Question

Prove that the points A(1, 1), B(–1, –1) and `C(sqrt(3) - sqrt(3))` are the vertices of an equilateral triangle.

Answer 1

Given: A(1, 1), B(–1, –1) and `C(sqrt(3), -sqrt(3))`

To Prove: AB = BC = CA i.e., the three sides are equal, so ΔABC is equilateral.

Proof [Step-wise]:

1. Note: I interpret the third point as `C(sqrt(3), -sqrt(3))`.

2. Use the distance squared formula to avoid square roots:

For points P(x₁, y₁) and Q(x₂, y₂)

PQ² = (x₁ – x₂)² + (y₁ – y₂)²

3. Compute AB²: 

AB² = (1 – (–1))² + (1 – (–1))²

= (2)² + (2)²

= 4 + 4

= 8

4. Compute AC²:

`AC^2 = (1 − sqrt(3))^2 + (1 - (-sqrt(3)))^2`

= `(1 - sqrt(3))^2 + (1 + sqrt(3))^2`

= `(1 - 2sqrt(3) + 3) + (1 + 2sqrt(3) + 3)` 

= `(4 - 2sqrt(3)) + (4 + 2sqrt(3))` 

= 8

5. Compute BC²:

`BC^2 = (-1 - sqrt(3))^2 + (-1 - (-sqrt(3)))^2`

=` (-1 - sqrt(3))^2 + (-1 + sqrt(3))^2` 

= `(1 + 2sqrt(3) + 3) + (1 - 2sqrt(3) + 3)` 

= `(4 + 2sqrt(3)) + (4 - 2sqrt(3))` 

= 8

6. From steps 3 – 5: 

AB² = AC² = BC² = 8

Hence, AB = AC = BC.

Since all three side lengths are equal, triangle ABC is equilateral.