Question

if(-4,3)and (4,3) are the vertices of an equilateral triangle , find the coordinates of the thrid vertex,given that the origin lies in the interior of the triangle

Answer 1

A(-4,3) (x1,y1)
B(4,3) (x2,y2)
C(x,y)
as ABC is equilateral
therefore AB = BC = AC
to find AB:
using distance formula,
AB = √[(x1-x2)²+(y1-y2)²]
= √[(-4-4)²+(3-3)²]
= √[64+0]
= 8 units
AB = BC = AC = 8 units

as ar(∆) = ½[x1(y2-y) + x2(y-y1) + x(y1-y2)]
and ar(eq. ∆) = √3/4 * a²
therefore,
$\frac{ \sqrt{3} }{4} 8 {}^{2} = \frac{1}{2} ( - 4(3 - y) + 4(y - 3) + x(3 - 3)) \\ 32 \sqrt{3} = - 12 + 4y + 4y - 12 + 0x \\ 32 \sqrt{3} = 8y - 24 \\ 4 \sqrt{3} = y - 3 \\ y = 3 + 4 \sqrt{3}$
using distance formula,
AC = √[(x1-x)²+(y1-y)²]
4 = √[16+x²+8x + 48]
4 = √[x²+8x + 64]
4 = √[x + 4]²
x = 0

but since origin is in interior of triangle
therefore y coordinate must be ( -3-4√3)

therefore the coordinates are (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).