2 answers:
Given the points (-5,5) and (1,-2):
Let (x1, y1) = (-5,5)
(x2,y2) = (1,-2)
Use the distance formula:
d = √[(x2 — x1)^2 + (y2 — y1)^2]
Plug in the values into the formula:
d = √[(x2 — x1)^2 + (y2 — y1)^2]
d = √[(1 — (-5))^2 + (-2 — 5)^2]
d = √[(1 + 5)^2 + (-7)^2]
d = √[(6)^2 + (-7)^2]
d = √[36 + 49]
d = √[85]
d = 9.2195 or 9.22
See the picture below for a way to construct a right triangle that makes that piece the longest side of the triangle your line segment.
With that setup, you can use they Pythagorean Theorem.
![7^2 + 6^2 = c^2](https://tex.z-dn.net/?f=7%5E2%20%2B%206%5E2%20%3D%20c%5E2)
And then solve that for c, keeping in mind c must be positive.
![\begin{aligned}49 + 36 &= c^2\\85 &= c^2\\\sqrt{85} &= c\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D49%20%2B%2036%20%26%3D%20c%5E2%5C%5C85%20%26%3D%20c%5E2%5C%5C%5Csqrt%7B85%7D%20%26%3D%20c%5Cend%7Baligned%7D)
If you want to use the distance formula, that is just the Pythagorean theorem solved for c:
![d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
Applying that to your situation
![d = \sqrt{(5-(-2))^2+(-5-1)^2} = \sqrt{(7)^2+(-6)^2} = \sqrt{49+36} = \sqrt{85}](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%7B%285-%28-2%29%29%5E2%2B%28-5-1%29%5E2%7D%20%3D%20%5Csqrt%7B%287%29%5E2%2B%28-6%29%5E2%7D%20%3D%20%5Csqrt%7B49%2B36%7D%20%3D%20%5Csqrt%7B85%7D)
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