Question

If $4a^2+9b^2-c^2+12ab=0$, then the family of straight lines $ax+by+c=0$ is concurrent at

• A. $(2,3)$ or $(-2,-3)$
• B. $(2,-3)$ or $(-2,6)$
• C. $(-2,-4)$ or $(-2,3)$
• D. $(2,5)$ or $(-1,-5)$

Answer 1

The correct option is A $(2,3)$ or $(-2,-3)$

We have,

$4a^2+9b^2-c^2+12ab=0$

Above equation can be written as

$(2a+3b{)}^2=c^2$

$\Rightarrow 2a+3b=c$ or $2a+3b=-c$

$2a+3b-c=0$ ...(1)

$2a+3b+c=0$ ...(2)

By comparing this to $ax+by+c=0$,

$x=-2,y=-3$ and $x=2,y=3$

$\therefore$ Two points where the family of straight lines $ax+by+c=0$ is concurrent is $(2,3)$ and $(-2,-3)$.

Hence, option A is the correct answer.