Question

Using determinant, show that the points \[A\left( {a,b + c} \right)\], \[B\left( {b,c + a} \right)\] and \[C\left( {c,a + b} \right)\] are collinear.

Answer 1

Hint: First, we will use the property that the three points \[\left( {{x_1},{y_1}} \right)\], \[\left( {{x_2},{y_2}} \right)\] and \[\left( {{x_3},{y_3}} \right)\] are collinear if \[\left| {\begin{array}{*{20}{c}}
  1&{{x_1}}&{{y_1}} \\
  1&{{x_2}}&{{y_2}} \\
  1&{{x_3}}&{{y_3}}
\end{array}} \right| = 0\]. Then we will take \[{C_2} \to {C_2} + {C_3}\] and use the rule of determinant that if any of the two rows or columns, the determinant is equal to zero to prove the required result.

Complete step-by-step answer:
We are given that the points are \[A\left( {a,b + c} \right)\], \[B\left( {b,c + a} \right)\] and \[C\left( {c,a + b} \right)\].
We are given that the three points \[\left( {{x_1},{y_1}} \right)\], \[\left( {{x_2},{y_2}} \right)\] and \[\left( {{x_3},{y_3}} \right)\] are collinear if \[\left| {\begin{array}{*{20}{c}}
  1&{{x_1}}&{{y_1}} \\
  1&{{x_2}}&{{y_2}} \\
  1&{{x_3}}&{{y_3}}
\end{array}} \right| = 0\].
So, we have the matrix from the three given point A, B and C, we get
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&a&{b + c} \\
  1&b&{c + a} \\
  1&c&{a + b}
\end{array}} \right|\]
Taking \[{C_2} \to {C_2} + {C_3}\] in the above equation, we get
\[
   \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&{a + b + c}&{b + c} \\
  1&{b + c + a}&{c + a} \\
  1&{c + a + b}&{a + b}
\end{array}} \right| \\
   \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&{a + b + c}&{b + c} \\
  1&{a + b + c}&{c + a} \\
  1&{a + b + c}&{a + b}
\end{array}} \right| \\
 \]
Taking \[a + b + c\] common from the second column of the above equation, we get
\[ \Rightarrow \left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}}
  1&1&{b + c} \\
  1&1&{c + a} \\
  1&1&{a + b}
\end{array}} \right|\]
Using the rule of determinant that if any of the two rows or columns, the determinant is equal to zero.
Since the first and second columns of the above matrix are the same, the determinant is 0.
Hence, the three points \[A\left( {a,b + c} \right)\], \[B\left( {b,c + a} \right)\] and \[C\left( {c,a + b} \right)\] are collinear.


Note: In these types of questions, the key concept is to know the basic properties of matrix and its determinant. This is a simple question once you know that the three points \[\left( {{x_1},{y_1}} \right)\], \[\left( {{x_2},{y_2}} \right)\] and \[\left( {{x_3},{y_3}} \right)\] are collinear if \[\left| {\begin{array}{*{20}{c}}
  1&{{x_1}}&{{y_1}} \\
  1&{{x_2}}&{{y_2}} \\
  1&{{x_3}}&{{y_3}}
\end{array}} \right| = 0\].