The equations of two sides of a square are y = 3x−1 and x+3y−6=0. If (0,-1) is one vertex of the square find the coordinates of the other vertices.

The first line ($L_1$) is given by $y = 3x - 1$. Its slope is $m_1 = 3$.
The second line ($L_2$) is given by $x + 3y - 6 = 0$. We can rewrite this in slope-intercept form:

Its slope is $m_2 = -\frac{1}{3}$.
Since $m_1 \times m_2 = 3 \times \left(-\frac{1}{3}\right) = -1$, the two lines are perpendicular. This means they represent adjacent sides of the square.

For $L_1: y = 3x - 1$:
Substitute $(0, -1)$: $-1 = 3(0) - 1 \implies -1 = -1$. So, $P(0, -1)$ lies on $L_1$.
For $L_2: x + 3y - 6 = 0$:
Substitute $(0, -1)$: $0 + 3(-1) - 6 = 0 \implies -3 - 6 = 0 \implies -9 = 0$. This is false. So, $P(0, -1)$ does not lie on $L_2$.

This point, let's call it $Q$, is another vertex of the square.
Substitute $y = 3x - 1$ from $L_1$ into the equation for $L_2$:

Now substitute $x = \frac{9}{10}$ back into $y = 3x - 1$:

So, the second vertex is $Q = \left(\frac{9}{10}, \frac{17}{10}\right)$.
Since $P$ lies on $L_1$ and $Q$ is the intersection of $L_1$ and $L_2$, $P$ and $Q$ are adjacent vertices, and the side $PQ$ lies on $L_1$.

The side length $s$ is the distance between $P(0, -1)$ and $Q\left(\frac{9}{10}, \frac{17}{10}\right)$.

Let the vertices of the square be $P, Q, R, S$ in counter-clockwise order.
We have $P = (0, -1)$ and $Q = \left(\frac{9}{10}, \frac{17}{10}\right)$.
The vector $\vec{PQ} = Q - P = \left(\frac{9}{10} - 0, \frac{17}{10} - (-1)\right) = \left(\frac{9}{10}, \frac{27}{10}\right)$.
The side $QR$ is perpendicular to $PQ$ and lies on $L_2$. A vector perpendicular to $(a, b)$ is $(-b, a)$ (for counter-clockwise rotation).
So, $\vec{QR} = \left(-\frac{27}{10}, \frac{9}{10}\right)$. (The slope of this vector is $\frac{9/10}{-27/10} = -\frac{1}{3}$, which matches $L_2$).

The third vertex $R$ is $Q + \vec{QR}$:

The fourth vertex $S$ can be found by adding $\vec{QR}$ to $P$, since $\vec{PS} = \vec{QR}$ (opposite sides of a square are parallel and equal in length).

The coordinates of the other vertices are:
$Q = \boxed{\left(\frac{9}{10}, \frac{17}{10}\right)}$
$R = \boxed{\left(-\frac{9}{5}, \frac{13}{5}\right)}$
$S = \boxed{\left(-\frac{27}{10}, -\frac{1}{10}\right)}$