Question

find a relation between x and y such that (x,y) is equidistance from (7,1) and (3,5)

Answer 1

Let the point be (x,y) which is equidistant form(7.1) and (3,5)=> we get
${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$
$\sqrt{ {(x - 3)}^{2} + {(y - 5)}^{2} } =$
$\sqrt{ {(x - 7)}^{2} + {(y - 1)}^{2} }$
${x}^{2} - 6x + 9 + {y}^{2} - 10y + 25 =$
${x}^{2} - 14x + 49 + {y}^{2} - 2y + 1$
$- 6x - 10y + 34 = - 14x - 2y + 50$
$= > 14x - 6x - 10y + 2y = 50 - 34 = 16$
$= > 8x - 8y = 16 = > x - y = 2$
is the equation of a point equidistant from(3,5)and (7,1).

Answer 2

Answer:

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