Find all points on the y-axis that are 6 units from the point (4, -3).

Expert Answers

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If we have two points (a,b) and (c,d) the distance(R) between them is given by;

`R = sqrt((a-c)^2+(b-d)^2)`


So it is given that the all the second points should be on y axis. This means the x coordinates are 0 at these places.

`R = 6`

`a = 4`

`b = -3`

`c = 0`

`d = ?`


`R = sqrt((a-c)^2+(b-d)^2)`

`6 = sqrt((4-0)^2+(-3-d)^2)`

`36 = 16+(-3-d)^2`

`20 = (3+d)^2`

`(3+d) = +-sqrt20`

`(3+d) = +-2sqrt5`

      `d = +-2sqrt5-3`


`d= 2sqrt5-3` OR `d = -sqrt5-3`


So the answers that satisfies the situation are `y = 2sqrt5-3` or y = `-2sqrt5-3` . The points are `(0,2sqrt5-3 )` and `(0,-2sqrt5-3)` .


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You should remember that all points found on y axis have x coordinate 0, hence, since the problem provides the value of distance beween the points, you may evaluate the missing coordinates using distance formula such that:

`d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)`

`6 = sqrt((4-0)^2 + (3-y_1)^2)`

You need to raise to square both sides such that:

`36 = 16 + (3-y_1)^2 => (3-y_1)^2 = 36-16 => (3-y_1)^2 = 20`

`3-y_1 = +-sqrt(20) => y_1 = 3-2sqrt5 or y_1 = 3+2sqrt5`

Hence, evaluating the missing coordinates of the points that follow the given conditions yields `(0,3+2sqrt5)`  and `(0,3-2sqrt5).`

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