Math, asked by anishmarchande5, 14 days ago

# Prove that (✓2, ✓2),(-✓2,-✓2) and (-✓6,✓6) are the vertices of an equilateral triangle​

10

Explanation,

Let, A(√2 , √2) , B(-√2 , -√2) & C(-√6 , √6).

Using distance formula,

d = (x2 - x1)² + (y2 - y1)²

Case (I) :- A(√2 , √2) , B(-√2 , -√2)

⇒ AB = √(-√2 - √2)² + (-√2 - √2)²

⇒ AB = √2(-√2 - √2)²

⇒ AB = √2(2√2)²

⇒ AB = √2 × 8

AB = 16 unitsㅤㅤㅤㅤeqn. (1)

Case (II) :- B(-√2 , -√2) & C(-√6 , √6)

⇒ BC = (-√6 + √2)² + (√6 + √2)²

⇒ BC = √(√2 - √6)² + 6 + 2√12 + 2

⇒ BC = √6 - 2√12 + 2 + 8 + 2√12

⇒ BC = √6 + 2 + 8

BC = 16 unitsㅤㅤㅤeqn.(2)

Case (III) :- A(√2 , √2) & C(-√6 , √6)

⇒ AC = √(-√6 - √2)² + (√6 - √2)²

⇒ AC = √6 + 2√12 + 2 + 6 - 2√12 + 2

⇒ AC = √6 + 2 + 6 + 2

AC = 16 units ㅤㅤㅤㅤeqn.(3)

From eqn. (1) , (2) & (3), We get ;

AB = BC = AC

.°. ΔABC is an equilateral triangle.

Hence,

(√2 , √2) , (-√2 , -√2) & (-√6 , √6) are the vertices of an equilateral triangle.

17

There are three points (√2 , √2), (-√2 , -√2) and (-√6 , √6).

Let

• A = (2 , 2)
• B = (-2 , -2)
• C = (-6 , 6)

If AB = BC = AC, then it will be an equilateral triangle.

Formula used :

• Distance = √[(x2 - x1)² + (y2 - y1)²]

As we know that, equilateral triangle has all sides equal.

In first case :

➟ AB = √[(-√2 - √2)² + (-√2 - √2)²]

➟ AB = √[(-2√2)² + (-2√2)²]

➟ AB = √[8 + 8]

➟ AB = √16

➟ AB = 4 units ...i)

In second case :

➟ BC = √[(-√6² + √2)² + (√6 + √2)²]

➟ BC = √[(-√6)² + (√2)² - 2.√6.√2 + (√6)² + (√2)² + 2.√6.√2]

➟ BC = √[6 + 2 - 2√12 + 6 + 2 + 2√12]

➟ BC = √[6 + 2 + 6 + 2 ]

➟ BC = √[8 + 8]

➟ BC = √16

➟ BC = 4 units ...ii)

In third case :

➟ AC = √[(-√6 - √2)² + (√6 - √2)²]

➟ AC = √[(-√6)² + (√2)² - 2.(-√6) .(√2) + (√6)² + (√2)² - 2.√6.√2]

➟ AC = √[ 6 + 2 + 2√12 + 6 + 2 - 2√12]

➟ AC = √[6 + 2 + 6 + 2]

➟ AC = √[8 + 8]

➟ AC = √16

➟ AC = 4 units ...iii)

From i), ii) and ii) -

• AB = BC = AC

Therefore, (2 , 2) , (-2 , -2) and (-6 , 6) are the vertices of an equilateral triangle.

Sen0rita: thanks
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