Question

If (-4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of
the third vertex, given that the origin lies in the interior of the triangle.

Answer 1

let the given vertices are A(-4,3) and B(4,3)
since y-axis is perpendicular bisector of AB,therefore point C lies on the y-axis.
Let the coordinates of C(0,y)
AB^2=(-4-4)^2+(3-3)^2
AB^2=8^2=64
SINCE ABC is an equilateral triangle
AC^2=AB^2=BC^2
SINCE AC^2=AB^2
(0+4)^2+(y-3)^2=64
16+y^2-6y+9=64
y^2-6y-39=0
solving the quadratic eqn (1) by the formula
      6+(or)-√36+156                   6+(or)-8√3                     
y=--------------------------------- = ----------------------------- =  3+(or)-4√3
             2                                    2
since the origin lies interior of the triangle therefore
y=3+(or)-4√3
hence the coordinates of third vertex is
(o,3+(or)-4√3)

Answer 2

Let the co-ordinate of third vertex be (x, y) 
Now Using Distance formula BC = [4 - (- 4)] 2 + (3 - 3) 2       = (4 + 4) 2 + 0 BC = 8 2= 8 Now , AB = [x - (- 4)] 2 + (y - 3) 2          AB = (x + 4) 2 + (y - 3) 2 and AC = (x - 4) 2+ (y - 3) 2 Given, ΔABC is equilateral triangle
∴ AB = AC = BC
Now, AB = AC ⇒ (x + 4) 2 + (y - 3) 2   = (x - 4) 2 + (y - 3) 2
On Squaring both sides, we get
(x + 4)2 + (y – 3)2 = (x – 4)2 + (y – 3)2
(x + 4)2 = (x – 4)2
or x 2 + 16 + 8x = x 2 + 16 – 8x
⇒ 16x = 0
x = 0  ....(1)
AC = BC implies that (x - 4) 2 + (y - 3) 2 = 8(0 - 4) 2 + (y - 3) 2 = 8                [from (1)]
On squaring both sides, we get
16 + y 2 + 9 – 6y = 64
y 2 – 6y – 39 = 0 y = -(-6) ± (- 6) 2 - 4(1)(-39) 2(1) y = 6 ± 36 + 156 2 = 6 ± 192 2 y =  6 ± 8 3 2 = 3 ± 4 3 ∴ y = 3 + 4√3 and 3 - 4√3 y ≠ 3 + 4 √3 , as origin lies in the interior of the triangle. Third vertex = (x, y) = (0, 3 - 4√3).