Question

If three lines \[ax+{{a}^{2}}y+1=0,bx+{{b}^{2}}y+1=0\] and \[cx+{{c}^{2}}y+1=0\] are concurrent, show that at least two of three constants a, b and c are equal.

Answer 1

Hint: From a \[3\times 3\] square matrix using the coefficients of x, y and the constant terms of the given equations. Take the determinant of this matrix and equate it with 0. Now, perform dome row operations and expand the determinant. Use the algebraic identity \[{{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)\] to get the expression in the form of \[\left( a-b \right)\left( b-c \right)\left( c-a \right)=0\] and substitute each term equal to 0 one by one to get the required proof.

Complete step by step answer:
Here, we have been provided with three concurrent lines \[ax+{{a}^{2}}y+1=0,bx+{{b}^{2}}y+1=0\] and \[cx+{{c}^{2}}y+1=0.\] We have to prove that either a = b, b = c or c = a. Now, we know that when three lines are given by the equations \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0,{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\] and \[{{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}=0\] are concurrent then the determinant of the matrix formed by the coefficients of x, y and the constant terms equals 0. Therefore, we have,
\[\left| \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|=0\]
So, applying the similar condition for the three lines given in the question, we get,
\[\Rightarrow \left| \begin{matrix}
   a & {{a}^{2}} & 1 \\
   b & {{b}^{2}} & 1 \\
   c & {{c}^{2}} & 1 \\
\end{matrix} \right|=0\]
Now, performing the raw operation, \[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}\] and \[{{R}_{2}}\to {{R}_{2}}-{{R}_{3}},\] we get,
\[\Rightarrow \left| \begin{matrix}
   a-b & {{a}^{2}}-{{b}^{2}} & 0 \\
   b-c & {{b}^{2}}-{{c}^{2}} & 0 \\
   c & {{c}^{2}} & 1 \\
\end{matrix} \right|=0\]
Using the algebraic identity, \[{{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right),\] we get,
\[\Rightarrow \left| \begin{matrix}
   a-b & \left( a-b \right)\left( a+b \right) & 0 \\
   b-c & \left( b-c \right)\left( b+c \right) & 0 \\
   c & {{c}^{2}} & 1 \\
\end{matrix} \right|=0\]
Taking (a – b) common from the first row and (b – c) common from the second row, we get,
\[\Rightarrow \left| \begin{matrix}
   1 & \left( a+b \right) & 0 \\
   1 & \left( b+c \right) & 0 \\
   c & {{c}^{2}} & 1 \\
\end{matrix} \right|=0\]
Now, expanding the simplified determinant, we get,
\[\Rightarrow \left( a-b \right)\left( b-c \right)\times \left[ 1\times \left\{ \left( b+c \right)\times 1-{{c}^{2}}\times 0 \right\}-\left( a+b \right)\times \left\{ 1\times 1-c\times 0 \right\}+0\times \left\{ 1\times {{c}^{2}}-c\times \left( b+c \right) \right\} \right]=0\]
\[\Rightarrow \left( a-b \right)\left( b-c \right)\times \left[ \left( b+c \right)-\left( a+b \right) \right]=0\]
\[\Rightarrow \left( a-b \right)\left( b-c \right)\times \left[ b+c-a-b \right]=0\]
Cancelling the like terms, we get,
\[\Rightarrow \left( a-b \right)\left( b-c \right)\times \left( c-a \right)=0\]
\[\Rightarrow \left( a-b \right)\left( b-c \right)\left( c-a \right)=0\]
Substituting each term equal to 0, we get,
\[\Rightarrow \left( a-b \right)=0;\left( b-c \right)=0;\left( c-a \right)=0\]
\[\Rightarrow a=b;b=c;c=a\]
Hence, it is proved that at least two of the three constants a, b, c are equal.

Note:
 One must remember the condition of three lines to be concurrent to solve the above question. Actually concurrent lines are defined as a set of lines passing through a common point which means these sets of lines must have a solution and hence the condition of determinant arises. Note that we cannot expand the determinant directly at the first step because this may lead us to some tough calculations that are the reason why row and column operations are performed. Note that these operations do not change the value of the determinant.