Question

Assertion (A) The points $A(a, b+c), B(b, c+a)$ and $C(c, a+b)$ are collinear.
Reason (R) Area of a triangle with three collinear points is zero.

• A. Both A and R are correct; R is the correct explanation of $A$
• B. Both $A$ and $R$ are correct; $R$ is not the correct explanation of $A$
• C. A is correct; R is incorrect
• D. $R$ is correct; $A$ is incorrect

Answer 1

Answer: (A)

We know that, the area of triangle with three collinear points is zero.
Now, consider
$\text { Area of } \triangle A B C=\frac{1}{2}\begin{vmatrix}a & b+c & 1 \\b & c+a & 1 \\c & a+b & 1\end{vmatrix}$
$=\frac{1}{2} \mid a\{(c+a) \times 1-(a+b) \times 1\}-(b+c)\{b \times 1-1 \times c\} $
$ +1\{b \times(a+b)-(c+a) \times c\} \mid $
$ =\frac{1}{2} \mid a(c+a-a-b)-(b+c)(b-c)$
$ +1\left(a b+b^2-c^2-a c\right) \mid $
$ =\frac{1}{2}\left|a c-a b-b^2+c^2+a b+b^2-c^2-a c\right|=\frac{1}{2} \times 0=0$
Since, area of $\triangle A B C=0$.
Hence, points $A(a, b+c), B(b, c+a)$ and $C(c, a+b)$ are collinear.