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in the equilateral triangle ABC,coordinates of A are (-4,0) and B are (6,0).the perpendicular from C to AB meets at P.
  • draw coordinate axes and rough sketch of triangle
  • find cordinates of P
  • find length of CP
  • find coordinates OF C

Dear Student,
Let the coordinates of C be (x,y)
AC = BC  = AB      (since, equilateral triangle)(x + 4)2 + (y - 0)2 = (x - 6)2 + (y - 0)2 = (6+ 4)2 + (0- 0)2Squaring we get (x + 4)2 + y2 = (x - 6)2 +y2 = (6+ 4)2 +0 (x + 4)2 + y2 = (x - 6)2 +y2(x + 4)2  = (x - 6)2 x2 + 8x + 16 = x2 - 12x +3620x = 20x = 1Putting x = 1 in the given equation we get(x + 4)2 + y2 = (6+ 4)2 +0(1 + 4)2 + y2 = (6+ 4)225 +   y2 = 100 y2 = 75y = ±75 = ±53Hence, the coordinates of C is (1, 53) or (1,-53)As, ABC is an equilateral triangle, and CP is a perpendicular from C to AB, So P bisects AB.Hence, P is the mid point of AB.P = 6-42,0 + 02 = (1,0)Length of CP = (1 - 1)2 + ( 53- 0)2 =  ( 53- 0)2 = 75 = 53


Please mention what point F is and ask again.

Regards,

 

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