Question

prove that the following points are colinear.
➜( a,b+c),b,c+a),(c,a+b)​

Answer 1

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

Given points are (a, b + c), (b, c + a) and (c, a + b).

Now, we have to prove that points (a, b + c), (b, c + a) and (c, a + b).

So, in order to prove that the points (a, b + c), (b, c + a) and (c, a + b) are collinear, we have to prove that area of triangle is 0.

Now, Consider area of triangle,

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

$\boxed{ \rm{ \:OP \: C_2 \: \to \: C_2 + C_1 \: }}$

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

Take out a + b + c common from column 2, we get

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

Since, second and third column are identical, so determinant value is 0.

So,

$\rm \: \triangle \: = \: \dfrac{1}{2}\: (a + b + c) \times 0$

$\rm\implies \:\triangle = 0$

$\color{blue}\rm\implies \:(a, b + c), (b, c + a) \: and \: (c, a + b) \: are \: collinear$

$\rule{190pt}{2pt}$

Additional Information :-

1. The determinant value remains unaltered if rows and columns are interchanged.

2. The determinant value is 0, if two rows or columns are identical.

3. The determinant value is multiplied by - 1, if successive rows or columns are interchanged.

4. The determinant value remains unaltered if rows or columns are added or subtracted.