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))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 (A2 + B2)
∴ 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 (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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