Question

If 4,3 and -4,3 are vertices of equilateral triangle, find coordinates of third vertex. Given origin lies inside triangle

Answer 1

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) 2(1) y = 6 ± 36 + 156 2 = 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).