Question

what are the coordinates of the points 3 units from the y-axis and at distance squareroot of 5 from (5,3)?

Answer 1

@AllTehMaffs

Answer 2

So this time, with the distance formula
\[d =\sqrt{ (x_2-x_1)^2+(y_2-y_1)^2} \]
you have one set of ordered pairs
\[ (5,3)=(x_1,y_1)\]
and a value for d
\[d=\sqrt {5}\]

What do you know about the other points?

Answer 3

3 units from y-axis = (3,y) ?

Answer 4

almost - (±3,y)

Answer 5

i dont knw that sample part of my reviewer

Answer 6

it doesn't say which direction the units are (that's the tricky part)

Answer 7

right

Answer 8

huh can you help me

Answer 9

So, @jacalneaila , It's like the other problem, except you're solving for y given a distance

\[\sqrt{5} =\sqrt{ (3-5)^2+(y-3)^2} \]

Answer 10

You know that the first part of the ordered pair (the x component) is either (+3,y) or (-3,y), so you solve for 2 solutions of y

Answer 11

\[\sqrt{5} =\sqrt{ (3-5)^2+(y-3)^2} \]
\[5 =(3-5)^2+(y-3)^2 \]
\[5 =4+y^2-6y+9 \]
\[y_2-6y+8=0\]

can you factor that?

Answer 12

factor?

Answer 13

what next

Answer 14

\[y^2-6y+8=0\]
\[(y-4)(y-2)=0\]
so at (+3,y) there are two values of y that are a distance of sqrt 5 from point (5,3)
y=4
y=2
So the two points would be (3,4) and (3,2)

Now we have to do the same thing for (-3,y)

\[\sqrt 5 =\sqrt{ (-3-5)^2+(y-3)^2}\]
\[5=(8)^2+(y^2-6y+9)\]
\[5=64+y^2-6y+9\]
\[y^2-6y-68=0\]
then the roots are
\[y=\frac{6±\sqrt{36+272}}2\]
\[y=3±\sqrt {77}\]

although I dunno if that's right

Answer 15

huh.. i will review that

Answer 16

next example

Answer 17

How many examples do you have?

Answer 18

in my reivewer its 50 examples .. i have exam tommorow its 200 question

Answer 19

I can help you with a few more I think ^^

Answer 20

I don't think I'll be able to last through all 50 though!

Answer 21

Just pick the one or two that you're having the most trouble with, unless you think that they're really short, then maybe a few more than 2.

Answer 22

ok thank you

Answer 23

can close this? then next sample

Answer 24

sure