Question

Use points A= (12,-5) and B= (-4,3)
a) find the distance between the points
b) find the midpoint

Answer 1

Do you know the distance formula?

Answer 2

sorta but not really.

Answer 3

\[\sqrt{(x2 - x1)^{2} + (y2-y1)^{2}}\]

Answer 4

That is your formula
you pick two points:
(x1, y1) and (x2, y2)
Plug in that to the formula

Answer 5

okay! thanks!

Answer 6

use distance formula in the first part ( as mentioned by @OrthodoxMan )
and in the second part use the formula
\[x = (x _{1} + x _{2})/ 2\]
and
\[y = (y _{1} + y _{2})/2 \]

and in this equation x1,x2 ,y1 and y2 are the co-ordinates of the given points :)

Answer 7

thanks guys!

Answer 8

Here is my first part...am I correct?
\[\sqrt{(12+4)^2 + (-5-3)^2} = \sqrt{(16)^2} = \sqrt{32 + 16} = \sqrt{48} = 6.928?\]

Answer 9

For part b: I got x=4 and y=-1...so how do I write my answer to the question for part b?

Answer 10

@Shizen @OrthodoxMan

Answer 11

for part b, it's just the coordinates
(x,y)

Answer 12

Shizen, you answer part a

Answer 13

@djibben615 you can't add the squares that ways!
cos
\[a ^{2} + b ^{2} = (a+b)^{2} - 2ab\]

Answer 14

in other words
Find out the sqaure of 16
and the square of 8

and then add them
and find out their square root

Answer 15

\[\sqrt{256 + 64 } = ?\]

find this question mark and thats your answer :)

Answer 16

And now lemme tell you about the mistake that you've been making

you added
the 16^2 + 8^2 in a 'wrong' way :)

Answer 17

so the answer would be 24?

Answer 18

No!
BTW, can you show me the procedure of getting 24..

Answer 19

look
square the 16
which would be 16*16 = 256
and square 8
which would be 8*8 = 64

and now add these
256 + 64
= 320

and now get its square root :)
which would be 17 something

Answer 20

and i repeat it again
\[16^{2} + 8^{2} \neq 24^{2}\]

and
\[16^{2} + 8^{2} = (16+8)^{2} - 2 (16)(8)\]

Answer 21

ah!!! LIGHTBULB! so its 17 then?

Answer 22

what about part 2? is the answer: x=4 and y=-1?

Answer 23

17.88 !
i.e. square root of 320

Answer 24

yup :)
you are correct with the second :)

Answer 25

Thanks!

Answer 26

Up to helping with another problem? It is attached...