Question

show that the points
the points (a, a) (-a-a) and
(-a√3,a√3) are the vertices of an aquilateral
trangle.​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Proved by using Distance formula

Step-by-step explanation:

Let vertices of the triangle be :

A(x1,y1)≡(a,a)

B(x2,y2)≡(-a,-a)

C(x3,y3)≡(-a√3,a√3)

Distance formula (for distance between two points): √((x2-x1)²+(y2-y1)²)

Now,

AB=√((-a-a)²+(-a-a)²)=√((-2a)²+(-2a)²)=√(4a²+4a²)=√8a²=2√2a

BC=√((-a√3-a)²+(a√3-a)²)=√((4a²+2√3a²)+(4a²-2√3a²))=√(4a²+4a²)=2√2a

AC=√((a+a√3)²+(a-a√3)²)=√((4a²+2√3a²)+(4a²-2√3a²))=√(4a²+4a²)=2√2a

(For obtaining BC and AC, use the following expansions:

• (a+b)²=a²+2ab+b²
• (a-b)²=a²-2ab+b²
• Note that the '-' sign can be taken out as common and then squaring can be done, so that the sign becomes positive '+' on squaring.)

Now, we know that for an equilateral triangle, all the sides should be equal.

Thus, from the above calculations, AB=BC=AC.

⇒The given triangle is an equilateral triangle.

Hence, proved.

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