Question

if(-4,3) and (4,3)are two vertics of equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle​

Answer 1

Step-by-step explanation:

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

Given :---

• 2 vertices of Equilateral ∆ = (-4,3) and (4,3)
• Origin lies in the interior of the ∆ .

Concept and Formula used :---

→ Since , all sides of Equilateral ∆ are Equal in Length , We will compare length of distance of all sides .

→ distance between (x1,y1) & (x2,y2) is = √(x2-x1)² + (y2-y1)²

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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)

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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

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

y ≠ 3+4√3 .

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