= 6, and 4x + y = 7 has vertices at (1,1), (2,2), and (5,-3).

To find the vertices of the triangle, we can solve the system of equations formed by the three lines. One way to do this is to solve for y in terms of x in each equation:

x y = 1 --> y = 1/x
2x 3y = 6 --> y = (6 - 2x)/3
4x + y = 7 --> y = 7 - 4x

Now we can set the expressions for y equal to each other and solve for x:

1/x = (6 - 2x)/3 --> 3 = x(6 - 2x)/3x --> 3x = 6 - 2x --> x = 2/5

Substituting this value of x into one of the expressions for y, we get:

y = 1/x = 5/2

So one vertex of the triangle is (2/5, 5/2). We can find the other vertices in the same way by solving for x and y:

2x 3y = 6 --> x = (6 - 3y)/2
4x + y = 7 --> x = (7 - y)/4

Setting these expressions for x equal to each other and solving for y, we get:

(6 - 3y)/2 = (7 - y)/4 --> 12 - 6y = 14 - 2y --> y = -3

Substituting this value of y into one of the expressions for x, we get:

x = (7 - y)/4 = 5/4

So another vertex of the triangle is (5/4, -3). Finally, we can find the third vertex by solving for y:

x y = 1 --> y = 1/x
4x + y = 7 --> y = 7 - 4x

Setting these expressions for y equal to each other and solving for x, we get:

1/x = 7 - 4x --> 4x^2 - 7x + 1 = 0

Using the quadratic formula, we find:

x = (7 ± √33)/8

Substituting these values of x into one of the expressions for y, we get:

y = 1/x = 8/(7 ± √33)

So the two possible third vertices of the triangle are ((7 + √33)/8, 8/(7 + √33)) and ((7 - √33)/8, 8/(7 - √33)). We can check that both of these points lie on the three lines given in the problem, so they are both vertices of the triangle.

Therefore, the triangle formed by the lines x y = 1, 2x 3y = 6, and 4x + y = 7 has vertices at (1,1), (2,2), and (5,-3).