Two objects of masses 9 kg and 4 kg are separated by distance of 10 m. Find the position of 1 kg mass for which the net gravitational force on 1 kg mass is zero. (Position of 1 kg mass should be in the line joining 9 kg and 4 kg masses)   

${\text{F}}_{\text{91}}=\text{G}\frac{\left(9\right)\left(1\right)}{x^2}=\frac{\text{G}.9}{x^2}\text{along}9\text{kg}$

${\text{F}}_{\text{41}}=\frac{\text{G}.4}{{\left(10-x\right)}^2}\text{along}4\text{kg}$

But the resultant force should be zero                       

$\therefore {\text{F}}_{\text{91}}={\text{F}}_{\text{41}}$

$\frac{9\text{G}}{x^2}=\frac{4\text{G}}{{\left(10-x\right)}^2}$

$9{\left(10-x\right)}^2=4x^2$

$9\left(100+x^2-20x\right)=4x^2$

$9x^2-180x+900=4x^2$

$5x^2-180x+900=0$

$x^2-36x+180=0$

$x^2-30x-6x+180=0$

$x\left(x-30\right)-6\left(x-30\right)=0$

$\left(x-6\right)\left(x-30\right)=0$

$x-6=0$

$x=6$

$x-30=0$

$x=30$

Since 30 m is beyond 4 kg hence 6 m should be considered

x = 6m (from 9 kg)