The points `(1, 1), (-1, -1)` and `(-sqrt3,sqrt3)` are the angular points of a triangle, then the triangle is

Options

A. right angled
B. isosceles
C. equilateral
D. None of these

Answer

To determine the type of triangle formed by the points \( A(1, 1) \), \( B(-1, -1) \), and \( C(-\sqrt{3}, \sqrt{3}) \), we will calculate the lengths of the sides of the triangle using the distance formula. The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

### Step 1: Calculate the length of side \( AB \)

Using points \( A(1, 1) \) and \( B(-1, -1) \):

\[ AB = \sqrt{((-1) - 1)^2 + ((-1) - 1)^2} \]
\[ = \sqrt{(-2)^2 + (-2)^2} \]
\[ = \sqrt{4 + 4} \]
\[ = \sqrt{8} \]
\[ = 2\sqrt{2} \]

### Step 2: Calculate the length of side \( BC \)

Using points \( B(-1, -1) \) and \( C(-\sqrt{3}, \sqrt{3}) \):

\[ BC = \sqrt{(-\sqrt{3} - (-1))^2 + (\sqrt{3} - (-1))^2} \]
\[ = \sqrt{(-\sqrt{3} + 1)^2 + (\sqrt{3} + 1)^2} \]
\[ = \sqrt{(1 - \sqrt{3})^2 + (1 + \sqrt{3})^2} \]
Calculating each square:
\[ (1 - \sqrt{3})^2 = 1 - 2\sqrt{3} + 3 = 4 - 2\sqrt{3} \]
\[ (1 + \sqrt{3})^2 = 1 + 2\sqrt{3} + 3 = 4 + 2\sqrt{3} \]
Now, adding these results:
\[ BC = \sqrt{(4 - 2\sqrt{3}) + (4 + 2\sqrt{3})} \]
\[ = \sqrt{8} \]
\[ = 2\sqrt{2} \]

### Step 3: Calculate the length of side \( CA \)

Using points \( C(-\sqrt{3}, \sqrt{3}) \) and \( A(1, 1) \):

\[ CA = \sqrt{(1 - (-\sqrt{3}))^2 + (1 - \sqrt{3})^2} \]
\[ = \sqrt{(1 + \sqrt{3})^2 + (1 - \sqrt{3})^2} \]
Calculating each square:
\[ (1 + \sqrt{3})^2 = 1 + 2\sqrt{3} + 3 = 4 + 2\sqrt{3} \]
\[ (1 - \sqrt{3})^2 = 1 - 2\sqrt{3} + 3 = 4 - 2\sqrt{3} \]
Now, adding these results:
\[ CA = \sqrt{(4 + 2\sqrt{3}) + (4 - 2\sqrt{3})} \]
\[ = \sqrt{8} \]
\[ = 2\sqrt{2} \]

### Conclusion

We have calculated the lengths of all three sides:
- \( AB = 2\sqrt{2} \)
- \( BC = 2\sqrt{2} \)
- \( CA = 2\sqrt{2} \)

Since all three sides are equal, the triangle is **equilateral**.