Question

In an triangle ABC is an equilateral triangle of side 3 find the co-ordinate of the other two vertices

Answer 1

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Answer 2

Answer:

The co-ordinates of C are  $\frac{7}{2} , \frac{3\sqrt{3} }{2}$, and A= (2,0), B= (5,0)

Step-by-step explanation:

Let co-ordinate of

$A= (2,0)$

Hence it is equilateral triangle of 3 unit. so,

$B= (5,0)$

Let co-ordinate of $C$ are ($x$, $y$)

$AC = BC\\Ac^{2} = Bc^{2} \\(x-2)^{2} +( y-0)^{2} = (x-5)^{2} +(y-0)^{2} \\x^{2} + 4 -4x + y^{2} = x^{2} + 25 - 10x + y^{2} \\-4x + 10x = 25 -4 \\6x = 21\\x = \frac{7}{2\\}$

now,

$AC= 3 \\Ac^{2} = 9 \\y = \sqrt{\frac{27}{4} }$$=\frac{3}{2} \sqrt{3}$

As y is in first quadrant, it cannot  be negative so the mentioned answer would be applicable.

To know more about quadrants, check out:

https://brainly.in/question/28784290

To know more about equilateral triangle, check out:

https://brainly.in/question/16653174

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