Question

If $a^2+9b^2-4c^2=6ab$ then the family of lines $ax+by+c=0$ are concurrent at

• A. $(\frac{1}{2},\frac{3}{2})$
• B. $(\frac{-1}{2},\frac{-3}{2})$
• C. $(\frac{-1}{2},\frac{3}{2})$
• D. $(\frac{1}{2},\frac{-3}{2})$

Answer 1

The correct option is D $(\frac{1}{2},\frac{-3}{2})$

Given $a^2+9b^2-4c^2=6ab$

$\Rightarrow (a-3b{)}^2=(2c{)}^2$

$\Rightarrow (a-3b)=\pm 2c$

$\therefore a-3b+2c=0$ $a-3b=2c$

$c=\frac{3b-a}{2}$ $c=\frac{a-3b}{2}$

family of lines is $ax+by+c=0$

$ax+by+\frac{3b-a}{2}=0$ $ax+by+\frac{a-3b}{2}=0$

if these two lines are concurrent then

$3b-a=0$

now, $ax+by+\frac{3b-a}{2}=0$

$3bx+by=0$ $y=3x$

$\therefore$ The points will be $(\frac{-1}{2},\frac{3}{2})$ and $(\frac{1}{2},\frac{-3}{2})$