Geometry! Please HELP ME!!!

- deerhunter15

Geometry! Please HELP ME!!!

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- deerhunter15

##### 1 Attachment

- deerhunter15

##### 1 Attachment

- deerhunter15

@Nnesha last one, need help, and fast because its due by 2:30, and having trouble with the whole question because its confusing.

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## More answers

- anonymous

whats the problem?

- deerhunter15

in the attatched files...

- anonymous

there are two witch one

- deerhunter15

##### 1 Attachment

- deerhunter15

the one i just posted is the one

- anonymous

what grade is this?

- deerhunter15

10th

- anonymous

by im in 6th

- Nnesha

alright then
i'm not good at geometry but lets try it out.
we need to use midpoint and slope formula to prove DE is parallel to AC

- deerhunter15

Ok

- Nnesha

|dw:1446580151082:dw|

- Nnesha

|dw:1446580309182:dw|
B point is at the origin that's why i put (0,0) order pair
now to find find D midpoint use the midpoint formula \[(\frac{ x_1 +x_2 }{ 2 } +\frac{y_1+y_2}{2})\]

- Nnesha

use point B and A to find midpoint D
B(0,0) A(0,a)

- deerhunter15

0,D

- Nnesha

|dw:1446580617770:dw|
first find midpoint D and E
then we can find slope of both line to prove they are parallel

- deerhunter15

how do you work m_1 and M_2

- Nnesha

m_1 means first midpoint which is D
and m_2 is 2nd mid point which is E

- Nnesha

do you mean D mid point is (0,D ) ??

- deerhunter15

Idk but i think maybe it is

- Nnesha

use point A and B to find midpoint D
formula is \[(\frac{ x_1 +x_2 }{ 2 }, \frac{y_1+y_2}{2})\]
x's and y's values are (0,0(0,a)

- Nnesha

and then use point A and C to find midpoint E
A(0,a) C(c,0)

- deerhunter15

|dw:1446581005508:dw|

- deerhunter15

so point e would be like 1,0

- Nnesha

we can't assume E (1,0)
it can be (2,0) we need to use given order pair

- Nnesha

use point A and B to find midpoint D
formula is \[(\frac{ x_1 +x_2 }{ 2 }, \frac{y_1+y_2}{2})\]
x's and y's values are (0,0(0,a)
\[(\color{ReD}{x_1} ,y_1)(\color{red}{x_2},y_2) \rightarrow (0,0)(0,a)\]

- Nnesha

plugin

- deerhunter15

D (0,a/2)

- Nnesha

yes right now find point E
use point C and A

- deerhunter15

e( C/2,0)

- Nnesha

ah ye that's correct
sorry i meant to say point C and B in my last comment
good job!

- Nnesha

now we have to use the slope formula to find slope of DE and slope of point AC

- Nnesha

\[\huge\rm slope =\frac{ y_2 -y_1}{ x_2 -x_1}\]|dw:1446581793811:dw|

- Nnesha

use the D and E point to find slope of DE
and use C and A point to find slope of AC

- Nnesha

let me know if you don't understand that ^^

- deerhunter15

I dont understand a^2-0^1
--------
0^2-c^1

- Nnesha

ohh it's not square
y_1 means first one

- Nnesha

like A(0,a) C(c,0)
so \[\rm (x_1 ,y_1)(x_2 ,y_2) \rightarrow (0,a)(c,0)\]

- deerhunter15

is that the answer?

- Nnesha

|dw:1446582619883:dw|

- Nnesha

no
we have to prove DE is parallel to AC
in order to prove we should find slope of both equation

- Nnesha

use the slope formula to find slope of AC
\[\huge\rm \frac{ y_2 -y_1 }{ x_2 -x_1 }\]
x and y's values are (0,a)(c,0)

- Nnesha

look at the drawing i've posted above
x_1 = 0
x_2 =c
y_1 = a
y_2=0

- deerhunter15

....im still not understanding and I have to go in about 10-20 minutes

- Nnesha

alright what part you don't understand ?

- deerhunter15

we are like just dividing a by c

- Nnesha

just find the sl0pe of both lines that's it

- deerhunter15

the slope of D and E will be half of A and C

- Nnesha

welll what about the sign
it should be like this \[(0,a)(c,0 ) \rightarrow \frac{ 0-a }{ c-0 }\]

- Nnesha

no no we are proving DE and AC are parallel lines
and parallel lines have the same slope

- deerhunter15

erm

- deerhunter15

so is it A=C

- deerhunter15

@Nnesha

- Nnesha

slope of both lines supposed to be the same

- Nnesha

definition of parallel lines: they both have the same slope

- Nnesha

read the direction
prove DE and AC are parallel lines
`state ` the coordinates and show all work
so you need slope of AC and slope of DE to prove they both are parallel lines

- Nnesha

\[(0,a)(c,0 ) \rightarrow \frac{ 0-a }{ c-0 }\] solve this

- Nnesha

In other words simplify what would be the final slope ?

- Nnesha

remember
if line is decreasing you will get negative slope
line increasing = positive slope

- Nnesha

for the 2nd part just use distance formula to prove DE is half of AC \[\large\rm \sqrt{(x_2 -x_1)^2 +(y_2 -y_1)^2}\]

- Nnesha

D(0,a/2) C (c/2 ,0)

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