Question

Equations $(b-c)x+(c-a)y+(a-b)=0$ and $(b^3-c^3)x+(c^3-a^3)y+a^3-b^3=0$ represents a line if

• A. $a=b=c$
• B. $b=c$
• C. $c=a$
• D. $a+b+c=0$

Answer 1

Answer: (A), (D)

$a=b=c$
$a+b+c=0$

The given equations are in the form $ax+by+c=0$ and $px+qy+r=0$ who represent the same line.

So, $\frac{a}{p}=\frac{b}{q}=\frac{c}{r}=k$ when, $k$ is a constant.

Here $a=b^3-c^3,b=c^3-a^3,c=a^3-b^3$ and $p=b-c,q=c-a,r=a-b$

$\therefore \frac{b^3-c^3}{b-c}=k$

$\Rightarrow (b-c)(b^2+c^2+bc)=(b-c)k$

$\Rightarrow (b-c)(b^2+c^2+bc-k)=0$

$\therefore$ either $b-c=0$

$\Rightarrow b=c$ .......(i)

or $(b^2+c^2+bc-k)=0$

$\Rightarrow b^2+c^2+bc=k$ ............(1)

Similarly $c=a$........(ii) and

$c^2+a^2+ca=k$ ............(2)

Also $a=b$ .......(iii) and

$a^2+b^2+ab=k$ ............(3)

$\therefore$ from (i), (ii) and (iii), we have

$a=b=c$ first condition

Again from (1) and (2), we have

$b^2+c^2+bc=c^2+a^2+ca$

$\Rightarrow b^2-a^2=c(a-b)$

$\Rightarrow (b-a)(b+a)=c(a-b)$

$\Rightarrow b+a=-c$

$\Rightarrow a+b+c=0$ second condition, we shall get the same considering (2) and (3).