Question

The equations $(b-c)x+(c-a)y+(a-b)=0$ and \[({{b}^{3}}-{{c}^{3}})x+({{c}^{3}}-{{a}^{3}})y+({{a}^{3}}-{{b}^{3}})=0\] will represent the same line, if
A. $b=c$
B. $c=a$
C. $a=b$
D. $a+b+c=0$
E. All the above

Answer 1

Hint: In this question, we are to find the conditions that lead the given equations to be the same line. By the appropriate formula, the required conditions are evaluated. Two equations that represent the same line have equal ratios of their coefficients.

Formula used: The condition for the lines ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ to represent the same line is
\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\]
I.e., the ratios of their respective coefficients are equal to one another.

Complete step by step solution: The given equations of the lines are
$(b-c)x+(c-a)y+(a-b)=0\text{ }...(1)$
\[({{b}^{3}}-{{c}^{3}})x+({{c}^{3}}-{{a}^{3}})y+({{a}^{3}}-{{b}^{3}})=0\text{ }...(2)\]
For the lines (1) and (2), to represent the same line we have the following condition
\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\]
Here for the given lines, the coefficients are
$\begin{align}
  & {{a}_{1}}=(b-c) \\
 & {{a}_{2}}=({{b}^{3}}-{{c}^{3}}) \\
 & {{b}_{1}}=(c-a) \\
 & {{b}_{2}}=({{c}^{3}}-{{a}^{3}}) \\
 & {{c}_{1}}=(a-b) \\
 & {{c}_{2}}=({{a}^{3}}-{{b}^{3}}) \\
\end{align}$
Then, on substituting these coefficients in the above condition, we get
\[\begin{align}
  & \dfrac{(b-c)}{({{b}^{3}}-{{c}^{3}})}=\dfrac{(c-a)}{({{c}^{3}}-{{a}^{3}})}=\dfrac{(a-b)}{({{a}^{3}}-{{b}^{3}})} \\
 & \dfrac{(b-c)}{(b-c)({{b}^{2}}+{{c}^{2}}+bc)}=\dfrac{(c-a)}{(c-a)({{c}^{2}}+{{a}^{2}}+ca)}=\dfrac{(a-b)}{(a-b)({{a}^{2}}+{{b}^{2}}+ab)} \\
 & \dfrac{1}{({{b}^{2}}+{{c}^{2}}+bc)}=\dfrac{1}{({{c}^{2}}+{{a}^{2}}+ca)}=\dfrac{1}{({{a}^{2}}+{{b}^{2}}+ab)} \\
\end{align}\]
From this, we can write
$\begin{align}
  & {{b}^{2}}+{{c}^{2}}+bc={{c}^{2}}+{{a}^{2}}+ca={{a}^{2}}+{{b}^{2}}+ab \\
 & \Rightarrow {{b}^{2}}+{{c}^{2}}+bc={{c}^{2}}+{{a}^{2}}+ca\text{ }...(3) \\
 & \Rightarrow {{c}^{2}}+{{a}^{2}}+ca={{a}^{2}}+{{b}^{2}}+ab\text{ }...(4) \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}+ab={{b}^{2}}+{{c}^{2}}+bc\text{ }...(5) \\
\end{align}$
Then, on simplifying these equations, we get
$\begin{align}
  & {{b}^{2}}+{{c}^{2}}+bc={{c}^{2}}+{{a}^{2}}+ca \\
 & \Rightarrow {{a}^{2}}-{{b}^{2}}+ac-bc=0 \\
 & \Rightarrow (a+b)(a-b)+c(a-b)=0 \\
 & \Rightarrow (a-b)(a+b+c)=0 \\
\end{align}$
Then,
\[\begin{align}
  & (a-b)=0;(a+b+c)=0 \\
 & \therefore a=b;a+b+c=0\text{ }...(6) \\
\end{align}\]
$\begin{align}
  & {{c}^{2}}+{{a}^{2}}+ca={{a}^{2}}+{{b}^{2}}+ab \\
 & \Rightarrow {{b}^{2}}-{{c}^{2}}+ab-ca=0 \\
 & \Rightarrow (b+c)(b-c)+a(b-c)=0 \\
 & \Rightarrow (b-c)(b+c+a)=0 \\
\end{align}$
Then,
$\begin{align}
  & (b-c)=0;(b+c+a)=0 \\
 & \therefore b=c;a+b+c=0\text{ }...(7) \\
\end{align}$
$\begin{align}
  & {{a}^{2}}+{{b}^{2}}+ab={{b}^{2}}+{{c}^{2}}+bc \\
 & \Rightarrow {{c}^{2}}-{{a}^{2}}+bc-ab=0 \\
 & \Rightarrow (c+a)(c-a)+b(c-a)=0 \\
 & \Rightarrow (c-a)(c+a+b)=0 \\
\end{align}$
Then,
$\begin{align}
  & (c-a)=0;(c+a+b)=0 \\
 & \therefore c=a;a+b+c=0\text{ }...(8) \\
\end{align}$
From (6), (7), and (8), we get
$a=b;b=c;c=a;a+b+c=0$

Thus, Option (E) is correct.

Note: Here we need to remember that, the ratio of coefficients of two lines is equal if they represent the same line. By this, we can get the required values that make them represent the same line.