Question

find the value of x, y if the distance of the point x, y from - 3, 0 as well as from 3, 0 are 4​

Answer 1

Let the points be P(x,y)

Q(-3,0)

R(3,0)

Distance PQ = 4

$\sqrt{(y - 0) {}^{2} + (x - ( - 3)) {}^{2} } \\ = \sqrt{ {y}^{2} + {(x + 3)}^{2} } = 4$

Distance PR = 4

$\sqrt{(y - 0) {}^{2} + (x - ( 3)) {}^{2} } \\ = \sqrt{ {y}^{2} + {(x - 3)}^{2} } = 4$

Since PQ = PR = 4, we equate both

$\sqrt{ {y}^{2} + {(x + 3)}^{2} } = \sqrt{ {y}^{2} + {(x - 3)}^{2} } \\ {y}^{2} + {(x + 3)}^{2} = {y}^{2} + (x - 3) {}^{2} \\ {x}^{2} + 9 + 6x = {x}^{2} + 9 - 6x \\ 12x = 0 \\ x = 0$

Substituting x = 0 in any equation,

$\sqrt{ {y}^{2} + {(0+ 3)}^{2} } = 4 \\ {y}^{2} + 9 = 16 \\ {y}^{2} = 7 \\ y = \frac{ + }{} \sqrt{7}$

Therefore the point P(x,y) = P(0,√7) or P(0,-√7)

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