Answers

2015-03-13T17:27:15+05:30

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)







2 5 2
if u cant understand it then iam very sorry
I understood it. Thank you
ur welcome