Question

Point X is located at (2, 4), and Y is located at (−6, 10). Find the y value of the point that lies halfway between X and Y.

Answer 1

2

7

-3

-8

Answer 2

@ganeshie8 @Mackenzie2013 @linda3 @tester97 @wolf1728 @deshawn1

Answer 3

@Awesome781 @Fire_Fighter_Of_Miami @tHe_FiZiCx99 @gabs15 @rockstar0765

Answer 4

wow lot of people

Answer 5

need help please !

Answer 6

midpoint formula...

Answer 7

midpoint formula forgot it

Answer 8

?idk now to do that can you please help @Orion1213

Answer 9

y = (y1 + y2)/2

Answer 10

two points X and Y are given with their y-coordinates... just plug-in...

Answer 11

so what numbers would i plug into whweer? @Orion1213

Answer 12

for X (2,4)... y1=4

Answer 13

for Y (-6,10)... y2=10

Answer 14

you can solve now the y-coordinate of the point half-way the two given points...

Answer 15

ok so y=(41+102)/2
y=52? @Orion1213

Answer 16

i am confused @Orion1213

Answer 17

no... it is y=(4+10)/2=14/2=7...

Answer 18

\[y = \dfrac{y_1+y_2}{2}\]

Answer 19

oh okay thanks @Orion1213

Answer 20

can u help with one more? @Orion1213

Answer 21

no problem @EmmaDrew94 ... :)

Answer 22

Calculate the area of triangle WXY with altitude YZ, given W (2, −1), X (6, 3), Y (7, 0), and Z (5, 2).

Answer 23

8 square units

9.4 square units

7.7 square units

12 square units

Answer 24

@Orion1213 can u help me with this one, please ?

Answer 25

let me recall the formula...

Answer 26

this might help.... http://www.mathguru.com/level2/application-of-coordinate-geometry-2007101600011139.aspx

Answer 27

ugh can u just please help me solve it?? @Orion1213

Answer 28

ok... for given vertices W, X and Y... we can use this formula...\[Area =\frac{ 1 }{ 2 }\left| (x_a-x_c)(y_b-y_a)-(x_a-x_b)(y_c-y_a) \right|\]

Answer 29

the a, b and c are simply the w, x and y....

Answer 30

ok so what numbers do i plug in, i am sorry i am just so confused math is so hard. @Orion1213

Answer 31

W or A (2,-1), X or B (6,3) and Y or C (7,0)

Answer 32

so would the answer be 9.4then

Answer 33

if you solve it already... cause i'm just starting...

Answer 34

oh ok sorry... hold on @Orion1213

Answer 35

why?

Answer 36

i am still here i am gonna try and solve it too @Orion1213

Answer 37

i think it's b. but i think i solved it wrong...

Answer 38

what did u get and how did you plug them into the equation? @Orion1213

Answer 39

ok we have it as \(x_a=2, y_a=-1\).... \(x_b=6, y_b=3\).... and \(x_c=7, y_c=0\)

Answer 40

so what would be the answer ? @Orion1213

Answer 41

\[Area=\frac{1}{2}\left|(2-7)(3-(-1))-(2-6)(0-(-1))\right|\]

Answer 42

i got 11?

Answer 43

bu that is not an answer choice so i know that's wrong

Answer 44

\[Area=\frac{1}{2}\left|(-5)(4)-(-4)(1)\right|=\frac{1}{2}\left|-20+4\right|=\frac{1}{2}\left|-16\right|\]
\[Area=\frac{16}{2}=8~ square~ units\]

Answer 45

oh okay! Thank you so it would -8 @Orion1213

Answer 46

no it's 8... we just take the absolute value of -16 which is 16...

Answer 47

but there is only a choice answer of -8 not 8 @Orion1213

Answer 48

it is 8 written in your selection above....

Answer 49

and it's asking for the area, and it can't be negative....

Answer 50

i have no idea where i got the negative from!! @Orion1213

Answer 51

remember the bracket in the formula... it denotes absolute value... |-16| = 16...

Answer 52

that is where you got the negative value...

Answer 53

hope it's clear already...

Answer 54

oh okay i see! Your so smart Thanks Orion! :)