find a point P on y axis which is equidistant from the point A (48) and B (minus 66) also find the distance of AP ​

1 thought on “find a point P on y axis which is equidistant from the point A (48) and B (minus 66) also find the distance of AP ​”

1. Answer:

   As the point is on y axis therefore let the coordinates of P be( 0,y)

   As it is equidistant from A(4,8) and B(-6,6),by distance formula

   (0-4)^2+(y-8)^2= (0+6)^2+(y-6)^2

   i.e. 16 + y^2 – 16y+ 64 = 36 + y^2 – 12y + 36

   i.e -16y + 80 = 72 – 12y

   i.e 80-72 = -12y + 16y

   i.e. 8 = 4y

   i.e y = 8/4

   i.e y = 2

   Therefore the coordinates of P are 0 and 2

   i.e P(0,2)

   Now distance between P(0,2) and A(4,8) by distance formula

   AP= √(4-0)^2+(8-2)^2

   i.e AP = √16+ 36

   i.e AP = √52

   i.e AP is √52= 2√13

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