned}
& =\frac{1}{2}[a(c+a)+b(a+b)+c(b+c)-b(b+c)+c(c+a)+a(a+b)] \\
& =\frac{1}{2}\left[a c+a^2+a b+b^2+b c+c^2-\left(b^2+b c+c^2+a c+a^2+a b\right)\right] \\
& =\frac{1}{2} \times 0 \\
& =0
\end{aligned}
$
Since the area of a triangle is 0 .
$\therefore$ The given points are collinear.
Area of the triangle $=\frac{1}{2}\left[\left(x_1 y_2+x_2 y_3+x_3 y_1\right)-\left(x_2 y_1+x_3 y_2+x_1 y_3\right)\right]$
$
\begin{aligned}
& =\frac{1}{2}[a(c+a)+b(a+b)+c(b+c)-b(b+c)+c(c+a)+a(a+b)] \\
& =\frac{1}{2}\left[a c+a^2+a b+b^2+b c+c^2-\left(b^2+b c+c^2+a c+a^2+a b\right)\right] \\
& =\frac{1}{2} \times 0 \\
& =0
\end{aligned}
$
Since the area of a triangle is 0 .
$\therefore$ The given points are collinear.
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