Question

If $4a^2+b^2+2c^2+4ab-6ac-3bc=0$, the family of lines $ax+by+c=0$ is concurrent at one or the other of the two points-

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

Answer 1

Answer: (A)

$4a^2+b^2+2c^2+4ab-6ac-3bc$
$\equiv (2a+b{)}^2-3(2a+b)c+2c^2=0$
$\Rightarrow (2a+b-2c)(2a+b-c)=0$
$\Rightarrow c=2a+b$ or $c=a+\frac{1}{2}b$
The equation of the family of lines is
$a(x+2)+b(y+1)=0$
or $a(x+1)+b\left(y+\frac{1}{2}\right)=0$
giving the point of concurrence $(-2,-1)$
or $\left(-1,-\frac{1}{2}\right)$
$a(x+2)+b(y+1)=0$
or $a(x+1)+b\left(y+\frac{1}{2}\right)=0$
giving the point of concurrence
$(-2,-1)$ or $\left(-1,-\frac{1}{2}\right)$