# Thread: Finding the distance between each pair of points

1. ## Finding the distance between each pair of points

Find the distance between each pair of points

(3,7) and (15,2)

I've been taught y=mx + b, slope, y-intercept, x-intercept - straight lines only, degree 1.

How do I solve the above? Since I know the values of x and y, is this finding the value of m and b?

Thanks.

2. Originally Posted by shenton
Find the distance between each pair of points

(3,7) and (15,2)

I've been taught y=mx + b, slope, y-intercept, x-intercept - straight lines only, degree 1.

How do I solve the above? Since I know the values of x and y, is this finding the value of m and b?

Thanks.
The points A=(3,7) and B=(15,2) together with C=(3,2) form a right triangle
with the segment AB forming the hypotenuse (sketch this out on paper
and you will see what is going on more clearly).

The length of AC is 5, and of BC is 12, so by Pythagoras's theorem:

AB^2 = 5^2 + 12^2 = 169,

so AB=sqrt(169) = 13.

Now that was the long way of doing it. If you look closly at what I did
you will see that the distance between any two points A=(a_1, a_2)
and B=(b_1, b_2) is:

d(A,B) = sqrt[(a_1-b_1)^2 + (a_2-b_2)^2].

RonL

3. Interesting, all the questions in the homework is about y=mx + b and this is one question that is different and I've not been taught the Pythagoras's theorem. But thanks for teaching me that.

4. Originally Posted by shenton
Interesting, all the questions in the homework is about y=mx + b and this is one question that is different and I've not been taught the Pythagoras's theorem. But thanks for teaching me that.
I'm sorry? How could you be in Pre-Calc and not have been taught the Pythagorean Theorem? You've had basic (7th grade or so) geometry haven't you?

-Dan

5. I never had high/middle school, just catching up what I've lost.

6. Originally Posted by shenton
I never had high/middle school, just catching up what I've lost.
No problem. I was just a little confused as to why this is in the "Pre-Calculus" room.

-Dan

7. Originally Posted by shenton
Interesting, all the questions in the homework is about y=mx + b and this is one question that is different and I've not been taught the Pythagoras's theorem. But thanks for teaching me that.
WHAT?!?!?!? I learned pythagoreum in 6th grade

Anyway, just to let you know, the theorum says that given a right triangle with sides a, b, and c, where c is the hypotenuse (c is the largest side) then: $a^2+b^2=c^2$

The distance formula makes it so that the distance between the two points is the hypotenuse, then the difference in height is a side, and the difference in width is a side. Thus the difference formula is: $d=\sqrt{a^2+b^2}$ (you can see it is the pythagoreum theorum solved for c, which is d for distance in this case)