Find all the points having a y-coordinate of -6 whose distance is from the point (1,2) is 17
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calendarEducator since 2012
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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
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calendarEducator since 2011
write5,348 answers
starTop subjects are Math, Science, and Business
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).`