# Find all the points having a y-coordinate of -6 whose distance is from the point (1,2) is 17

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

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

For our question;

R = 17

a = 1

b = 2

c = ?

d = -6

`17 = sqrt[(1-c)^2+(2+6)^2]`

`17^2 = (1-c)^2+(8)^2`

`0 = c^2-2c-224`

Solution to the above quadratic equation is given by;

`c = ((-(-2)+-sqrt((-2)^2-4*1*(-224)))/(2*1))`

c = 16 or c = -14

*So the x coordinates that satisfies the situation will be x = 16 and x = -14*

**Sources:**

You should use distance formula such that:

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

Notice that the problem provides the both y coordinates `y_1 = 2` , `y_2 = -6` , one x coordinate, `x=1` , and the distance between the points, `d = 17` , hence, you should sbstitute these values in the formula of distance such that:

`17 = sqrt((x_2 - 1)^2 + (2 + 6)^2)`

`17 = sqrt((x_2 - 1)^2 + 64)`

You need to raise to square both sides such that:

`289 = (x_2 - 1)^2 + 64 =>(x_2 - 1)^2 = 289 - 64`

`(x_2 - 1)^2 = 225 => x_2 - 1 = +-sqrt(225) => x_2 - 1 = +-15`

`x_2 = 15 + 1 => x_2 = 16`

`x_2 = -15 + 1 => x_2 = -14`

**Hence, evaluating the x coordinate of the points that follow the given conditions yields `(16,-6)` and `(-14,-6).` **