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Mathematics 21 Online
OpenStudy (anonymous):

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)

OpenStudy (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)

OpenStudy (anonymous):

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

OpenStudy (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)

OpenStudy (danjs):

The distance is figured using the pythagorean theorem

OpenStudy (anonymous):

ive never heard of that

OpenStudy (danjs):

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

OpenStudy (anonymous):

okay then what

OpenStudy (danjs):

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

OpenStudy (anonymous):

okay

OpenStudy (jdoe0001):

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

OpenStudy (danjs):

|dw:1438902788817:dw|

OpenStudy (anonymous):

ohhh f.o.i.l right?

OpenStudy (danjs):

Right, that is just memorizing the formula

OpenStudy (anonymous):

ya

OpenStudy (anonymous):

is that how i do it using foil?

OpenStudy (danjs):

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

OpenStudy (anonymous):

this is way to complicated

OpenStudy (danjs):

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

OpenStudy (anonymous):

its going way over my head

OpenStudy (danjs):

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

OpenStudy (anonymous):

ya

OpenStudy (anonymous):

but still really confused

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (danjs):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so all i do is foil pretty much right?

OpenStudy (danjs):

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

OpenStudy (danjs):

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

OpenStudy (anonymous):

oh

OpenStudy (anonymous):

well guess i dont get it

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

OpenStudy (danjs):

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

OpenStudy (anonymous):

oh isee what u did there

OpenStudy (danjs):

coo, try the next one

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