The hypotenuse of a right-angled triangle has its endpoints at the points $(1, 3)$ and $(-4, 1)$. Find the equations of the legs (perpendicular sides) of the triangle.

(N/A) Let the vertices of the right-angled triangle be $A(1, 3)$,$B(-4, 1)$,and $C(x, y)$,where $\angle C = 90^{\circ}$.
Since $C$ lies on a circle with diameter $AB$,the locus of $C$ is $(x - 1)(x + 4) + (y - 3)(y - 1) = 0$.
There are infinitely many such triangles depending on the choice of point $C$ on this circle.
Let the slope of leg $AC$ be $m$. Then the slope of leg $BC$ is $-\frac{1}{m}$ (since $AC \perp BC$).
The equation of line $AC$ passing through $(1, 3)$ is $y - 3 = m(x - 1)$.
The equation of line $BC$ passing through $(-4, 1)$ is $y - 1 = -\frac{1}{m}(x + 4)$.
For any real value of $m$,these two lines represent the legs of a right-angled triangle with the given hypotenuse.