Two charges Q_1 = q and Q_2 = mq are placed at the points P_1(a, b)...

The potential at a point $(x, y)$ due to charge $Q_1$ is $V_1 = \dfrac{1}{4\pi\epsilon_0} \dfrac{q}{\sqrt{(x-a)^2 + (y-b)^2}}$.

The potential at $(x, y)$ due to charge $Q_2$ is $V_2 = \dfrac{1}{4\pi\epsilon_0} \dfrac{mq}{\sqrt{(x-ma)^2 + (y-mb)^2}}$.

Given $|V_1| = |V_2|$, we have:

$\dfrac{1}{(x-a)^2 + (y-b)^2} = \dfrac{m^2}{(x-ma)^2 + (y-mb)^2}$

Cross-multiplying and expanding both sides:

$(x-ma)^2 + (y-mb)^2 = m^2[(x-a)^2 + (y-b)^2]$

$x^2 + m^2a^2 - 2max + y^2 + m^2b^2 - 2mby = m^2(x^2 + a^2 - 2ax + y^2 + b^2 - 2by)$

$x^2 + y^2 - 2m(ax+by) + m^2(a^2+b^2) = m^2(x^2+y^2) - 2m^2(ax+by) + m^2(a^2+b^2)$

Canceling $m^2(a^2+b^2)$ from both sides and rearranging:

$(m^2-1)(x^2+y^2) - 2m(m-1)(ax+by) = 0$

Since $m \neq 1$, dividing by $m-1$ gives the general equation of the locus:

$(m+1)(x^2+y^2) - 2m(ax+by) = 0$

For $m = -1$:

$0 - 2(-1)(ax+by) = 0 \Rightarrow ax+by = 0$. This represents a straight line.

For $m = 2$:

$3(x^2+y^2) - 4(ax+by) = 0 \Rightarrow x^2+y^2 - \dfrac{4}{3}ax - \dfrac{4}{3}by = 0$.

This represents a circle with center \left(\dfrac{2}{3}a, \dfrac{2}{3}b\right)$and radius$\sqrt{\left(\dfrac{2}{3}a\right)^2 + \left(\dfrac{2}{3}b\right)^2} = \dfrac{2}{3}\sqrt{a^2+b^2}$.<br><br>For$m = -2$:<br><br>$(-1)(x^2+y^2) - 2(-2)(ax+by) = 0 \Rightarrow x^2+y^2 - 4ax - 4by = 0$.<br><br>This represents a circle with center$(2a, 2b)$and radius$\sqrt{(2a)^2 + (2b)^2} = 2\sqrt{a^2+b^2}$.<br><br>For$m = -3$:<br><br>$(-2)(x^2+y^2) - 2(-3)(ax+by) = 0 \Rightarrow x^2+y^2 - 3ax - 3by = 0$.<br><br>This represents a circle, not a straight line.<br><br>Answer: For$m = -1$, locus of these points is$ax + by = 0$.; For$m = 2$, the locus of these points is a circle of radius$\dfrac{2}{3}\sqrt{a^2 + b^2}$centered at \left(\dfrac{2}{3}a, \dfrac{2}{3}b\right)$; For $m = -2$, the locus of these points is a circle of radius $2\sqrt{a^2 + b^2}$ centered at $(2a, 2b)$