Question

-4,3&4,3are two vertices of an equilateral triangle,find the coordinates of the 3rd vertex,given that the origins lies in the 1.interior,2.exterior of triangle

Answer 1

Step-by-step explanation:

where is the diagram

please show the diagram since diagram is required.

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).