The points (a, a), (–a, –a) and `(-sqrt(3)a, sqrt(3)a)`  are the vertices of

Options

A. right triangle
B. scalene triangle
C. equilateral triangle
D. isosceles triangle

Answer

To determine the type of triangle formed by the points (a, a), (–a, –a), and (–√3a, √3a), we will calculate the lengths of the sides using the distance formula and then classify the triangle based on the lengths of its sides.

### Step-by-Step Solution:

1. **Identify the Points**:
- Let \( A = (a, a) \)
- Let \( B = (-a, -a) \)
- Let \( C = (-\sqrt{3}a, \sqrt{3}a) \)

2. **Use the Distance Formula**:
The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

3. **Calculate the Length of Side AB**:
\[ AB = \sqrt{((-a) - a)^2 + ((-a) - a)^2} \]
\[ = \sqrt{(-2a)^2 + (-2a)^2} = \sqrt{4a^2 + 4a^2} = \sqrt{8a^2} = 2\sqrt{2}a \]

4. **Calculate the Length of Side BC**:
\[ BC = \sqrt{((-√3a) - (-a))^2 + ((√3a) - (-a))^2} \]
\[ = \sqrt{(-\sqrt{3}a + a)^2 + (\sqrt{3}a + a)^2} \]
\[ = \sqrt{(-(\sqrt{3} - 1)a)^2 + ((\sqrt{3} + 1)a)^2} \]
\[ = \sqrt{((\sqrt{3} - 1)^2 + (\sqrt{3} + 1)^2)a^2} \]
\[ = \sqrt{(3 - 2\sqrt{3} + 1 + 3 + 2\sqrt{3} + 1)a^2} = \sqrt{(6)a^2} = \sqrt{6}a \]

5. **Calculate the Length of Side CA**:
\[ CA = \sqrt{((-√3a) - a)^2 + ((√3a) - a)^2} \]
\[ = \sqrt{(-\sqrt{3}a - a)^2 + (\sqrt{3}a - a)^2} \]
\[ = \sqrt{(-(\sqrt{3} + 1)a)^2 + ((\sqrt{3} - 1)a)^2} \]
\[ = \sqrt{((\sqrt{3} + 1)^2 + (\sqrt{3} - 1)^2)a^2} \]
\[ = \sqrt{(3 + 2\sqrt{3} + 1 + 3 - 2\sqrt{3} + 1)a^2} = \sqrt{(8)a^2} = 2\sqrt{2}a \]

6. **Compare the Lengths**:
- \( AB = 2\sqrt{2}a \)
- \( BC = \sqrt{6}a \)
- \( CA = 2\sqrt{2}a \)

7. **Determine the Type of Triangle**:
Since \( AB = CA \) and \( AB \neq BC \), the triangle is **isosceles**.

### Conclusion:
The triangle formed by the points (a, a), (–a, –a), and (–√3a, √3a) is an **isosceles triangle**.