Question

show that these points are collinear​

Answer 1

Answer:

Ummm I'm confused. Please tell me what is collinear.

Answer 2

$\large\underline{\sf{Solution-}}$

We have to prove that, the points (a, b + c), (b, c + a) and (c, a + b) are collinear.

So, its enough to show that area of triangle formed by these coordinates is 0.

Consider area of triangle formed by these coordinates,

$\bf :\longmapsto\:Area_{(\triangle ABC)}$

$\: \rm \: = \: \: \dfrac{1}{2} \: \begin{gathered}\sf \left | \begin{array}{ccc}a&b + c&1\\b&c + a&1\\c&a + b&1\end{array}\right | \end{gathered}$

$\red{\rm :\longmapsto\:OP \: C_2 \: \to \: C_2 \: + \: C_1}$

$\: \rm \: = \: \: \dfrac{1}{2} \: \begin{gathered}\sf \left | \begin{array}{ccc}a&a + b + c&1\\b&b + c + a&1\\c&c + a + b&1\end{array}\right | \end{gathered}$

Take out a + b + c common from Second column, we get

$\: \rm \: = \: \dfrac{1}{2} \: (a + b + c) \: \begin{gathered}\sf \left | \begin{array}{ccc}a&1&1\\b&1&1\\c&1&1\end{array}\right | \end{gathered}$

$\rm \: \: = \: 0 \: \: \: \: \: \: \: \{ \because \: C_2 \: and \: C_3 \: are \: identical \: \}$

[ If any two row or column are identical, determinant value is 0. ]

$\bf :\longmapsto\:Area_{(\triangle ABC)} = 0$

Hence,

$\underbrace{ \boxed{ \sf \: (a, b + c), (b, c + a) \: and \: (c, a + b) \: are \: collinear}}$

Additional Information :-

1. The determinant remains unaltered if its rows are changed into columns and the columns into rows.

2. If all the elements of a row (or column) are zero, then the determinant is zero.

3. If the all elements of a row (or column) are proportional (identical) to the elements of some other row (or column), then the determinant is zero.

4. The interchange of any two rows (or columns) of the determinant changes its sign.

5. If all the elements of a determinant above or below the main diagonal consist of zeros, then the determinant is equal to the product of diagonal elements