Find the co-ordinates of the incentre of the triangle whose vertices are the points (4,-2) , (5,5) and (-2,4) :

Options

A. a. `(7/3 , 2/3)`
B. b. `(5/2 , 5/2)`
C. c. `(6/5 , 5)`
D. d. none of these

Answer

To find the coordinates of the incenter of the triangle with vertices at points \( A(4, -2) \), \( B(5, 5) \), and \( C(-2, 4) \), we will follow these steps:

### Step 1: Calculate the lengths of the sides of the triangle
We need to find the lengths of the sides opposite to each vertex. The lengths of the sides \( a \), \( b \), and \( c \) can be calculated using the distance formula.

- **Length \( a \)** (opposite vertex \( A \)):
\[ a = BC = \sqrt{(5 - (-2))^2 + (5 - 4)^2} = \sqrt{(5 + 2)^2 + (5 - 4)^2} = \sqrt{7^2 + 1^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} \]

- **Length \( b \)** (opposite vertex \( B \)):
\[ b = AC = \sqrt{(4 - (-2))^2 + (-2 - 4)^2} = \sqrt{(4 + 2)^2 + (-2 - 4)^2} = \sqrt{6^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \]

- **Length \( c \)** (opposite vertex \( C \)):
\[ c = AB = \sqrt{(4 - 5)^2 + (-2 - 5)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \]

### Step 2: Use the incenter formula
The coordinates of the incenter \( (I_x, I_y) \) can be calculated using the formula:
\[ I_x = \frac{a \cdot x_1 + b \cdot x_2 + c \cdot x_3}{a + b + c} \]
\[ I_y = \frac{a \cdot y_1 + b \cdot y_2 + c \cdot y_3}{a + b + c} \]
where \( (x_1, y_1) = (4, -2) \), \( (x_2, y_2) = (5, 5) \), and \( (x_3, y_3) = (-2, 4) \).

### Step 3: Substitute the values into the formulas
Substituting the values we calculated:
\[ I_x = \frac{(5\sqrt{2}) \cdot 4 + (6\sqrt{2}) \cdot 5 + (5\sqrt{2}) \cdot (-2)}{5\sqrt{2} + 6\sqrt{2} + 5\sqrt{2}} \]
\[ = \frac{20\sqrt{2} + 30\sqrt{2} - 10\sqrt{2}}{16\sqrt{2}} = \frac{40\sqrt{2}}{16\sqrt{2}} = \frac{40}{16} = 2.5 \]

For \( I_y \):
\[ I_y = \frac{(5\sqrt{2}) \cdot (-2) + (6\sqrt{2}) \cdot 5 + (5\sqrt{2}) \cdot 4}{5\sqrt{2} + 6\sqrt{2} + 5\sqrt{2}} \]
\[ = \frac{-10\sqrt{2} + 30\sqrt{2} + 20\sqrt{2}}{16\sqrt{2}} = \frac{40\sqrt{2}}{16\sqrt{2}} = \frac{40}{16} = 2.5 \]

### Step 4: Final coordinates of the incenter
Thus, the coordinates of the incenter \( I \) of the triangle are:
\[ I(2.5, 2.5) \]