Question

In an equilateral triangle ABC if the coordinates of B and C are (3, 0) and (-3, 0) then the vertex A will be?

please answer with steps​

Answer 1

Answer:

Step-by-step explanation:

here we have to find the coordinates vertex of A. since A is in the Y axis. the x coordinate is 0. so, vertex of A is (0,y).now we have find the value of y. given the triangle is equilateral. hence

distance of BC= distance of AB

=> (3--3)^2+(0-0)^2=(3-0)^2+(0-y)^2

=> 36= 9+ y^2

=> y^2= 25

=> y =5

hence coordinate vertex of A is (0,5)

hope this helps you.

Answer 2

$\huge\rm\underline\pink{\underline{Solution:-}}$

Given vertices of B and C of an equilateral triangle are (3,0) , (-3,0)

Let coordinates of A be (x₃,y₃)

Since the given vertex is equilateral length of every side (or) Distance between the vertices is equal .

Distance between the two points (x₁,y₁) , (x₂,y₂) is given by ,

$\large\rm\bold{\boxed{D\:=\:\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}}$

Distnace between the Vertex B (3,0) and C (-3,0) is

$\large\rm{D\:=\:\sqrt{(-3-3)^2+(0-0)^2}}$

$\large\rm{\implies{D\:=\:\sqrt{36}\:=\:6\:units}}$

This mean the distance between all vertices is 3 units

So distance between AB = Distance between BC = Distance between CA = 6 units

Distance between C(-3,0) and A(x₃,y₃) is

$\large\rm{D\:=\:\sqrt{(x_3-(-3))^2+(y_3-0)^2}}$

$\large\rm{\implies{3\:=\:\sqrt{(x_3+3)^2+(y_3)^2}\rightarrow\:eq(1)}}$

Distance between A(x₃,y₃) and B(3,0) is

$\large\rm{6\:=\:\sqrt{(3-x_3)^2+(0-y_3)^2}}$

$\large\rm{\implies{6\:=\:\sqrt{(3-x_3)^2+(y_3)^2}\rightarrow\:eq(2)}}$

Since in both equations LHS is equal RHS can be equated,

$\large\rm{\implies{\sqrt{(x_3+3)^2+(y_3)^2}\:=\:\sqrt{(3-x_3)^2+(y_3)^2}}}$

$\large\rm\bold{\boxed{(a+b)^2\:=\:a^2+b^2+2ab}}$

$\large\rm\bold{\boxed{(a-b)^2\:=\:a^2+b^2-2ab}}$

$\large\rm{\implies{(x_3)^2+9+6x_3+(y_3)^2\:=\:(x_3)^2+9-6x_3+(y_3)^2}}$

$\large\rm{\implies{(x_3)^2+(y_3)^2-(x_3)^2-(y_3)^2+6x_3+6x_3\:=\:0}}$

$\large\rm{\implies{12x_3\:=\:0}}$

$\large\rm{\implies{x_3\:=\:0}}$

From eq(1) ,

$\large\rm{3\:=\:\sqrt{(x_3+3)^2+(y_3)^2}}$

Squaring on both sides ,

$\large\rm{\implies{6\:=\:(x_3+3)^2+(y_3)^2}}$

$\large\rm{\implies{36\:=\:9+(y_3)^2}}$

$\large\rm{\implies{(y_3)^2\:=\:27}}$

$\large\rm{\implies{y_3\:=\:3\sqrt{3}}}$

∴ The coordinates of vertex A are (0,3√3)