Expert: Sherman D. - 9/30/2006

Question
Find the relationship relating x and y if C is the point (1,7), D is the point (1,13), Q is the point with coordinates (x,y), we know that
abs(DQ)=2*abs(CQ).

Answer
DQ = sqrt((x - 1)^2 + (y - 13)^2)
CQ = sqrt((x - 1)^2 + (y - 7)^2)

so

|sqrt((x - 1)^2 + (y - 13)^2)| = 2|sqrt((x - 1)^2 + (y - 7)^2)|

this is the same as saying

|sqrt((x - 1)^2 + (y - 13)^2) = |sqrt(4((x - 1)^2 + (y - 7)^2))|

now lets just get rid of the absolute value marks, because we all know that it needs to be a positive distance.

sqrt((x - 1)^2 + (y - 13)^2) = sqrt(4((x - 1)^2 + (y - 7)^2))

get rid of the sqrt since both of them have sqrt

(x - 1)^2 + (y - 13)^2 = 4((x - 1)^2 + (y - 7)^2)
(x - 1)^2 + (y - 13)^2 = 4(x - 1)^2 + 4(y - 7)^2
(x - 1)^2 - 4(x - 1)^2 = 4(y - 7)^2 - (y - 13)^2
(1 - 4)(x - 1)^2 = 4(y - 7)^2 - (y - 13)^2
-3(x - 1)^2 = 4(y - 7)^2 - (y - 13)^2

-3(x - 1)(x - 1) = 4((y - 7)(y - 7)) - ((y - 13)(y - 13))
-3(x^2 - x - x + 1) = 4(y^2 - 7y - 7y + 49) - (y^2 - 13y - 13y + 169)

-3(x^2 - 2x + 1) = 4(y^2 - 14y + 49) - (y^2 - 26y + 169)
-3x^2 + 6x - 3 = 4y^2 - 56y + 196 - y^2 + 26y - 169
-3x^2 + 6x - 3 = (4 - 1)y^2 + (-56 + 26)y + (196 - 169)
-3x^2 + 6x - 3 = 3y^2 - 30y + 27
3(-x^2 + 2x - 1) = 3(y^2 - 10y + 9)
-x^2 + 2x - 1 = y^2 - 10y + 9
-(x^2 - 2x + 1) = y^2 - 10y + 9
-(x - 1)(x - 1) = (y - 9)(y - 1)
-(x - 1)^2 = (y - 9)(y - 1)
(x - 1)^2 = -(y - 9)(y - 1)
x - 1 = sqrt(-(y - 9)(y - 1))
x = sqrt(-(y - 9)(y - 1)) + 1
x = sqrt(-(y^2 - 10y + 9)) + 1
x = sqrt(-y^2 + 10y - 9 + 1)

x = sqrt(-y^2 + 10y - 8)

Using the quadratic formula

y = 5 ± sqrt(17)

By adding the y-intercept, you get 10, and if you divide that by 2, you get 5, thats just the maximum y value, you plug that in for y, and you get sqrt(17), which would be the maximum domain.

Domain : 0 < x < sqrt(17)
Range : 5 - sqrt(17) < y < 5 + sqrt(17)

for a graph, go to www.quickmath.com, click on Plot under Equations, then type in y = sqrt(-x^2 + 10y - 8), because using y instead of x, it won't show up like its suppose to.

You should get like a half circle shape.

the domain might be -sqrt(17) < x < sqrt(17) instead, but the range remains the same.