Question

plz help me out fast...
If (-4,3)and (4,3) are two vertices of an equilateral triangle, find the coordinate of third vertex, given that the origin lies (1)interior and (2)exterior of the triangle.​

Answer 1

Answer:

hear is the answer

Step-by-step explanation:

Let the co-ordinate of third vertex be (x, y)

Now Using Distance formula

BC =  √(4-(-4)^2 +(3-3)^2

√ (4 + 4) ^2 + 0

BC =  √ 8 ^2 = 8

Now , AB =  √ [x - (- 4)]^ 2 + (y - 3) ^2

AB =  √ (x + 4)^ 2 + (y - 3)^ 2  

and AC =  √ (x - 4)^ 2 + (y - 3)^ 2  

Given, ΔABC is equilateral triangle

∴ AB = AC = BC  

Now, AB = AC

⇒  √ (x + 4) ^2 + (y - 3) ^2 =  √ (x - 4) ^2 + (y - 3) ^2

On Squaring both sides, we get

(x + 4)^2 + (y – 3)^2 = (x – 4)^2 + (y – 3)^2

(x + 4)^2 = (x – 4)^2

or x 2 + 16 + 8x = x 2 + 16 – 8x

⇒ 16x = 0

x = 0  ....(1)

AC = BC implies that

√ (x - 4) ^2 + (y - 3) ^2 = 8

√ (0 - 4) ^2 + (y - 3) ^2 = 8                        [from (1)]

On squaring both sides, we get

16 + y^2 + 9 – 6y = 64

y 2 – 6y – 39 = 0

y =  -(-6) ±  √ (- 6) ^2 - 4(1)(-39)     by

                    2(1)

y =  6 ± √ 36 + 156

              2

Unknown node type: br = 6 ± √ 192

                                              2

y =  6 ± 8 √ 3

         2

=  3 ± 4 √ 3

∴ y = 3 + 4√3 and 3 - 4√3

y ≠ 3 + 4 √3 , as origin lies in the interior of the triangle.

Third vertex = (x, y) = (0, 3 - 4√3).

Answer 2

Solution :---

let the Third vertices be (x,y)

then Distance between (x,y) & (4,3) is :--

→ √(x-4)² + (y-3)² ---------------- Equation (1)

and Distance between (x,y) & (-4,3) is :-----

→ √(x+4)² + (y-3)² ---------------- Equation (2)

Distance between (4,3) &(-4,3) is :-------

→ √(4+4)² + (3-3)² = 8 units. ---------------- Equation (3)

Now, since, Distance Between them all is Equal , as it is Equaliteral ∆.

so, Equation (1) = Equation (2)

→ √(x-4)² + (y-3)² = √(x+4)² + (y-3)²

→ (x-4)² = (x+4)²

→ x² - 8x + 16 = x² +8x +16

→ 16x = 0

→ x = 0

And, also , Equation (1) = Equation (3)

→ √(x-4)² + (y-3)² = 8

Squaring both sides

→ (x-4)² + (y-3)² = 64

Putting value of x = 0, now,

→ (y-3)² = 64-16

→ (y-3)² = 48

Square - root both sides now,

→ (y-3) = ±4√3

→ y = ±4√3 + 3

Now, as origin lies in the interior of the triangle,

y ≠ 3+4√3 .

∴ Third vertex = (x, y) = (0, 3 - 4√3).