Question

Show that the points (2,2) , (-2,-2) and $(2\sqrt{3} , -2\sqrt{3})$ are the vertices of an equilateral triangle.

Answer 1

Answer:

ABC IS AN EQUILATERAL TRIANGLE.

Step-by-step explanation:

$a(2 \: \: \: 2) \: \: \: b( - 2 \: \: \: \: \: - 2) \: \: \: \: c(2 \sqrt{3} \: \: \: \: \: \: - 2 \sqrt{3} ) \\ \\ therefore \: \: \: ab = \sqrt{ {(2 + 2)}^{2} + {(2 + 2)}^{2} } = 4 \sqrt{2} \: \: units \\ \\ bc = \sqrt{ {(2 \sqrt{3} + 2) }^{2} + {(2 \sqrt{3 } - 2)}^{2} } = 2 \sqrt{3 + 1 + 2 \sqrt{3} + 3 + 1 - 2 \sqrt{3} } \\ \\ = 2 \sqrt{8} = 4 \sqrt{2} \: \: units \\ \\ ca = \sqrt{ {(2 - 2 \sqrt{3} )}^{2} + {(2 + 2 \sqrt{3} )}^{2} } \\ \\ = 2 \sqrt{1 + 3 + 2 \sqrt{3} + 1 + 3 + 2 \sqrt{3} } \\ \\ = 2 \sqrt{8} \\ \\ = 4 \sqrt{2 } \: \: units \\ \\ therefore \: \: ab \: = \: bc \: = \: ca \: = \: 4 \sqrt{2} \: \: units \: . \\ \\ therefore \: \: abc \: is \: an \: equileteral \: triangle.$