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Mathematics 13 Online
OpenStudy (anonymous):

help geomtry

OpenStudy (anonymous):

OpenStudy (anonymous):

Can we work through it step by step how you got midpint 7 please?

OpenStudy (john_es):

Do you know the formula to obtain the middlepoint?

OpenStudy (anonymous):

uh no.

OpenStudy (john_es):

Well, it's easy, I write you, \[PB_m=\frac{P+B}{2}\]

OpenStudy (anonymous):

ok got that

OpenStudy (john_es):

You can also make it graphically. In a straight line divide in parts from -2 to 12. You'll see that the midpoint is in 5.

OpenStudy (anonymous):

John_ES is the formula the same when both points are positive?

OpenStudy (anonymous):

Can you like do it step by step?

OpenStudy (john_es):

Yes, the formula is the same no matter what sign the points have. In this case, \[P=-2,\ \ B=12\Rightarrow PB_m=\frac{12-2}{2}=10/2=5\]

OpenStudy (john_es):

Also, graphically, -2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12 You'll see that the point that is in the middle is 5 (from -2 to 5 there are 7 steps, from 5 to 12 there are also 7 steps).

OpenStudy (anonymous):

so 5 is my answer?

OpenStudy (john_es):

Yes, because is the number that is in the middle.

OpenStudy (anonymous):

can u help me with 2 more please?

OpenStudy (john_es):

Ok, let's go.

OpenStudy (anonymous):

OpenStudy (john_es):

You should apply the same formula, but in this case you want one of the points. \[(-1,0)=\frac{(x,y)+(5,2)}{2}\]Could you finish form this point?

OpenStudy (anonymous):

well im little confused.

OpenStudy (anonymous):

can u walk me thru it?

OpenStudy (john_es):

You need the point of one endpoint, we named (x,y). Using the equation I wrote before, you'll have, \[(-1,0)\ \text{the middlepoint}\]\[(x,y)\ \text{our wanted endpoint}\]\[(5,2)\ \text{our given endpoint}\] Then we apply the formula, \[\frac{(x,y)+(5,2)}{2}=(-1,0)\Rightarrow {(x,y)+(5,2)}=(-2,0)\Rightarrow\\\Rightarrow(x,y)=(-2,0)-(5,2)=(-7,-2)\]

OpenStudy (anonymous):

is (-7,-2 the answer?

OpenStudy (anonymous):

????

OpenStudy (john_es):

Sorry, I went out. Yes this is the answer ;).

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