Question

if (-4,0)and (4,0) are two vertices of an equilateral find coordinates of its third vertex.​

Answer 1

Answer:

Let C(x,y) be the third vertex of triangle ABC having two vertices at A(-4,0) and B(4,0).Since △ABC is equilateral. Therefore,

AC=BC=AB

Now,AC=BC

⇒

(x+4)

2

+(y−0)

2

=

(x−4)

2

+(y−0)

2

⇒(x+4)

2

+y

2

=(x−4)

2

+y

2

⇒16x=0

⇒x=0

Again

,AC=BC=AB

⇒AC=AB

⇒

(x+4)

2

+(y−0)

2

=

(4+4)

2

+0

2

⇒(x+4)

2

+y

2

=64

⇒(0+4)

2

+y

2

=64

⇒y

2

=48

⇒y=±4

3

Hence, the coordinates of the third vertex are C(0,4

3

)andD(0,−4

3

).

Step-by-step explanation:

hope it helped

Answer 2

GiveN:-

• Two verices of triangle are (-4,0) and (4,0) .
• The triangle is a equilateral triangle.

To FinD:-

• Coordinate of third vertex .

SolutioN :-

Let us name the points as A(-4,0) and B (4,0)

We know that in a equilateral triangle all sides are equal ,hence distance of all three points from each other will also be equal . Let us take third vertex be C( x , y ) . Now , we will use Distance Formula :-

$\boxed{\boxed{\red{\sf Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}}}$

Distance between A and B :-

⇒ D = √[ (x2 - x1)² + (y2 - y1)² ]

⇒ D = √ [ (4+4)² + (0-0)² ]

⇒ D = √ 8² + 0²

⇒D = √64

⇒ D = 8

Distance between B and C :-

⇒ D' = √ [ ( x2 - x1 )² + (y2 - y1)² ]

⇒ D' = √ [ (4 - x)² + ( y -0)² ]

⇒ D' = √ [ 4² + x² -2*4*x + y² ]

⇒ D' = √ [ 16 + x² - 8x + y²]

⇒ 8 = √ [ 16 + x² - 8x + y² ]

⇒ x² + y² - 8x + 16 = 64

⇒ x² + y² - 8x = 64 - 16

⇒ x² + y² - 8x = 48 . .......................[ i ]

Distance between C and A :-

⇒ D" = √ [ (x2 - x1 )² + (y2 - y1 )² ]

⇒ D" = √ [ ( x + 4)² + (y - 0)² ]

⇒ D" = √ [ x² + 16 + 2*4*x + y² ]

⇒ D" = √ [ x² + 16 + 8x + y² ]

⇒ 8 = √ [ x² + 16 + 8x + y² ]

⇒ 64 = x² + 16 + 8x + y²

⇒ x² + y² + 8x = 64 - 16

⇒ x² + y² + 8x = 48. ......................[ii]

Subtracting equations (i) and (ii) :-

⇒ ( x² + y² - 8x ) - ( x² + y² + 8x ) = 48 - 48

⇒ x² + y² - 8x - x² - y² - 8x = 0

⇒ -16x = 0

⇒ x = 0/(-16)

⇒ x = 0

Putting this value in (ii) :-

⇒ x² + y² + 8x = 48

⇒ 0² + y² + 8*0 = 48

⇒ y² = 48

⇒ y² = 16*3

⇒ y = √16 * √3

⇒ y = 4√3 .

$\Large\boxed{\pink{\sf Third\: coordinate=C(0,4\sqrt 3)}}$