# Thread: Quick Finding Y Values Question (algebra 1)

1. ## Quick Finding Y Values Question (algebra 1)

I have one homework question I can't seem to get it!! I'm looking for the answers thank you.

It says, Find all values of Y such that the distance between (9,y) and (-4,6) is 17.

Y=???

THANK YOU SO MUCH

2. use the distance formula ...

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

plug in your known values and solve for the missing value of y ... note that you will get two solutions.

3. I'm looking for the answers thank you.

4. We use the distance formula to find the distance between two points:
$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given the distance, a point (name it A), and the x-coordinate of another point (name it B), we can find the y-coordinate of B using the distance formula.

d = 17
A(-4,6) = $(x_2, y_2)$
B(9, y) = $(x_1, y_1)$

$(17)^2 = (-4 - 9)^2 + (6 - y)^2$

$289 = 169 + 36 - 12y + y^2$

$y^2 - 12y - 84 = 0$

Solve for y now. I will complete the square.

$y^2 - 12y + 36 - 36 - 84 = 0$

$(y - 6)^2 = 120$

$y - 6 = \pm\sqrt{120}$

$y = 6\pm\sqrt{120}$

EDIT: It's all good.

5. k thats wrong it said

6+2[30

6-2[30

damnit.

6. No it is not wrong.

$\sqrt{120} = \sqrt{4\cdot30} = 2\sqrt{30}$