To solve the problem step by step, we will find the centroid D of triangle ABC, the point P of intersection of the two lines, and then determine the equation of the line joining D and P. Finally, we will check which of the given points lies on that line.
### Step 1: Find the Centroid D of Triangle ABC
The vertices of triangle ABC are:
- A(3, -1)
- B(1, 3)
- C(2, 4)
The formula for the centroid \(D\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
\[
D\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)
\]
Substituting the coordinates of points A, B, and C:
\[
D\left(\frac{3 + 1 + 2}{3}, \frac{-1 + 3 + 4}{3}\right) = D\left(\frac{6}{3}, \frac{6}{3}\right) = D(2, 2)
\]
### Step 2: Find the Point of Intersection P of the Lines
The equations of the lines are:
1. \(x + 3y - 1 = 0\)
2. \(3x - y + 1 = 0\)
We can solve these equations simultaneously. First, rearranging the first equation for \(x\):
\[
x = 1 - 3y
\]
Now, substitute \(x\) in the second equation:
\[
3(1 - 3y) - y + 1 = 0
\]
\[
3 - 9y - y + 1 = 0
\]
\[
4 - 10y = 0 \implies y = \frac{4}{10} = \frac{2}{5}
\]
Now substituting \(y\) back to find \(x\):
\[
x = 1 - 3\left(\frac{2}{5}\right) = 1 - \frac{6}{5} = \frac{-1}{5}
\]
Thus, the point of intersection \(P\) is:
\[
P\left(-\frac{1}{5}, \frac{2}{5}\right)
\]
### Step 3: Find the Equation of the Line Joining D and P
We have:
- \(D(2, 2)\)
- \(P\left(-\frac{1}{5}, \frac{2}{5}\right)\)
To find the slope \(m\) of the line joining D and P:
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{2}{5} - 2}{-\frac{1}{5} - 2} = \frac{\frac{2}{5} - \frac{10}{5}}{-\frac{1}{5} - \frac{10}{5}} = \frac{-\frac{8}{5}}{-\frac{11}{5}} = \frac{8}{11}
\]
Using the point-slope form of the line equation:
\[
y - y_1 = m(x - x_1)
\]
Substituting \(m\), \(x_1\), and \(y_1\):
\[
y - 2 = \frac{8}{11}(x - 2)
\]
Rearranging to standard form:
\[
11(y - 2) = 8(x - 2)
\]
\[
11y - 22 = 8x - 16
\]
\[
8x - 11y + 6 = 0
\]
### Step 4: Check Which Given Point Lies on the Line
Now we need to check which of the given points satisfies the equation \(8x - 11y + 6 = 0\).
Let's say we have a point \( (x_0, y_0) \) to check:
1. Substitute \(x_0\) and \(y_0\) into \(8x_0 - 11y_0 + 6\) and see if it equals 0.
