Question

The midpoint of a line segment with end points as (2, 3) and (4, y2) is (3, 2). What is the value of Y2?

7

-2

1

-1

Answer 1

Midpoint is given by the average of the two points

\[ \frac{1}{2} \left(\begin{matrix}2+4 \\ 3+2y\end{matrix}\right)=\left(\begin{matrix}3 \\ 3/2 + y\end{matrix}\right) \]

Answer 2

so you have an equation for y there already

\[ \frac{3}{2}+y=2 \]

Answer 3

So its -2? Im confused.

Answer 4

y=1 I would say from here.

Answer 5

oh wait, is that y^2=

Answer 6

I read it as 2y

Answer 7

No is y2

Answer 8

So you think its 1?

Answer 9

y=1 is not the solution what did I not see?

Answer 10

\[\frac{3+y_2}{2} = 2\]
solve for y_2

Answer 11

y=1 is right @Spacelimbus why do you say it's not the solution?

Answer 12

oh god, excuse me, I misread my own equation, I do that from time to time to make the problems more challenging! @lgbasallote

Answer 13

So its 1?

Answer 14

no that makes no sense. @jhonathanceballos @lgbasallote

again my equation above

\[ \frac{3}{2}+y=2

\\3+2y=4 \\2y=1
\\y=0.5 \]

Answer 15

that's the correct answer, because 3+2y=2 which is the condition above,

Answer 16

im confused.. I cant figure out the answer either.

Answer 17

it's correct I am pretty sure, maybe @Neemo can double check my algebra, I am an insomniac at the moment.

Answer 18

Its looking for Y2 though..

Answer 19

I asked you above, y^2 and 2y is not the same thing, but y2 and 2y is the same thing, one is multiplication by a scalar component, one is multiplication times itself, therefore exponent.

Answer 20

if you look for 2y rather than for y just multiply my result times two, then you get 2y=1, but that is rather unusual to write a result like that.

Answer 21

Yeah your right @Spacelimbus So it should be 1 right?.

Answer 22

\[ 2y=1 \\y= \frac{1}{2} \]
But lets wait for @Neemo 's response.

Answer 23

I think he meant--- \[y_2\] because in the answer they ask for y2 wich can be \[y_2\] not 2y or y2 @Spacelimbus

Answer 24

then the answer I developed isn't correct @Neemo , but it seems algebraically logic to me, however it's not in the solution set.

Answer 25

\[y_2=1 \] sorry, I have trouble with the subscripts today (-: it's what we derived already above, just that I understood your y2 as 2 times y, but it's a subscript.

Answer 26

Oh Alright,@Neemo @Spacelimbus Thanks for your help!

Answer 27

yeah ! Y2=1 ! is the correct answer ! I confirm :) you're welcome !