ABC is a equilateral triangle with vertex A fixed and B moving in a given straight line. Find the locus of C.
Though it is clear that being an equilateral triangle, the size of the triangle must not change but still I am not able to visualize the exact conditions. One more thing, is there any mathematical way (other than just visualizing) through which the problem can be solved.
First, it should be obvious that if $A$ is a fixed vertex and $B$ moves in a straight line, then $AB$ cannot be constant. Equivalently stated, if $A$ is fixed and $AB$ is constant, then the locus of $B$ is a circle with $AB$ as radius.
We conclude that the only reasonable interpretation of the question is to suppose $A$ is fixed and $B$ moves in a straight line, allowing the size of equilateral $\triangle ABC$ to vary as $B$ varies.
Under such an interpretation, the locus is one of two lines (depending on which triangle you choose), either of which forms an angle of $\pi/3$ with the given line along which $B$ moves.
To see why, we may assume without loss of generality that $A = (0,0)$ and $B = (2,2t)$ for some parameter $t \in \mathbb R$. Then the locus of point $C = (x,y)$ corresponds to the solution of the system $$\begin{align*} x^2 + y^2 &= 4(1+t^2) \\ (x-2)^2 + (y-2t)^2 &= 4(1+t^2). \end{align*}$$ The difference of these equations gives $$4(x-1) + 4t(y-t) = 0,$$ or $$x = 1+t^2-ty.$$ Substituting into the first equation then gives $$4(1+t^2) = (1+t^2-ty)^2 + y^2 = (1+t^2)(1+(y-t)^2),$$ or $$(y-t)^2 = 3,$$ which gives us the pair of solutions $$(x(t),y(t)) = \left(1 \pm t \sqrt{3}, t \mp \sqrt{3} \right).$$ This of course describes a parametrization of the aforementioned pair of lines, which has Cartesian coordinates $$y = \frac{1}{\sqrt{3}}(x-4), \quad y = -\frac{1}{\sqrt{3}}(x + 4).$$