∆ABC has vertices A(0, 0), B(2, 4), and C(4, 2). ∆RST has vertices R(0, 3), S(-1, 5), and T(-2, 4). Prove that ∆ABC ∼ ∆RST. (Hint: Graph ∆ABC and ∆RST in the coordinate plane.)

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∆ABC has vertices A(0, 0), B(2, 4), and C(4, 2). ∆RST has vertices R(0, 3), S(-1, 5), and T(-2, 4). Prove that ∆ABC ∼ ∆RST. (Hint: Graph ∆ABC and ∆RST in the coordinate plane.)

Mathematics
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@steve816 he can help you i think.
Well, do you know how to plot points on the coordinate plane? All you do is draw the points and connect the dots.
|dw:1443590087569:dw|

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omg thank you
No problem :)
so how do i do this problem?
ok so as steve put it do you have a graph?
ive already graphed the points but i dont know how to prove it similar
If the triangles are the same size once you graphed, then they are similar.
i thought you had to do the distance formula?
they werent the same size
Yes, you can use the distance formula
For each of the sides
okay how can i do it with the distance formula?
Plug in your (x, y) coordinates into this formula: http://2.bp.blogspot.com/-GMrbM3fg4NU/VHljne7DleI/AAAAAAAABec/3emecyWG4ew/s1600/Slide1.jpg
ok give me a second
wait which numbers do i put into it?
i did AB and got 20 is that correct?
?
@steve816 is that correct?
hello?
i need help fast can someone help?
ok now what?
is that correct?
i got the squareroot of 20 for AB
yes
okay thanks
yep
for rs i got the sqaureroot of 5 is that correct?
yes
and for CT i got the sqareroot of 40 is that correct?
yes
ok what do i do now?
no that one would have to be up to Steve because he gave you the square root idea but i might be able to get you someone else that can help because Steve is helping other pp at the time.
okay thanks!
yep
Hai, how can i help you ?
i was doing the distance formula for the problem and dont know what to do next
|dw:1443591908295:dw| According to your question..
yes
i did Ab and got the squareroot of 20 then i did RS and got the sqareroot of 5 then i did CT and got the squareroot of 40
hmm... better, this user @dan815 can help you more clearly.
this is cheating a bit becasue this program gives the lentgh of the sides but you need to use the SSS postulate and this is easier because 2 side length are the same \[\frac{ BC}{ EF} = \frac{ AB }{DF} \]
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