Let's first ignore the isoceles condition and find all right triangles that satisfy the remaining conditions.
We have $(4,2) = K$. Let us denote line $y=2x$ as $l$
- Find any two points $A, B \in l$ such that $|AK|=|BK|$. You can find them by drawing a line orthogonal to $l$ passing through $K$, and then choosing two points of $l$ that have equal distance from this line.
- Draw a line going through $B$ and $K$. Draw a line orthogonal to $l$ passing through point $A$. Define point $C$ as the intersection of these two lines.
You can check that points $A$, $B$, $C$ defined in this way create a right triangle, with point $K$ being the middle of hypotenuse.
The formuale for $A$,$B$,$C$ are $$ A = (1.6-t,3.2-2t)$$
$$ B = (1.6+t,3.2+2t) $$ $$C = (6.4-t,0.8-2t) $$
for some $t\in\mathbb R$.
Since we want the triangle be an isosceles, we need $|AB|=|AC|$. Solving this equation gives us $t=\pm1.2$ and we have two solutions:
$$ A = (0.4,0.8),\qquad B = (2.8,5.6) \qquad C = (5.2,-1.6) $$
$$ A = (2.8,5.6),\qquad B = (0.4,0.8) \qquad C = (7.6,3.2) $$