Question

ABC is an equilateral triangle with coordinates of B and C as B(-4,0) and C(4,0). Find the coordinate of vertex A.

Answer 1

Answer:

clearly A lies on y-axis. because mid point of BC is origin.

Now length of BC = 8 = AB = AC

So AO =

$\sqrt{ {8}^{2} - {4}^{2} } = \sqrt{64 - 16} = \sqrt{48} = 4 \sqrt{3}$

so coordinate of A is (0 , 4√3) or (0 , -4√3)

Thanks for the question.

mark BRAINLIEST if this is helpful to you.

Answer 2

Given : -

ABC is an equilateral triangle .

The co-ordinates of B , C are B(-4,0) & C(4,0)

Required to find : -

• Co-ordinates of the vertex A ?

Formula used : -

Here , we need to use the distance formula to solve this question .

Distance Formula

Distance between 2 points = √[(x1 - x)² + (y1 - y)²]

Solution : -

∆ ABC is an equilateral triangle .

The co-ordinates of B , C are B(-4,0) & C(4,0)

Recall the properties of an equilateral triangle .

According to which ;

In a equilateral ∆ , all sides are equal .

=> AB = BC = AC

Using this logic let's solve this question !

The co-ordinates of the 3 vertices are ;

• A = (x,y)

• B = (-4,0) => B = (x1,y1)

• C = (4,0) => B = (x2,y2)

According to problem <<

Length of BC =

☛ √[ (x2 - x1)² + (y2 - y1)² ]

☛ √[ (- 4 - 4)² + (0 - 0)² ]

☛ √[ (-8)² + (0)² ]

☛ √[ 64 + 0 ]

☛ √[64]

☛ ± 8 units

Since, this distance is taken with respective to quadrant 1 .

so,

☛ Length of BC = 8 units

Recalling the above conclusions ;

• Length of AB = Length of AC

Length of AB = √[ (x1 - x)² + (y1 - y)² ]

Length of AB = √[ (-4 - x)² + (0 - y)² ]

Length of AB = √[ (-4)² + (x)² - 2(-4)(x) + (0)² + (y)² - 2(0)(y) ]

☛ Length of AB = √[ 16 + x² + 8x + y² ] units

Similarly,

Length of AC = √[ (x2 - x)² + (y2 - y)² ]

Length of AC = √[ (4 - x)² + (0 - y)² ]

Length of AC = √[ (4)² + (x)² - 2(4)(x) + (0)² + (y)² - 2(0)(y) ]

☛ Length of AC = √[ 16 + x² - 8x + y² ] units

This implies;

AB = AC

☛ √[ 16 + x² + 8x + y² ] = √[ 16 + x² - 8x + y² ]

Square root get's cancelled on both sides

☛ 16 + x² + 8x + y² = 16 + x² - 8x + y²

☛ 16 + x² + 8x + y² - 16 - x² + 8x + y² = 0

☛ 16x = 0

☛ x = 0/16

☛ x = 0

• Value of x = 0

However,

AB = BC

Length of AB = √[ 16 + x² + 8x + y² ] units

Length of BC = 8 units

☛ √[ 16 + x² + 8x + y² ] = 8

☛ √[ 16 + (0)² + 8(0) + y² ] = 8

☛ √[ 16 + 0 + 0 + y²] = 8

☛ √[ 16 + y²] = 8

Squaring on both sides

☛ ( √[ 16 + y² ] )² = (8)²

☛ 16 + y² = 64

☛ y² = 64 - 16

☛ y² = 48

☛ y = √(48)

☛ y = √(16 x 3)

☛ y = 4√3

• Value of y = 4√3

Therefore,

Co-ordinates of the vertex A = (0,4√3)

From the above we can also conclude that ;

The point A lies on the y - axis because the value of the x co-ordinate is 0 ( zero ) .