**Q) Show that the points (-3, – 3), (3, 3) and (-3√3, 3√3) are the vertices of an equilateral triangle.**

**Ans: **

Let’s consider the points given (-3, – 3) is A, (3, 3) is B and (-3√3,3√3) is C.

Now for a triangle to be an equilateral triangle, required condition is that its all three sides should be equal.

therefore, AB = AC = BC

Step 2: Let’s calculate the lengths of each of the three sides:

We know that the distance between two points (X_{1}, Y_{1}) and (X_{2}, Y_{2}) is given by:

S = **√ **(X_{2} – X_{1})^{2 } + (Y_{2} – Y_{1})^{2 })

∴ AB = **√** (3 – (- 3))^{2 }+ (3 – (- 3))^{2} ) = **√**(36 + 36) =** 6√2**

Similarly, BC = √ (- 3 √3 – 3)^{2 }+ (3 √3 – 3)^{2} ) = √ (3 √3 + 3)^{2 }+ (3 √3 – 3)^{2} )

Since (A + B)^{2} + (A – B)^{2} = 2 (A^{2} + B^{2})

∴ BC = √ 2((3 √3)^{2} + (3)^{2 }) = √(36 + 36) = **6√2**

Similarly, AC = √ (- 3 √3 – (- 3))^{2 }+ (3 √3 – (- 3)^{2} ) = √ (3 √3 – 3)^{2 }+ (3 √3+ 3)^{2} )

Since (A + B)^{2} + (A – B)^{2} = 2 (A^{2} + B^{2})

∴ BC = √ 2((3 √3)^{2} + (3)^{2 }) = √(36 + 36) = **6√2**

Since, AB = AC = BC

Hence, Δ ABC is an equilateral triangle

**Therefore, given 3 points are vertices of an equilateral triangle **

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