Question

Which of the following points is an equal distance (equidistant) from A(−4, 1) and B(2, −3)?

J(−3, −4)
K (- 3/2 , - 1/2)
L(−1, −2)
N(4, 1)

Answer 1

Your looking for midpoint on this one do you know the formula?

Answer 2

i dont i have been looking and can't find one

Answer 3

so i need help

Answer 4

Are you sure this is typed correctly the question and answers?

Answer 5

Cause I did the math in my head and none of the answers are correct

Answer 6

And the midpoint formula is (X1+x2)/2 and (y1+y2)/2

Answer 7

@lydiasantos

Answer 8

yeah i know its so weird

Answer 9

show me what the equation looks like with the coordinates plugged in

Answer 10

Cause if we would plug in our points (-4,1) (-2,3 we could call point a (x1,y1) and point b (x2,y2) do you want to try? I think you can do it :)

Answer 11

Is that one point supposed to be (-4,-1)?

Answer 12

A(−4, 1) and B(2, −3)

Answer 13

Okay

Answer 14

So A will be (X1,y1) (x2,y2) so you can say that -4 will be X1 and 1 will be y1 make sense?

Answer 15

Sorry I forgot to mention B will be (x2,y2)

Answer 16

Confused?

Answer 17

so i just have to plug in those values to the equation

Answer 18

Correct but don't forgot to divide by two after adding the x's and y's so what will you have for the x part?

Answer 19

(-4+1)/2 and (2+(-3)/2
is the equation right?

Answer 20

No

Answer 21

You have to keep your x's and y's together

Answer 22

None of the answers on here are right :/

Answer 23

(X1+x2)/2 and (y1+y2)/2
-4+2 /2 and 1+(-3) / 2

Answer 24

Yes that's right

Answer 25

Simplify you get?

Answer 26

Cause I got -1,-1

Answer 27

me too

Answer 28

And I know that isn't the wrong process

Answer 29

Your finding what point will be equal distance away from both points

Answer 30

And both points would be 3 units awAy on the x axis and 2 units away from the y

Answer 31

@Directrix

Answer 32

well thanks for all your help anyways, it was a big help

Answer 33

No problem

Answer 34

Points that are equidistant from points A and B will lie on the line that is the perpendicular-bisector of segment AB.