math (algebra) help!!! how to do this problem???

Question

how would i go about to complet the problem???

it states find the value of p so that each pair is 25 units apart....
1. (8, p) , (-13,2)
2. (p, -280) , (-29, -35)
3. (-6.8, p) , (3.7, -8.9)

Answer 1

What's going to be involved in this question is the distance formula, which goes as follows:

d = √( (x2 - x1)² + (y2 - y1)² )

In this case, we're given d, and will have to solve for p.

1. (8, p) , (-13, 2). d = 25

25 = √( (x2 - x1)² + (y2 - y1)² )
25 = √( (-13 - 8)² + (2 - p)² )

Square both sides,

625 = (-13 - 8)² + (2 - p)²
625 = (-21)² + (2 - p)²
625 = 441 + (2 - p)²
184 = (2 - p)²

Taking the square root of both sides (Note: don't forget to include a ± when doing this).

±√184 = 2 - p.

Multiply both sides by (-1), to get

±√184 = p - 2

Therefore

p = 2 ± √184
p = 2 ± 2√46

This means we have *two* values for p.

2) (p, -280) , (-29, -35)

25 = √( (x2 - x1)² + (y2 - y1)² )
25 = √( (-29 - p)² + ( -35 - (-280) )² )
25 = √( (-29 - p)² + ( -35 + 280 )² )
25 = √( (-29 - p)² + (245)² )

Squaring both sides,

625 = (-29 - p)² + (245)²
625 = (-29 - p)² + 60025
-59400 = (-29 - p)²

We don't even have to continue from here, because NO value squared (the right hand side is squared) is going to equal a negative number [in the domain of real numbers]. We're not going to have any real solutions.

3. I'll get you to do #3 on your own. It's very similar to #1 and #2.

Answer 2

Substitute the values into "the distance formula"

Answer 3

The length of a line segment with end points (a1,b1) and (a2,b2) is given by sqrt{(a1-a2)^2+(b1-b2)^2}.

Just subsitute the values so that the length is 25
i.e., sqrt{(a1-a2)^2+(b1-b2)^2}=25

Answer 4

the formula for the distance between two points is:
{(x2-x1)^2+(y2-y1)^2}^1/2