1. I can't troubleshoot this

I'm trying to solve for the coordinate that is 1/3 of the way between point (1,7) and point (10,4) USING the distance formula.

This is what I had but I am stuck.

Distance between point A and B:
Distance = Sqrt ((Xsub2 - Xsub1)^2 + (Ysub2 - Ysub1)^2)

Find the distance, multiply it by 1/3 - we'll call this distance ab.

ab = sqrt ((Xsub2 - Xsub1)^2 + (Ysub2 - Ysub1)^2); Simplify and solve by pluging in the values.

ab^2 = (Xsub2 - 1)^2 + (Ysub2 - 7)^2; where Xsub2 and Ysub2 are the coordinates you are looking for.

Note: Keep in mind that point A is (1,7) and B is (10,4).

2. Originally Posted by Masterthief1324
I'm trying to solve for the coordinate that is 1/3 of the way between point (1,7) and point (10,4) USING the distance formula.
note the horizontal distance between the x-values ...

x = 1 and x = 10 ... 9 units apart

x=4 and x = 7 would be 1/3 and 2/3 of the way respectively.

do the same in the vertical direction ...

y = 7 and y = 4 ... 3 units apart

y = 6 and y = 5 would be 1/3 and 2/3 of the way respectively

so ,,,

(1,7) (4,6) (7,5) (10,4)

3. Originally Posted by Masterthief1324
I'm trying to solve for the coordinate that is 1/3 of the way between point (1,7) and point (10,4) USING the distance formula.

This is what I had but I am stuck.

Distance between point A and B:
Distance = Sqrt ((Xsub2 - Xsub1)^2 + (Ysub2 - Ysub1)^2)

Find the distance, multiply it by 1/3 - we'll call this distance ab.

ab = sqrt ((Xsub2 - Xsub1)^2 + (Ysub2 - Ysub1)^2); Simplify and solve by pluging in the values.

ab^2 = (Xsub2 - 1)^2 + (Ysub2 - 7)^2; where Xsub2 and Ysub2 are the coordinates you are looking for.

Note: Keep in mind that point A is (1,7) and B is (10,4).
$D = \sqrt{(10 - 1)^2 + (4 - 7)^2}$

$= \sqrt{9^2 + (-3)^2}$

$= \sqrt{81 + 9}$

$= \sqrt{90}$

$= 3\sqrt{10}$.

Since there is some point, call it $(x, y)$ which is between $(1, 7)$ and $(10, 4)$, which divides the distance into the ratio $1:2$.

That would mean that one length has magnitude $\sqrt{10}$ and the other has magnitude $2\sqrt{10}$.

So $\sqrt{10} = \sqrt{(x - 1)^2 + (y - 7)^2}$

$10 = (x - 1)^2 + (y - 7)^2$.

Also $2\sqrt{10} = \sqrt{(10 - x)^2 + (4 - y)^2}$

$40 = (10 - x)^2 + (4 - y)^2$.

Now you have two equations in two unknowns you can solve simultaneously.