Question

If $6a^2-3b^2-c^2+7ab-ac+4bc=0$, then the family of lines $ax+by+c=0$ is concurrent at

• A. (-2,-3)
• B. (3,-1)
• C. (2,3)
• D. (-3,1)

Answer 1

The correct option is D (-3,1)

$\text{Given,}6a^2-3b^2-c^2+7ab-ac+4bc=0$

$\Rightarrow 6a^2+a(7b-c)-3b^2-c^2+4bc=0$

$\text{It is Quadratic in a so solving}$

$\text{of value of a we get,}$

$a=\frac{(c-7b)\pm \sqrt{49b^2+c^2-14bc+72b^2+24c^2-96bc}}{12}$

$\Rightarrow a=\frac{(c-7b)\pm \sqrt{121b^2+25c^2-110bc}}{12}$

$\Rightarrow a=\frac{(c-7b)\pm \sqrt{(11b-5c{)}^2}}{12}$

$\Rightarrow a=\frac{(c-7b)\pm (11b-5c)}{12}$

on taking positive sign

$a=\frac{c-7b+11b-5c}{12}=\frac{b-c}{3}$

on taking negative sign

$a=\frac{c-7b-11b+5c}{12}=\frac{-3b+c}{2}$

$\Rightarrow 3a-b+c=0$ (i)

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

To find the point of concurrency

we have to solve $ax+by+c=0$

and equation (i) simultaneously

on compairing we get,

$\frac{x}{3}=\frac{y}{-1}=\frac{1}{1}$

for another point on concurrent

on comparing $ax+by+c=0$ with

equation (ii) we get,

$\frac{x}{2}=\frac{y}{3}=\frac{1}{-1}$

$\Rightarrow x=-2,y=-3$

So, (3,-1) and (-2,-3) are

the point of Concurrency.