In a triangle PQR, the co- ordinates of points P, Q and R are (3, 2) , (6, 4) and (9, 3) respectively . Find the co-ordinates of centroid G. Also find areas of `Delta PQG` and `Delta PRG` .

Answer

To solve the problem step by step, we will first find the coordinates of the centroid G of triangle PQR, and then we will calculate the areas of triangles PQG and PRG.

### Step 1: Finding the Centroid G

The coordinates of points P, Q, and R are given as:
- P(3, 2)
- Q(6, 4)
- R(9, 3)

The formula for the centroid (G) of a triangle with vertices at (x1, y1), (x2, y2), and (x3, y3) is given by:

\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]

Substituting the coordinates of P, Q, and R into the formula:

\[ G\left(\frac{3 + 6 + 9}{3}, \frac{2 + 4 + 3}{3}\right) \]

Calculating the x-coordinate:

\[ \frac{3 + 6 + 9}{3} = \frac{18}{3} = 6 \]

Calculating the y-coordinate:

\[ \frac{2 + 4 + 3}{3} = \frac{9}{3} = 3 \]

Thus, the coordinates of the centroid G are:

\[ G(6, 3) \]

### Step 2: Finding the Area of Triangle PQG

To find the area of triangle PQG, we will use the formula for the area of a triangle given by vertices (x1, y1), (x2, y2), (x3, y3):

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

For triangle PQG, the coordinates are:
- P(3, 2)
- Q(6, 4)
- G(6, 3)

Substituting into the area formula:

\[ \text{Area}_{PQG} = \frac{1}{2} \left| 3(4 - 3) + 6(3 - 2) + 6(2 - 4) \right| \]

Calculating each term:

\[ = \frac{1}{2} \left| 3(1) + 6(1) + 6(-2) \right| \]
\[ = \frac{1}{2} \left| 3 + 6 - 12 \right| \]
\[ = \frac{1}{2} \left| -3 \right| = \frac{3}{2} \]

### Step 3: Finding the Area of Triangle PRG

Now, we will find the area of triangle PRG. The coordinates are:
- P(3, 2)
- R(9, 3)
- G(6, 3)

Using the same area formula:

\[ \text{Area}_{PRG} = \frac{1}{2} \left| 3(3 - 3) + 9(3 - 2) + 6(2 - 3) \right| \]

Calculating each term:

\[ = \frac{1}{2} \left| 3(0) + 9(1) + 6(-1) \right| \]
\[ = \frac{1}{2} \left| 0 + 9 - 6 \right| \]
\[ = \frac{1}{2} \left| 3 \right| = \frac{3}{2} \]

### Final Results

- The coordinates of the centroid G are **(6, 3)**.
- The area of triangle PQG is **1.5 square units** (or \(\frac{3}{2}\) square units).
- The area of triangle PRG is also **1.5 square units** (or \(\frac{3}{2}\) square units).