© All Rights reserved @ LearnWithDash
Step 1: Find the centroid of the triangle
The vertices of the triangle are $(3, -1)$, $(1, 3)$, and $(2, 4)$.
The centroid $C$ has coordinates
$$
\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right).
$$
Substituting the given points:
$$
x\text{-coordinate} = \frac{3 + 1 + 2}{3} = \frac{6}{3} = 2,
\quad
y\text{-coordinate} = \frac{-1 + 3 + 4}{3} = \frac{6}{3} = 2.
$$
Hence,
$$
C = (2, 2).
$$
Step 2: Find the point of intersection of the two lines
The lines are
$$
x + 3y - 1 = 0
\quad \text{and} \quad
3x - y + 1 = 0.
$$
From the first line, $x = 1 - 3y.$
Substitute $x = 1 - 3y$ into the second line:
$$
3(1 - 3y) - y + 1 = 0
\quad \Longrightarrow \quad
3 - 9y - y + 1 = 0
\quad \Longrightarrow \quad
4 - 10y = 0
\quad \Longrightarrow \quad
y = \frac{2}{5}.
$$
Then
$$
x = 1 - 3\left(\frac{2}{5}\right) = 1 - \frac{6}{5} = \frac{5 - 6}{5} = -\frac{1}{5}.
$$
Thus,
$$
P = \left(-\frac{1}{5}, \frac{2}{5}\right).
$$
Step 3: Determine the line passing through C and P
Let the coordinates of $C$ be $(x_1, y_1) = (2, 2)$ and the coordinates of $P$ be $(x_2, y_2) = \left(-\frac{1}{5}, \frac{2}{5}\right).$
The two-point form of the line is:
$$
\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}.
$$
Substituting:
$$
\frac{y - 2}{x - 2} = \frac{\frac{2}{5} - 2}{-\frac{1}{5} - 2}.
$$
Simplifying eventually leads to the linear equation
$$
8x - 11y + 6 = 0.
$$
Step 4: Check which option lies on this line
We test each given option in the equation $8x - 11y + 6 = 0$.
Checking the point $(-9, -6)$:
$$
8(-9) - 11(-6) + 6 = -72 + 66 + 6 = 0,
$$
so $(-9, -6)$ satisfies the line equation.
Therefore, the line passing through $C$ and $P$ also passes through $(-9, -6).$
Final Answer
The line passing through the centroid of the triangle and the intersection of the given lines also passes through the point $(-9, -6).$
