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 (X1, Y1) and (X2, Y2) is given by:

S = (X2 – X1)2 + (Y2 – Y1)2 )

∴  AB = (3 – (- 3))+ (3 – (- 3))2 ) = (36 + 36) = 6√2

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

Since (A + B)2 + (A – B)2 = 2 (A2 + B2)

∴  BC = √ 2((3 √3)2 + (3)2 ) = √(36 + 36) = 6√2

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

Since (A + B)2 + (A – B)2 = 2 (A2 + B2)

∴  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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