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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!

Sep 18, 2018

#1
+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?

Sep 18, 2018
edited by TheXSquaredFactor  Sep 18, 2018
#2
+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?

Sep 18, 2018
#3
+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
+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)

Sep 18, 2018
#5
+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
+2

WOOHOOOO!!!!

Now I can sleep  ;-)

Thanks!

Sep 18, 2018