Question

Prove that the points $A(a,b+c),B(b,c+a)$ and $C(c,a+b)$ are collinear (By determinant)

Answer 1

Three point are collinear if they lie on same line

$\Rightarrow$They do not form a triangle

$\Rightarrow$Area of triangle=0

We know that ,Area of triangle

$△=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 2\\ x_3& y_3& 3\end{array}\right|$

Here,

$x_1=a,y_1=b+c$

$x_2=b,y_2=c+a$

$x_3=c,y_3=a+b$

$△=\frac{1}{2}\left|\begin{array}{ccc}a& b+c& 1\\ b& c+a& 2\\ c& a+b& 3\end{array}\right|$

Applying $C_1\to C_1+C_2$

$△=\frac{1}{2}\left|\begin{array}{ccc}a+b+c& b+c& 1\\ b+c+a& c+a& 2\\ c+a+b& a+b& 3\end{array}\right|$

Taking $(a+b+c)$ common from $C_1$

$△=\frac{1}{2}(a+b+c)\left|\begin{array}{ccc}1& b+c& 1\\ 1& c+a& 2\\ 1& a+b& 3\end{array}\right|$

Here 1st and 3rd Column are identical

Hence value of determinant is zero

$△=\frac{1}{2}(a+b+c)\times 0=0$

So, $△=0$

Hence points A, B & C are collinear.