Find the distance between the points on the coordinate plane. points I and C A. –2 B. 2 C. 4 D. 5
@Mehek14 @LoveIt
@Awolflover1
can u help please
Okay. Do you know where is Point C at?
(-1,1)
Correct, What about I?
(4,1)
Wait for c isn't it (1,-1)?
no
No..
brb
Okay, so count to C to I
ok @Awolflover1
@Awolflover1
is it 5 ?
Can you tell me the coordinates of the 2 points first?
ok (-1,1) for l and (4,1) for c
perfect have you ever heard of the distance formula it looks like this: \[\huge d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]
that equation thing does not make sence
I'll teach it to you :P \[\huge d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\] We know our coordinates C comes before I so (4,1) <---coordinate 1 (-1,1) <---coordinate 2 Plug in the numbers. \[\huge d=\sqrt{(-1-4)^2+(1-1)^2}\] -1-4=?
Pooja is it (x,y) or (y,x) because I always thought it was (x,y)
(x,y)
So then for c should it be (1,-1)?
You're right! \[\huge d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\] We know our coordinates C comes before I so (4,1) <---coordinate 1 (1,-1) <---coordinate 2 Plug in the numbers. \[\huge d=\sqrt{(1-4)^2+(-1-1)^2}\] so...1-4=?
@daisyduck04
That kind of goes for 1 too right? So (1,4)?
I
We literally flipped the equation ,-, it wouldn't have mattered but ur correct \[\huge d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\] We know our coordinates C comes before I so (1,4) <---coordinate 1 (1,-1) <---coordinate 2 Plug in the numbers. \[\huge d=\sqrt{(1-1)^2+(-1-4)^2}\] \[\huge d=\sqrt{(0)^2+(-5)^2}\] 0^2=0 -5^2=25 \[\huge d=\sqrt{25}\] \[\huge d=5\]
You had it correct just showed u a different way of getting the answer :-)
Oh sorry
Thanks @LoveIt <3 don't be sorry ;)
@LoveIt @@pooja195 sorry internet problems
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