Question

In a equalitaral triangle two vertices are (0,3),(4,3) find the co-ordinates of the third vertice.​

Answer 1

Given

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).

Answer 2

$\bold{all \: side \: of \: the \: equilateral \: triangle \: is \: the \: same, \:}$

So,

$\bold{\sqrt{(x - 3 {)}^{2} + (y - 4 {)}^{2} } = }$

$\bold{ \sqrt{(x + 2 {)}^{2} + (y - 3 {)}^{2} } = } \bold{\sqrt{(5 {)}^{2} + (1 {)}^{2} } }$

(x−3)² + (y−4)² =(x+2)² + (y−3)² = 26

x² + y² − 6x − 8y = 1

x² + y² = 6x + 8y + 1 -------------(1)

------------------------------------

26 = x² + y² + 4x − 6y + 11

x² + y² + 4x − 6y = 15

x² + y² = −4x + 6y + 15 ----------(2)

-----------------------------------

x² + 9 −6x + y² + 16 −8y = x² +y² + 4x −6y + 11

10x + 3y = 12 ----------------(3)

-----------------------------------

In equation (1) & (2) RHS is equal , therefore ;

6x + 8y + 1 = −4x + 6y + 15

10x + 2y = 14--------------(4)

__________________

Solving equation (3) & (4)

y = −2

x=18

==============================

$\bold{Hence \: the \: third \: vertex \: of \: the \: triangle \: is}\bold\red{ (−2,18)}$