If (-5,3) and (5,3) are two vertices of an equilateral . . .

Question : If (-5,3) and (5,3) are two vertices of an equilateral triangle, then find the coordinates of the third vertex, given that origin lies inside the triangle. (Take √3=1.7)

Doubt by Bhavya

Solution : 
Let A, B and C are three vertices of an equilateral triangle. 
So, AB=BC=CA

A(-5,3) and B(5,3)
Let C(x,y) 

Using Distance Formula 

AB=√[(x₂-x₁)²+(y₂-y₁)²]
AB=√[(5+5)²+(3-3)²]
AB=√[(10)²+(0)²]
AB=√[100]
AB=10 units

Also, 
AC=BC
Squaring both sides
AC²=BC²
(x+5)²+(y-3)²=(x-5)²+(y-3)²
(x+5)²=(x-5)²
x²+10+10x=x²+10-10x
10x=-10x
10x+10x=0
20x=0
x=0/20
x=0

Also 
AC=10 
√[(x+5)²+(y-3)²]=10
Squaring both sides
(x+5)²+(y-3)²=100
(0+5)²+(y-3)²=100 [∵x=0]
25+y²+9-6y=100
y²-6y+34-100=0
y²-6y-66=0

Here
a=1
b=-6
c=-66

D=b²-4ac
D=(-6)²-4(1)(-66)
D=36+264
D=300
Using Quadratic Equation
y=[-b±√D]/2a
y=[-(-6)±√300]/2(1)
y=[6±10√3]2
y=3±5√3
It is given that origin lies inside the triangle, so y-coordinate of c must be negative.
so, 
y=3-5√3
y=3-(5×1.7)
y=3-8.5
y=-5.5

Hence, the required coordinate third vertex of the triangle is (0,-5.5)