Question

determine if a triangle DEF with coordinates D(2,1), E(3,5), and F(6,2) is an equilateral triangle. Use evidence to support your claim. If it is not an equilateral triangle, what changes can be made to make it equilateral ? be specific.

Answer 1

will be given medal

Answer 2

please someone help me!!

Answer 3

whoever helps will be given a medal!!

Answer 4

@zepdrix please help

Answer 5

It's equilateral if all sides are the same length.
So let's use the `distance formula` to figure out the length of each side.

Calculating the distance from D to E,\[\Large\rm d_{1}=\sqrt{(3-2)^2+(5-1)^2}\]Understand how I plugged those values in?

Answer 6

yes i do

Answer 7

So what do you get for the length of side DE?

Answer 8

sqrt 17

Answer 9

\[\Large\rm \overline {DE}=\sqrt{17}\]Ok great!
Understand how to check the lengths of the other sides? :o

Answer 10

yes i get that but then how do i answer the second part of the question?

Answer 11

Well you have to answer the first part before you can answer the second part.
If they're all the same length, then you're done.

If they're not, we can think about it.
But figure out the lengths first! :o

Answer 12

ok im doing that now :)

Answer 13

ok so i got e to f sqrt 18
and d to f sqrt 17

Answer 14

so its not equilateral

Answer 15

\[\Large\rm \overline {DE}=\sqrt{17},\qquad \overline{EF}=\sqrt{18},\qquad \overline{FD}=\sqrt{17}\]Ok cool I got the same values. So it's currently Isosceles?

Answer 16

So we need to fix this, hmmm

Answer 17

yea the second part is where i was having a difficult time with

Answer 18

This is the problem,
When you `move a point`, it changes the length of `two sides`, not just one.

Answer 19

So we don't want to try and turn the sqrt(18) into sqrt(17) by moving a point.
We actually want to try and change the sqrt(17)'s into sqrt(18).

Answer 20

One sec I gotta think about this one >.< hehe

Answer 21

lol its fine no rush :)

Answer 22

So this is what I'm thinking...
We want to move our point E somewhere else to change our lengths of DE and FD.
Let's call this new point (x,y).

Again using the distance formula,\[\Large\rm \overline{DE}=\sqrt{(x-2)^2+(y-1)^2}\]And then set that equal to the length we want,\[\Large\rm \sqrt{18}=\sqrt{(x-2)^2+(y-1)^2}\]We can do the same with FD,\[\Large\rm \sqrt{18}=\sqrt{(x-6)^2+(y-2)^2}\]And that gives us a system of two equations.
Mmmmmm that might not be the right approach, thinkinggggg >.<

Answer 23

can we just change one coordinate to make it equal

Answer 24

So the old coordinate was (3,5),
you mean like leave the y at 5 and solve for the new x?

I was trying to do that... I can't seem to make it work :(
Grrr so confusingggg

Answer 25

@jdoe0001 @dan815 @phi @campbell_st

Answer 26

I would use the fact that an equilateral has 60 degree angles.
we want to "rotate" radius EF to form a 60 degree angle at point E

Answer 27

i dont understand..what do i do after

Answer 28

@jdoe0001 can you try please

Answer 29

one way to find where to move point E is to form the perpendicular bisector of side DF

Answer 30

you can use algebra and find where that line intersects the circle.

or (easier) we can use trigonometry to find the (x,y) values

Answer 31

yes lets use trigonometry

Answer 32

hmm

Answer 33

the mid point of line segment DF is
\[ (2,1) + \sqrt{17} (6-2,2-1) \\ (2,1) + \sqrt{17} (4,1) \]
or, letting D be the origin for this exercise,
\[ (4\sqrt{17},\sqrt{17}) \]

the angle is inverse tangent of y/x = 1/4

Answer 34

we get 14.036 degrees. Add 60º to get 74.036

now find the x and y coordinates (with D the origin) as
sqr(17) cos(74.036) , sqr(17) sin(74.036)

then add the coords of D to get the (x,y) values relative to the true origin.

Answer 35

dont we find the measure of each angle and then shorten angle D?

Answer 36

each angle is 60 degrees by defintion

Answer 37

ok thank you guys so much

Answer 38

There are different ways of doing this.
Here is the idea I am using

Answer 39

im doing this exam as the beginning of my geometry course and i dont think the answer should be that complex

Answer 40

the idea is point G is point E moved a little bit (along the circle) so its length DG is sqr(17)

we want to find the x and y offsets from pt G using R cos(angle) and R sin(angle)
where R is sqr(17)
and the angle is 60+ little angle

Answer 41

If there is a short easy way to get a numerical answer, I am all ears.

Perhaps they want a more "hand-waving" answer. e.g. rotate side DE so it forms a 60º angle with side DF

Answer 42

maybe you're not required to get the (x,y) coordinates for F.... maybe you're just asked to give an explanation on what needs to be done to be an equilateral

Answer 43

yea thats probably what they want

Answer 44

well ill just put what @phi said thank you so much

Answer 45

In geometry they teach you how to construct an equilateral triangle. Perhaps they want you to use this technique. See
http://www.mathopenref.com/constequilateral.html