This problem is nearly one dimensional. In this solution when we talk about the distance from one point to another we mean the horizontal distance except when we specify otherwise. The first step will be to locate the center of mass of the system.

Because the rods are uniform, we can say that the center of mass of each rod is at its geometric center. Because we have an equilateral triangle, the center of the 4 kg rod is 7 cm to the left of the hinge and the center of the 3 kg rod is 7 cm to the right of the hinge. We use a weighted average to find the horizontal location of the center of mass of the system.

We know that the center of mass of the system is 1 cm to the left of the hinge. Now we will impose a coordinate system centered at the horizontal location of the center of mass. In this system, the hinge is originally 1 cm to the right of the origin.

What is the distance from the center of mass of the system to the hinge along the length of the 4 kg rod? The right triangle drawn is a 30-60-90 triangle whose base is 1 cm and whose hypotenuse must be 2 cm. When the system has slid to the ground with the center of mass remaining in the same horizontal position the hinge will be 2 cm to the right of the origin. Thus, the hinge will have moved 1 cm to the right.