The vertices of a triangle are the points `P(-26, 17), Q(30, 17) and R(10, 2)`. If G and I be the centroid and incentre of the triangle PQR, then the value of `(GI)^(2)` is equal to

Options

A. `(205)/(9)`
B. `(sqrt(205))/(3)`
C. `3sqrt3`
D. 27

Answer

To solve the problem, we need to find the squared distance between the centroid (G) and the incenter (I) of triangle PQR with vertices at P(-26, 17), Q(30, 17), and R(10, 2).

### Step 1: Calculate the lengths of the sides of the triangle
Using the distance formula, we can find the lengths of the sides PQ, QR, and PR.

- **Length PQ**:
\[ PQ = \sqrt{(30 - (-26))^2 + (17 - 17)^2} = \sqrt{(30 + 26)^2} = \sqrt{56^2} = 56 \]

- **Length QR**:
\[ QR = \sqrt{(10 - 30)^2 + (2 - 17)^2} = \sqrt{(-20)^2 + (-15)^2} = \sqrt{400 + 225} = \sqrt{625} = 25 \]

- **Length PR**:
\[ PR = \sqrt{(10 - (-26))^2 + (2 - 17)^2} = \sqrt{(10 + 26)^2 + (-15)^2} = \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39 \]

### Step 2: Find the coordinates of the centroid (G)
The coordinates of the centroid G of triangle PQR can be calculated using the formula:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
Substituting the coordinates of P, Q, and R:
\[ G\left(\frac{-26 + 30 + 10}{3}, \frac{17 + 17 + 2}{3}\right) = G\left(\frac{14}{3}, \frac{36}{3}\right) = G\left(\frac{14}{3}, 12\right) \]

### Step 3: Find the coordinates of the incenter (I)
The coordinates of the incenter I can be calculated using the formula:
\[ I\left(\frac{a \cdot x_1 + b \cdot x_2 + c \cdot x_3}{a + b + c}, \frac{a \cdot y_1 + b \cdot y_2 + c \cdot y_3}{a + b + c}\right) \]
Where \(a\), \(b\), and \(c\) are the lengths of the sides opposite to vertices P, Q, and R respectively.

Substituting the values:
- \(a = QR = 25\)
- \(b = PR = 39\)
- \(c = PQ = 56\)

Calculating the coordinates of I:
\[ I\left(\frac{25 \cdot (-26) + 39 \cdot 30 + 56 \cdot 10}{25 + 39 + 56}, \frac{25 \cdot 17 + 39 \cdot 17 + 56 \cdot 2}{25 + 39 + 56}\right) \]
Calculating the x-coordinate:
\[ = \frac{-650 + 1170 + 560}{120} = \frac{1080}{120} = 9 \]
Calculating the y-coordinate:
\[ = \frac{425 + 663 + 112}{120} = \frac{1200}{120} = 10 \]
Thus, \(I(9, 10)\).

### Step 4: Calculate the squared distance \(GI^2\)
Now, we find the squared distance \(GI^2\):
\[ GI^2 = \left(\frac{14}{3} - 9\right)^2 + (12 - 10)^2 \]
Calculating each term:
\[ \frac{14}{3} - 9 = \frac{14}{3} - \frac{27}{3} = \frac{-13}{3} \]
Thus,
\[ GI^2 = \left(\frac{-13}{3}\right)^2 + (2)^2 = \frac{169}{9} + 4 = \frac{169}{9} + \frac{36}{9} = \frac{205}{9} \]

### Final Answer
The value of \(GI^2\) is \(\frac{205}{9}\).

---