## anonymous one year ago PLEASE SOMEONE HELP ITS AN SOS!!! WILL MEDAL AND FAN Find the distance between these points. C(0, 4), T(-6, -3) √(37) √(85) √(109)

1. anonymous

A parallelogram whose vertices have coordinates R(1, -1), S(6, 1), T(8, 5), and U(3, 3) has a shorter diagonal of ___ . 5 √(13) √(97)

2. anonymous

Find the distance between these points. W(-6, -8), X(6, 8) 20 10 √8

3. anonymous

Find the other endpoint of a line segment with one endpoint at (-3, 5) and whose midpoint is (5, 2). (7, 2) (1, 7/2) (13, -1)

4. DanJS

The distance is figured using the pythagorean theorem

5. anonymous

ive never heard of that

6. DanJS

(0,4) and (-6,-3) |dw:1438902675608:dw|

7. anonymous

okay then what

8. DanJS

you can form a right triangle to get the components of a right triangle

9. anonymous

okay

10. anonymous

$$\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) C&({\color{red}{ 0}}\quad ,&{\color{blue}{ 4}})\quad % (c,d) T&({\color{red}{ -6}}\quad ,&{\color{blue}{ -3}}) \end{array}\qquad % distance value d = \sqrt{({\color{red}{ x_2}}-{\color{red}{ x_1}})^2 + ({\color{blue}{ y_2}}-{\color{blue}{ y_1}})^2}$$

11. DanJS

|dw:1438902788817:dw|

12. anonymous

ohhh f.o.i.l right?

13. DanJS

Right, that is just memorizing the formula

14. anonymous

ya

15. anonymous

is that how i do it using foil?

16. DanJS

If you know how to get to it, more likely to remember even if you forget the formulas

17. anonymous

this is way to complicated

18. DanJS

That forms a right triangle, with the property d^2 = (side1)^2 + (side2)^2

19. anonymous

its going way over my head

20. DanJS

solving for Distance d, you get what @jdoe0001 typed above

21. anonymous

ya

22. anonymous

but still really confused

23. DanJS

$distance = \sqrt{(\Delta x)^2 + (\Delta y)^2}$ Distance is square root of the sum of the difference in x coordinates and the difference in y coordinates of the two points

24. anonymous

um i dont get it still but thank you sooo much for your trouble

25. DanJS

(0,4) and (-6 , -3) $distance = \sqrt{(-6 - 0)^2 + (-3 - 4)^2} = \sqrt{36 + 49} = \sqrt{85}$

26. anonymous

oh so it was how i thought. I dont know why but for some reason its harder then it should be

27. anonymous

so all i do is foil pretty much right?

28. DanJS

i was just explaining what the distance formula comes from. ha sorry

29. DanJS

no, no Foil. You just put the numbers in and calculate it

30. anonymous

oh

31. anonymous

well guess i dont get it

32. DanJS

I guess all you have to take note of is, if you given 2 points (x , y) and (x1 , y1) The distance between them is calculated by.. $distance = \sqrt{(x - x1)^2+(y-y1)^2}$

33. DanJS

$distance = \sqrt{(-6 - 0)^2 + (-3 - 4)^2} = \sqrt{(-6)^2+(-7)^2 } = \sqrt{85}$

34. anonymous

oh isee what u did there

35. DanJS

coo, try the next one