Question

If two vertices of an equilateral triangle are (3,0) and (6,0),find the third vertex.

Answer 1

Let ABC be the equilateral triangle such that,
A=(3,0),B=(6,0) and C=(x,y)
Distance formula:
√{(x2-x1)^+(y2-y1)^2}
we know that ,
AB=BC=AC
By distance formula we get,
AB=BC=AC=3units
AC=BC
√{(3-x)^2+y^2}=√{(6-x)^2+y^2}
9+x^2-6x+y^2=36+x^2-12x+y^2
6x=27
x=27/6=9/2
BC=3units
√{(6-27/6)^2+y^2}=3
{(36-27)/6}^2+y^2=9
(9/6)^2+y^2=9
(3/2)^2+y^2=9
9/4+y^2=9
9+4y^2=36
4y^2=27
y^2=27/4
y=√(27/4)
y=3√3/2
(x,y)=(9/2,3√3/2)
Hence third vertex of equilateral triangle=C=(9/2,3√3/2)

Answer 2

Let A (3,0) , B(6,0) and C(x,y)
ABC will be triangle when
AB = BC = CA
use distance formula,
AB = √{(6-3)²+0}= 3
BC = √{(x-6)² + y²}
CA = √{(x -3)² + y²}

Now, AB = BC
then, 3 = √{(x -6)²+y²}
9 = (x -6)² + y² ----(1)

BC = CA
Then, √{(x-6)²+y²}=√{(x-3)²+y²}
(x-6)² + y² = (x-3)² + y²
x² -12x + 36 = x² - 6x + 9
-12x + 36 = -6x + 9
-6x = -27
x = 9/2

now, put x in equation (1)

9 = (9/2-6)² + y²
9 = 9/4 + y²
27/4 = y²
y = ±3√3/2

Hence, C(x,y) =(9/2,3√3/2) and (9/2,-3√3/2)