Given three points $$P$$, $$Q$$, $$R$$ with $$P(5, 3)$$ and $$R$$ lies on the $$x$$-axis. If the equation of $$RQ$$ is $$x - 2y = 2$$ and $$PQ$$ is parallel to the $$x$$-axis, then the centroid of $$\Delta PQR$$ lies on the line:

We are given three points P, Q, and R. Point P is at (5, 3). Point R lies on the x-axis, so its y-coordinate is 0. Let the coordinates of R be (h, 0). The equation of line RQ is given as $$x - 2y = 2$$. Also, PQ is parallel to the x-axis. Since PQ is parallel to the x-axis and P has y-coordinate 3, the y-coordinate of Q must also be 3. Let the coordinates of Q be (a, 3).

Since R lies on the line RQ, we substitute R(h, 0) into the equation $$x - 2y = 2$$:

$$h - 2 \times 0 = 2$$

$$h = 2$$

So, R is at (2, 0).

Now, Q(a, 3) must also lie on the line RQ because RQ is the line joining R and Q. Substitute Q(a, 3) into the equation $$x - 2y = 2$$:

$$a - 2 \times 3 = 2$$

$$a - 6 = 2$$

$$a = 8$$

Therefore, Q is at (8, 3).

We now have the coordinates:

P(5, 3), Q(8, 3), R(2, 0)

The centroid of triangle PQR is found by averaging the x-coordinates and y-coordinates of the vertices. Let the centroid be G(x_g, y_g).

$$x_g = \frac{x_P + x_Q + x_R}{3} = \frac{5 + 8 + 2}{3} = \frac{15}{3} = 5$$

$$y_g = \frac{y_P + y_Q + y_R}{3} = \frac{3 + 3 + 0}{3} = \frac{6}{3} = 2$$

So, the centroid G is at (5, 2).

We need to determine which line this centroid lies on by substituting (5, 2) into each option.

Option A: $$x - 2y + 1 = 0$$

Substitute x = 5, y = 2:

$$5 - 2 \times 2 + 1 = 5 - 4 + 1 = 2 \neq 0$$

Not satisfied.

Option B: $$2x + y - 9 = 0$$

Substitute x = 5, y = 2:

$$2 \times 5 + 2 - 9 = 10 + 2 - 9 = 3 \neq 0$$

Not satisfied.

Option C: $$2x - 5y = 0$$

Substitute x = 5, y = 2:

$$2 \times 5 - 5 \times 2 = 10 - 10 = 0$$

Satisfied.

Option D: $$5x - 2y = 0$$

Substitute x = 5, y = 2:

$$5 \times 5 - 2 \times 2 = 25 - 4 = 21 \neq 0$$

Not satisfied.

Hence, the centroid lies on the line given in Option C: $$2x - 5y = 0$$. So, the answer is Option C.