Question

Really need help solving this problem.

A thousand points are graphed in a coordinate plane. Explain why it is possible to draw a straight line in the plane so that half of the points are on one side of the line and half are on the other.
(Consider the slopes of the lines determined by each pair of points.)

Answer 1

Really need help solving this step by step. I know the answer is 2 but i dont know how to get the answer by doing it on paper\[\frac{\sqrt{2}+\sqrt{6}}{\sqrt{2+\sqrt{3}}}\]

Answer 2

you can easily draw a straight line through the pts by finding semi averages......the standard deviation of both sides of the line must be the same if it the line of best fit.......remember if it is a liner relationship it is possible to draw a straight line the fluctuations in pts might be due to experimental errors by drawing the line of best fit u are actually trying to rule out these errors

Answer 3

so would you only be able to do this with linear equations?

Answer 4

Let \[y = \frac{ \sqrt{2} + \sqrt{6}}{\sqrt{2 + \sqrt{3}\]

Answer 5

oops !

Answer 6

\[y = \frac{\sqrt2 + \sqrt3}{\sqrt{2 +\sqrt3}}\]

Answer 7

then square them

Answer 8

oh no typo again its 6 not 3 in numerator

Answer 9

\[y^2 = \frac{ 2 + 6 - 2 \times 2 \times \sqrt3}{2 + \sqrt3}\]

Answer 10

it gives \[y^2 = 4 \] \[y = \pm 2\] ...But you take y = 2

Answer 11

To answer the first question ie
"A thousand points are graphed in a coordinate plane. Explain why it is possible to draw a straight line in the plane so that half of the points are on one side of the line and half are on the other. (Consider the slopes of the lines determined by each pair of points.)"

If u imagine an arbitrary line in the plane and then for each point make a point on the line at the perpendicular distance, then mark a point on the line such that half of the points are on one side of it, then the perpendicular line at that point will divide all the points into 2 equal halves (see pic)

Might be some odd possibilities, lines in arow, that sort of thing, u can just select a different arbitrary line to get round it, the principle involved is still valid.

Answer 12

ishaan94- howd did you get that the y^2=4? i typed it into the calculator and i didnt get four

Answer 13

if not linear then you can make them linear by linear law

Answer 14

Courtney16
top is 8 + 4 root3 = 4(2 + root 3) so 4.

(ishaan94 made a typo with the sign)

Answer 15

estudia-does this answer have anything to do with the slopes though?

Answer 16

and thank you!

Answer 17

oh yeah typo sorry thanks estudier

Answer 18

Julia asked me the same thing:-)
U can see the answer I gave to her (basically the "helpful hint" is not helpful(to me) since I can't thin of any rule to associate pairs of points ie if I choose one pair, I then have 999 candidates for the other).

Doesn't mean there is not some method, I am thinking about it...

Answer 19

can i see the link to see what you told her?

Answer 20

Besides, I like my answer....:-)

Answer 21

Not Julia (smacks head) tulip

http://openstudy.com/users/tulip235/updates/4e32e0770b8ba7b2da418624

Answer 22

ha its ok.. for some reason its not clicking with me!so you basically draw a line and then every point you have...i dont understand what you do to have half the points on each side

Answer 23

Did u look at the pic?

U have projected all the points perpendicularly (a projection) onto an arbitrary line.

This gives them a linear order and u can count them.

So u can select (on the arbitrary line) a point between the 2 middle points and draw a perpendicular to divide the points in two halves.

This demonstrates that it can be done although I haven't considered some edge cases (u can always shift the arbitrary line)

Answer 24

Try putting all of the points on one side of the arbitrary line and follow the procedure...

Answer 25

ohh okay..i think!! so if you graphed a thousand points, you would just draw an arbitrary line and then draw a perpendicular line inbetween 500 and 501?

Answer 26

Yay!

Answer 27

thank you soo much haha!! i dont understand why it took me that long to figure that out! and for that graphing question you answered for tulip,is it really just y=5 y=-x and y=x?

Answer 28

Looks that way...

Answer 29

Cubic in y so 3 roots.

Answer 30

ok! and i just saw another person ask the exact same question as what i did haha!

Answer 31

\[\frac{ \sqrt{2}+\sqrt{6} }{ \sqrt{2+\sqrt{3}} }=\frac{ \sqrt{2}+\sqrt{6} }{ \sqrt{\frac{ 1 }{ 4 }\left( 8+4\sqrt{3} \right)} }=\frac{ 2\left( \sqrt{2}+\sqrt{6} \right) }{ \sqrt{2+2\sqrt{12}+6} }=\frac{ 2\left( \sqrt{2}+\sqrt{6} \right) }{ \sqrt{\left( \sqrt{2}+\sqrt{6} \right)^{2}} }=\frac{ 2\left( \sqrt{2}+\sqrt{6} \right) }{ \sqrt{2}+\sqrt{6} }=2\]
This is the correct approach to the problem that was given, using the Perfect Square Special Factoring Formula.