Check out some similar questions!

daddysredneck24 · Jan 18, 2011, 09:34 PM

I'm having trouble understanding how to get an answer for this problem, I got the answer (I looked in the back of the book for help, hoping It would help me) but I don't know how to get that answer... OH by the way its distance formula... etc..

Problem:
Find all poings having an x-coordinate of 2 whose distance from the point (-2, -1) is 5.

the right answer is:
92, 2); (2, -4)

Unknown008 · Jan 18, 2011, 10:40 PM

Let a point have the coordinates (x, y) which you don't know.

However, you know that the x coordinate is 2, hence, the coordinates of the point(s) become (2, y).

Now, the distance formula is:

$Distance = \sqrt{(x - x_o)^2 + (y - y_o)^2}$

This distance is 5, x = 2, y =? xo = -2, yo = -1. Substitute those into the distance formula:

$5 = \sqrt{(2 - (-2))^2 + (y - (-1))^2}$

Can you complete it now? Solve for y!

daddysredneck24 · Jan 19, 2011, 02:52 PM

Thanks so much... but I'm still a little bit confused, every time I try to solve it I end up with a weird answer.. .

Unknown008 · Jan 19, 2011, 11:12 PM

Would you be so kind as to show what you did? We can help you stop your mistakes so that you know where to be more careful next time :)

daddysredneck24 · Jan 19, 2011, 11:54 PM

I finally figured out half of it I think. Here's what I put:

5 =sqrt(2-(-2))^2+(y-(-1))^2)
5 =sqrt(4)^2+(y+1)^2
5 =sqrt(16+y^2+2y+1)
5 =sqrt(17+y^2+2y)
25 =17+y^2+2y -------- (I squared both sides to cancel out the sqrt)
8 =y^2+2y
4y =y^2
y=2
(2,2)
that's how I figured it... Please correct my mistakes if you can

Unknown008 · Jan 20, 2011, 12:01 AM

Those are good procedures up to this line:

$8 =y^2+2y$

This is a quadratic equation, and so, it cannot become

$4y =y^2$

This, instead becomes:

$0 =y^2+2y - 8$

Then, factorise:

$0 =(y - 2)(y +4)$

This done, you get:

$y - 2 = 0$ or $y + 4 = 0$

This gives you two answers, which are what the given answers say (You already know the x coordinate, which is x = 2) :)

daddysredneck24 · Jan 20, 2011, 06:47 PM

OH! Ok! Thank you SO much... I completely understand now... thanks again!

Unknown008 · Jan 20, 2011, 09:31 PM

You're welcome :)