Question

show that the following points from a equilateral triangle A(a,0)B(-a,0)C(0,a√3)​

Answer 1

Step-by-step explanation:

The given points are vertices of an equilateral triangle because the length of each sides are same.

Step-by-step explanation:

The given vertices are A(a,0), B(-a,0), C(0,a√3).

All sides of a equilateral triangle are same.

The given vertices form an equilateral triangle if

AB=BC=ACAB=BC=AC

The distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}d= </p><p>(x </p><p>2</p><p> </p><p> −x </p><p>1</p><p> </p><p> ) </p><p>2</p><p> +(y </p><p>2</p><p> </p><p> −y </p><p>1</p><p> </p><p> ) </p><p>2</p><p> </p><p> </p><p> </p><p></p><p>AB=\sqrt{(-a-a)^2+(0-0)^2}=2aAB= </p><p>(−a−a) </p><p>2</p><p> +(0−0) </p><p>2</p><p> </p><p> </p><p> =2a</p><p></p><p>BC=\sqrt{(0-(-a))^2+(a\sqrt{3}-0)^2}=\sqrt{a^2+3a^2}=2aBC= </p><p>(0−(−a)) </p><p>2</p><p> +(a </p><p>3</p><p> </p><p> −0) </p><p>2</p><p> </p><p> </p><p> = </p><p>a </p><p>2</p><p> +3a </p><p>2</p><p> </p><p> </p><p> =2a</p><p></p><p>AC=\sqrt{(0-(a))^2+(a\sqrt{3}-0)^2}=\sqrt{a^2+3a^2}=2aAC= </p><p>(0−(a)) </p><p>2</p><p> +(a </p><p>3</p><p> </p><p> −0) </p><p>2</p><p> </p><p> </p><p> = </p><p>a </p><p>2</p><p> +3a </p><p>2</p><p> </p><p> </p><p> =2a</p><p>$

Since AB=BC=CA, therefore given points are vertices of an equilateral triangle.