Question

plz anyone answer my question

Answer 1

Answer:  The coordinates of B = ( 4,4 )

The coordinates of C = ( -4,4 )

The coordinates of A = (0, 4√3)

Step-by-step explanation:

Here ABC is an equilateral triangle,

In which  AB = BC = CA = 8 units,

Let D is the mid point of the base BC.

According to the question,

D≡(0,0)

Let the coordinates of B = (b,0) and the coordinates of C = (c,0) ( because BC is in x-axis thus, in both cases the y-coordinate will be 0)

Since, DB = 4 and DC = 4 ( Because DB = DC = BC/2)

By distance formula,

$\sqrt{(0-b)^2+(0-0)^2} = 4$ and $\sqrt{(0-c)^2+(0-0)^2} = 4$

⇒ b = -4 or 4  and c = -4 or 4

But point B is in negative x-axis and C is in positive x-axis,

Thus, b = - 4 and c = 4

⇒ Coordinates of B = (-4,0) and Coordinates of C = (4,0)

Now, Let the coordinates of A = (0,a) ( Since A is in y-axis, therefore the x-coordinate of A will be zero)

Since, AD is the median of the equilateral triangle ABC of side 8.

Thus, AD = 8√3/2 = 4√3

Again by the distance formula,

$\sqrt{(0-0)^2+(0-a)^2} = 4\sqrt{3}$

⇒ a = -4√3 or 4√3

But point A is in positive y-axis,

Thus, the coordinates of A = ( 0, 4√3)