Question

If $ab + bc + ca = 0,$ show that the lines $\dfrac{x}{a} + \dfrac{y}{b} = \dfrac{1}{c},\;\dfrac{x}{b} + \dfrac{y}{c} = \dfrac{1}{a}\;{\text{and}}\;\dfrac{x}{c} + \dfrac{y}{a} = \dfrac{1}{b}$ are concurrent.

Answer 1

Hint:
To prove the given condition for concurrency of the given three lines, first of all make or express the lines in matrix form and then use the property that if three lines are concurrent or have a common point and their coefficients are linearly independent then the determinant of their matrix should be equals to zero. So equate the determinant with zero and then prove the given condition is required for the determinant to be zero and eventually the three lines to be concurrent.

Complete step by step solution:
In order to prove that the given condition $ab + bc + ca = 0,$ is required for the lines $\dfrac{x}{a} + \dfrac{y}{b} = \dfrac{1}{c},\;\dfrac{x}{b} + \dfrac{y}{c} = \dfrac{1}{a}\;{\text{and}}\;\dfrac{x}{c} + \dfrac{y}{a} = \dfrac{1}{b}$ to be concurrent, we will first simplify the lines in a way such that all its terms will only be on the left hand side and only zero is on right hand side as follows
$
   \Rightarrow \dfrac{x}{a} + \dfrac{y}{b} = \dfrac{1}{c},\;\dfrac{x}{b} + \dfrac{y}{c} = \dfrac{1}{a}\;{\text{and}}\;\dfrac{x}{c} + \dfrac{y}{a} = \dfrac{1}{b} \\
   \Rightarrow \dfrac{x}{a} + \dfrac{y}{b} - \dfrac{1}{c} = 0,\;\dfrac{x}{b} + \dfrac{y}{c} - \dfrac{1}{a} = 0\;{\text{and}}\;\dfrac{x}{c} + \dfrac{y}{a} - \dfrac{1}{b} = 0 \\
 $
Now, expressing the coefficients of these lines in matrix form, we will get
$\left( {\begin{array}{*{20}{c}}
  {\dfrac{1}{a}}&{\dfrac{1}{b}}&{ - \dfrac{1}{c}} \\
  {\dfrac{1}{b}}&{\dfrac{1}{c}}&{ - \dfrac{1}{a}} \\
  {\dfrac{1}{c}}&{\dfrac{1}{a}}&{ - \dfrac{1}{b}}
\end{array}} \right)$
We know that if three lines are concurrent or have a common point and their coefficients are linearly independent then the determinant of their matrix should be equals to zero,
So finding and equating determinant of the above matrix, we will get
$
   \Rightarrow \det \left( {\begin{array}{*{20}{c}}
  {\dfrac{1}{a}}&{\dfrac{1}{b}}&{ - \dfrac{1}{c}} \\
  {\dfrac{1}{b}}&{\dfrac{1}{c}}&{ - \dfrac{1}{a}} \\
  {\dfrac{1}{c}}&{\dfrac{1}{a}}&{ - \dfrac{1}{b}}
\end{array}} \right) = 0 \\
   \Rightarrow \dfrac{1}{{{a^3}}} + \dfrac{1}{{{b^3}}} + \dfrac{1}{{{c^3}}} - \dfrac{3}{{abc}} = 0 \\
 $
Simplifying the equation, we will get
$ \Rightarrow \dfrac{{(ab + ac + bc)\left( {{a^2}{b^2} - {a^2}bc - a{b^2}c + {a^2}{c^2} - ab{c^2} + {b^2}{c^2}} \right)}}{{{a^3}{b^3}{c^3}}} = 0$
And we can see that the above equation holds true only for $ab + bc + ca = 0,$ hence proved.

Note:
When simplifying the equations in order to express them in matrix form, take care of the sign of each term, because an incorrect sign (positive or negative) will lead you to have wrong determinant and through which you cannot get to the required condition.