Question

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

• A. $(1,2)(-1,-2)$
• B. $(7,8)(-7,-8)$
• C. $(2,3)(-2,-3)$
• D. $Noneofthese$

Answer 1

The correct option is C $(2,3)(-2,-3)$

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

$(2a{)}^2+(3c{)}^2+2\left(2a\right)\left(3b\right)-c^2=0$

$(2a+3b{)}^2-c^2=0$

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

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

$ax+by+c=0$

$ax+by+2a+3b=0$

$a(x+2)+b(y+3)=0$

Hence at $x=-2,y-3$ $(-2,-3)$

or

$ax+by+(-2a-3b)=0$

$a(x-2)+b(y-3)=0$

at point $(2,3)$.