SOLUTION: Find all points having an​ x-coordinate of 2 whose distance from the point (−2,−4) is 5. I know we must use distance formula, d=√(x2-x1)^2+(y2-y1)^2

Algebra ->  Length-and-distance -> SOLUTION: Find all points having an​ x-coordinate of 2 whose distance from the point (−2,−4) is 5. I know we must use distance formula, d=√(x2-x1)^2+(y2-y1)^2       Log On


   

Question 1070513: Find all points having an​ x-coordinate of 2 whose distance from the point (−2,−4) is 5.
I know we must use distance formula, d=√(x2-x1)^2+(y2-y1)^2
I though the points would be (-2,-4) and (2,y)
It would look like 5=√(2-(-2))^2+(-4-y)^2
simplifies 5 = √(4)^2+(-4-y)^2
then 5 = √16+(-4-y)^2
then square to get rid of radicals (5)^2 = ( √16+(-4-y)^2 )^2
25 = 16 + (-4-y)^2
9 = (-4 -y)^2
then square root to get rid of power of 2
√9 = √(-4-y)^2
3 = -4-y
But because +/- from squaring I would get -4-y=3 and -4-y=-3
Solving I get -7 and -1
Which would be (2,-7) and (2,-1)

I'm not sure if I'm right or messed up entirely
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
x coordinate of 2 and any y coordinate, 5 units away from the point (-2,-4).

You want some points (2,y) which is 5 units distance from (-2,-4).
sqrt%28%282-%28-2%29%29%5E2%2B%28y-%28-4%29%29%5E2%29=5
Work with this equation, to solve for y.

sqrt%28%284%29%5E2%2B%28y%2B4%29%5E2%29=5
sqrt%2816%2B%28y%2B4%29%5E2%29=5

square both sides,
16%2B%28y%2B4%29%5E2=25

y%2B4=0%2B-+3
system%28y=-4-3%2Cor%2Cy=-4%2B3%29

system%28y=-7%2Cor%2Cy=-1%29
For the points (2,-7) and (2,-1).