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Suppose C(-4,5) is the midpoint of Line AB and the coordinates of A are (2,17). Find the coordinates of B.

 

Thanks!

Mr.Owl  Sep 18, 2018
 #1
avatar+2248 
+1

Hello, Mr.Owl. I hope you are doing well.

 

Generally, when we face problems that involve the midpoint, we are generally tasked with identifying the midpoint. This is not the case here. I have provided a nice visual of the midpoint formula in action. 

 

 

 

Coordinate A with coordinates \((x_1,y_1)\) on the above diagram has \((2,17)\). Therefore, \(x_1=2\text{ and }y_1=17\). The midpoint, coordinate C, represented as \(\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\) on the visual, has coordinates \((-4,5)\). This means that \(\frac{x_1+x_2}{2}=-4\text{ and }\frac{y_1+y_2}{2}=5\). Does this give you an idea on how to proceed?

TheXSquaredFactor  Sep 18, 2018
edited by TheXSquaredFactor  Sep 18, 2018
 #2
avatar+443 
+2

Is there a formula for finding half of the line?

 

x1= 2  x2= 17 

y1=5   y2=5

A=P

C=M

Q=B

Q= (x2,y2)

 

From what you have said, I gather that point B is 5 and 17= (5,17)

Is that right or did I just drive off the cliff?

Mr.Owl  Sep 18, 2018
 #3
avatar+2248 
+1

Be careful! I know that this is a lot of information to take in at once.

 

\(x_1=2\text{ and }x_2=?\\ y_1=17\text{ and }y_2=?\\ A=P\\ C=M\\ Q=B=(x_2,y_2)\\ \)

 

We know that \(\frac{x_1+x_2}{2}=-4\text{ and }\frac{y_1+y_2}{2}=5\). We just have to solve for the missing variables. That's all.

TheXSquaredFactor  Sep 18, 2018
edited by TheXSquaredFactor  Sep 18, 2018
edited by TheXSquaredFactor  Sep 18, 2018
 #4
avatar+443 
+2

Alright, so I have:

 

2+x2 divided by 2 = -4  so 2+ -10 divided by 2 = -4

17+ y2 divided by 2 = 5   so 17 +  -7 divided by 2 = 5

 

So x2 = -10  and  y2 = -7

 

So the answer should be (-10,-7)

Mr.Owl  Sep 18, 2018
 #5
avatar+2248 
+1

Nice job, Mr.Owl.

 

You deserve a round of applause! Woohoo!

 

You asked about a formula regarding how to find the other endpoint when an endpoint and midpoint are given in the problem. I believe it exists, just for your knowledge. 

 

If A is the endpoint at \((x_1,y_1)\) and C is at the midpoint at \((x_2,y_2)\), then B, the opposite endpoint, should be located at \((2x_2-x_1,2y_2-y_1)\). You will see that it works with the coordinates we have. 

TheXSquaredFactor  Sep 18, 2018
edited by TheXSquaredFactor  Sep 18, 2018
 #6
avatar+443 
+2

WOOHOOOO!!!! 

 

Now I can sleep  ;-)

 

Thanks!

Mr.Owl  Sep 18, 2018

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