Question

If $a^2-b^2-c^2-2bc=0,$ then the family of lines $ax+by+c=0$ are concrrent at the points

• A. $(1,-1)$
• B. $(-1,1)$
• C. $(1,1)$
• D. $(-1,-1)$

Answer 1

Answer: (B), (C)

The correct options are
B $(-1,1)$
C $(1,1)$

$a^2-b^2-c^2-2bc=0$
$\Rightarrow a^2-(b+c{)}^2=0$
$\Rightarrow (a-b-c)(a+b+c)=0$
$\Rightarrow a-b-c=0\text{or}a+b+c=0$
$\Rightarrow -a+b+c=0\text{or}a+b+c=0$

So, the family of lines $ax+by+c=0$ passing through $(1,1)$ and $(-1,1).$