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)

$\left( - 1, - \frac{1}{2}\right), \left(-2,-1\right)$

Answer 2

$4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c$
$\equiv(2 a+ b)^{2}-3(2 a+ b) c+2 c^{2}=0$
$\Rightarrow(2 a+b-2 c)(2 a+ b -c)=0$
$\Rightarrow c=2 a+b$ 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 $\left(-1,-\frac{1}{2}\right)$