Question

prove that the following points are the vertices of an equilateral triangle A(0, 0), B(3,√3), C(3, -√3) ​

Answer 1

Answer:

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Point A( 0, 0)
• Point B ( 3, √3)
• Point C (3, -√3)

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

• The points are vertices of an equilateral triangle

$\Large{\underline{\underline{\bf{Proof:}}}}$

→ Here we have to prove that all the sides of the triangle are equal,

  AB = BC = CA

→ By distance formula we know that distance between 2 points is given by

  $\sf {Distance=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2} } }$

→ First we have to find the distance of AB

  where x₁ = 0, x₂ = 3, y₁ = 0, y₂ = √3

  $\sf{AB=\sqrt{(3-0)^{2}+(\sqrt{3}-0)^{2} } }$

  $\sf{AB=\sqrt{9+3}}$

  AB = √12-------(1)

→ Now finding the distance of BC

  where x₁ = 3, x₂ = 3, y₁ = √3, y² = -√3

  $\sf{BC=\sqrt{(3-3)^{2}+(-\sqrt{3}-\sqrt{3} )^{2} } }$

  $\sf{BC=\sqrt{0^{2}+(-2\sqrt{3} )^{2} } }$

  $\sf{BC=\sqrt{4\times 3} }$

  BC = √12------(2)

→ Now find the distance between CA

  where x₁ = 3, x₂ = 0, y₁ = -√3, y₂ = 0

  $\sf{CA=\sqrt{(3-0)^{2}+(-\sqrt{3}-0)^{2} } }$

  $\sf{CA=\sqrt{9+3} }$

  CA = √12------(3)

→ From equations 1 , 2 , 3 we can see that RHS are equal, hence LHS must also be equal.

→ Therefore AB = BC = CA

→ Hence ABC is an equilateral triangle

→ Hence proved.

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The distance between two points is given by,

   $\sf {Distance=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2} } }$

Answer 2

Answer:

l think you get your answer