The points $(1,3)$ and $(5,1)$ are two opposite vertices of a rectangle. The other two vertices are lie on the line $\mathrm{y}=2 x+\mathrm{c}$ where c is the constant, then co-ordinates of other two vertices are

Given the rectangle with opposite vertices at $A = (1, 3)$ and $C = (5, 1)$, the other two vertices lie on the line $y = 2x + c$.

The diagonal $AC$ has midpoint $M = \left(\frac{1+5}{2}, \frac{3+1}{2}\right) = (3, 2)$. Since $M$ lies on the line, $2 = 2(3) + c$ implies $c = -4$, so the line is $y = 2x - 4$.

The diagonal length is $AC = \sqrt{(5-1)^2 + (1-3)^2} = \sqrt{20} = 2\sqrt{5}$. The distance from $M$ to any vertex is half this, $\sqrt{5}$.

Let $B = (x, 2x - 4)$. The squared distance $MB^2 = (x - 3)^2 + (2x - 6)^2 = 5$. Simplifying, $(x - 3)^2 + 4(x - 3)^2 = 5$, so $5(x - 3)^2 = 5$ and $(x - 3)^2 = 1$.

Solving, $x - 3 = \pm 1$. For $x = 4$, $y = 2(4) - 4 = 4$; for $x = 2$, $y = 2(2) - 4 = 0$. The vertices are thus $(4, 4)$ and $(2, 0)$.

Option A matches the result.