Three charges -q₁,+q₂ and-q₃ are placed as shown in the figure. The x-component of the force on -q₁ is proportional to
 

The force on -q₁ due to charge + q₂ along X - axis is given by
${\mathrm{F}}_{1\mathrm{x}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{b}}^2}$
The force on -q₁, due to charge -q₃ along the direction -q₃ to -q₁ is given by
${\mathrm{F}}_2=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(-{\mathrm{q}}_1\right)\left(-{\mathrm{q}}_3\right)}{{\mathrm{a}}^2}\times (\sin \mathrm{\theta i}-\cos \mathrm{\theta j})$
or   ${\mathrm{F}}_{2\mathrm{x}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{{\mathrm{q}}_1{\mathrm{q}}_3}{{\mathrm{a}}^2}(\sin \mathrm{\theta })$
Resultant force along X-axis is given by
$\begin{array}{l}{\mathrm{F}}_{\mathrm{x}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{b}}^2}+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_3}{{\mathrm{a}}^2}\sin \mathrm{\theta }\\ =\frac{{\mathrm{q}}_1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{{\mathrm{q}}_3}{{\mathrm{b}}^2}+\frac{{\mathrm{q}}_1}{{\mathrm{a}}^2}\sin \mathrm{\theta }\right]\propto \left[\frac{{\mathrm{q}}_2}{{\mathrm{b}}^2}+\frac{{\mathrm{q}}_3}{{\mathrm{a}}^2}\sin \mathrm{\theta }\right]\end{array}$