1.

Find the coordinates of the point equidistant from three given points `A(5, 1), B(-3,-7)` and `C(7,-1).`

Answer» Let the required point be P(x, y). Then,
`PA = PB = PC rArr PA^(2) = PB^(2) = PC^(2)`
`rArr PA^(2) = PB^(2) "and "PB^(2) = PC^(2)`
`rArr {((x-5)^(2) +(y-1)^(2) = (x+3)^(2) + (y+7)^(2)), ((x+3)^(2) + (y+7)^(2) = (x-7)^(2) + (y+1)^(2)):}`
`rArr {(x^(2)+y^(2)-10x-2y+26 =x^(2) +y^(2) +6x +14y +58), (x^(2)+ y^(2) +6x +14y +58 = x^(2) +y^(2)-14x +2y +50):}`
Now, `x^(2) + y^(2) -10x -2y +26 =x^(2) + y^(2) +6x +14y +58`
`rArr 16x+ 16y =-32 rArr x+y = -2 " "...(i)`
And, `x^(2) +y^(2) +6x +14y +58 =x^(2) +y^(2) -14x +2y+50`
`rArr 20x+12y = -8 rArr 5x +3y =-2 " "..(ii)`
Multiplying (i) by 5 and subtracting (ii) from the result, we get
2y =-8 `rArr` y = -4
Putting y = -4 (i), we get x = 2.
`therefore` x = 2 and y = -4.
Hence, the required point is P(2, -4).


Discussion

No Comment Found

Related InterviewSolutions