Question

Prove that \[\left( {a,b + c} \right)\left( {b,c + a} \right)\left( {c,a + b} \right)\] are collinear points.

Answer 1

Hint: We use the concept of collinear points and that three points in general form a triangle and if the points are collinear then the area of the triangle formed by collinear points is equal to zero.
* Three points are said to be collinear if they lie on the same line.
* Area of a triangle formed by points \[({x_1},{y_1}),({x_2},{y_2}),({x_3},{y_3})\] is given by \[\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  {{x_1}}&{{y_1}}&1 \\
  {{x_2}}&{{y_2}}&1 \\
  {{x_3}}&{{y_3}}&1
\end{array}} \right|\]
* Determinant of a matrix \[\left[ {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right] = a(ei - hf) - b(di - fg) + c(dh - eg)\]

Complete step-by-step solution:
We are given three points \[\left( {a,b + c} \right);\left( {b,c + a} \right);\left( {c,a + b} \right)\]
Let us assume three points form a triangle.
Then we can write the area of triangle formed by the three points as \[\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  a&{b + c}&1 \\
  b&{c + a}&1 \\
  c&{a + b}&1
\end{array}} \right|\]
We calculate the determinant value using column transformations
\[ \Rightarrow \]Area of triangle\[ = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  a&{b + c}&1 \\
  b&{c + a}&1 \\
  c&{a + b}&1
\end{array}} \right|\]
Applying column operation \[{C_1} \to {C_1} + {C_2}\] to the determinant
\[ \Rightarrow \]Area of triangle\[ = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  {a + b + c}&{b + c}&1 \\
  {a + b + c}&{c + a}&1 \\
  {a + b + c}&{a + b}&1
\end{array}} \right|\]
Since all elements of first column are same we can take out the same element
\[ \Rightarrow \]Area of triangle\[ = \dfrac{1}{2}(a + b + c)\left| {\begin{array}{*{20}{c}}
  1&{b + c}&1 \\
  1&{c + a}&1 \\
  1&{a + b}&1
\end{array}} \right|\]
Applying column operation \[{C_1} \to {C_1} - {C_3}\] to the determinant
\[ \Rightarrow \]Area of triangle\[ = \dfrac{1}{2}(a + b + c)\left| {\begin{array}{*{20}{c}}
  0&{b + c}&1 \\
  0&{c + a}&1 \\
  0&{a + b}&1
\end{array}} \right|\]
Since one whole column of the determinant is zero, then the value of the determinant is zero.
\[ \Rightarrow \]Area of triangle\[ = \dfrac{1}{2}(a + b + c) \times 0\]
We know when zero multiplied by any other number is always zero
\[ \Rightarrow \]Area of triangle\[ = 0\]
Since the area of the triangle is zero, then the three points which are said to be vertices of the triangle are collinear points.

\[\therefore \]Three points \[\left( {a,b + c} \right)\left( {b,c + a} \right)\left( {c,a + b} \right)\] are collinear points.

Note: Alternate method:
We can also prove the determinant equal to zero by showing the two columns identical to each other. Since we know area of triangle is given by
\[ \Rightarrow \]Area of triangle\[ = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  a&{b + c}&1 \\
  b&{c + a}&1 \\
  c&{a + b}&1
\end{array}} \right|\]
Applying column operation\[{C_1} \to {C_1} + {C_2}\]to the determinant
\[ \Rightarrow \]Area of triangle\[ = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  {a + b + c}&{b + c}&1 \\
  {a + b + c}&{c + a}&1 \\
  {a + b + c}&{a + b}&1
\end{array}} \right|\]
Since all elements of first column are same we can take out the same element
\[ \Rightarrow \]Area of triangle\[ = \dfrac{1}{2}(a + b + c)\left| {\begin{array}{*{20}{c}}
  1&{b + c}&1 \\
  1&{c + a}&1 \\
  1&{a + b}&1
\end{array}} \right|\]
Here first and the third columns are identical to each other, so the value of determinant is zero
\[ \Rightarrow \]Area of triangle\[ = \dfrac{1}{2}(a + b + c) \times 0\]
\[ \Rightarrow \]Area of triangle\[ = 0\]