Let the vertices of a triangle ABC be (4,4) , (3,5) , (-1,-1) , then the triangle is :

Options

A. a. Scalene
B. b. equilateral
C. c. right angled
D. d. none of (a) , (b) ,(c)

Answer

To determine the type of triangle formed by the vertices A(4, 4), B(3, 5), and C(-1, -1), we will calculate the lengths of the sides using the distance formula and then check the properties of the triangle.

### Step 1: Calculate the distance AB
The distance between two points (x1, y1) and (x2, y2) is given by the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
For points A(4, 4) and B(3, 5):
\[ AB = \sqrt{(3 - 4)^2 + (5 - 4)^2} = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 2: Calculate the distance BC
For points B(3, 5) and C(-1, -1):
\[ BC = \sqrt{(-1 - 3)^2 + (-1 - 5)^2} = \sqrt{(-4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} \]

### Step 3: Calculate the distance CA
For points C(-1, -1) and A(4, 4):
\[ CA = \sqrt{(4 - (-1))^2 + (4 - (-1))^2} = \sqrt{(4 + 1)^2 + (4 + 1)^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} \]

### Step 4: Compare the lengths of the sides
Now we have the lengths:
- \(AB = \sqrt{2}\)
- \(BC = \sqrt{52}\)
- \(CA = \sqrt{50}\)

Since \(AB \neq BC\), \(BC \neq CA\), and \(CA \neq AB\), the triangle is not equilateral.

### Step 5: Check if it is a right triangle
To check if the triangle is a right triangle, we can use the Pythagorean theorem. We need to see if the square of the longest side is equal to the sum of the squares of the other two sides.

The longest side is \(BC\):
\[ BC^2 = 52 \]
Now we calculate \(AB^2 + CA^2\):
\[ AB^2 = 2, \quad CA^2 = 50 \]
So,
\[ AB^2 + CA^2 = 2 + 50 = 52 \]

Since \(AB^2 + CA^2 = BC^2\), the triangle ABC is a right triangle.

### Conclusion
The triangle formed by the vertices A(4, 4), B(3, 5), and C(-1, -1) is a right triangle.

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