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)

- anonymous

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- 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)

- anonymous

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

- 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)

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## More answers

- DanJS

The distance is figured using the pythagorean theorem

- anonymous

ive never heard of that

- DanJS

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

- anonymous

okay then what

- DanJS

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

- anonymous

okay

- jdoe0001

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

- DanJS

|dw:1438902788817:dw|

- anonymous

ohhh f.o.i.l right?

- DanJS

Right, that is just memorizing the formula

- anonymous

ya

- anonymous

is that how i do it using foil?

- DanJS

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

- anonymous

this is way to complicated

- DanJS

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

- anonymous

its going way over my head

- DanJS

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

- anonymous

ya

- anonymous

but still really confused

- 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

- anonymous

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

- DanJS

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

- anonymous

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

- anonymous

so all i do is foil pretty much right?

- DanJS

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

- DanJS

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

- anonymous

oh

- anonymous

well guess i dont get it

- 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..
\[distance = \sqrt{(x - x1)^2+(y-y1)^2}\]

- DanJS

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

- anonymous

oh isee what u did there

- DanJS

coo, try the next one

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