Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2005-05-17 07:07:33

Milos
Member

Offline

Analytic geometry

Find equation of a straight line which passes thought O point with coordinates (0,0) if it is cut by two straight lines with equations:
2x+5-y=0; 2x-y+10=0; and length of the part which is cut by these two straight lines is sqr10.

I think that this one is interesting. Try it.

#2 2005-05-18 21:13:37

MathsIsFun

Offline

Re: Analytic geometry

Rearrange into classic y=mx+c:

y=2x+5
y=2x+10

Hey, they have the same slope (parallel) !

The line we are trying to find has the equation y=mx+c, but c is 0 because you said it passes through (0,0) so it is really just:

y=mx (where m is the slope we want to find)

This should intersect the first line at 2x' + 5 = mx'  ==> (2-m)x' + 5 = 0  ==> x'= 5 / (m-2)
... and the y value is y'=2x'+5 = 2 (5 / (m-2)) + 5 = 10/(m-2) + 5

and should intersect the second line at 2x'' + 10 = mx''  ==> (2-m)x'' + 10 = 0  ==> x''= 10 / (m-2)
... and the y value is y'=2x'+10 = 2 (10 / (m-2)) + 10 = 20/(m-2) + 10

Because we have a LOT of 1/(m-2) terms lets call that "k"

So, the intersection points are:
1) x'=5k and y'=10k+5
2) x''=10k and y''=20k+10

The distance between these two points is

d = sqrt( (x''-x')^2 + (y''-y') ) which you say is also equal to sqrt(10)

sqrt(10) = sqrt( (x''-x')^2 + (y''-y') ) = sqrt( (10k-5k)^2 + (20k+10 - 10k -5)^2 ) = sqrt ( (5k)^2 + (10k+5)^2 )

10 = (5k)^2 + (10k+5)^2 = 25 k^2 + 25 (2k+1)^2 = 25 k^2 + 25 (4k^2 + 4k + 1)

Subtract 10 and gather terms: 0 = 25 (1+4) k^2 + 100 k + 25 - 10

Neaten Up: 125 k^2 + 100 k + 15 = 0

Divide by 5: 25 k^2 + 20 k + 3 = 0

Factor: (5k+3)(5k+1) = 0

The roots are: k = -(3/5) and -(1/5)  (if you don't believe me just try them)

... but we didn't want k we wanted m ... !!

k=1/(m-2), so we solve that knowing k and get: m = 1/3 and m = -3

So there are two slopes that work!

THE TWO SOLUTIONS ARE:

y = (1/3)x

y = -3x

NOTE 1: because this is a long solution, we skipped over some minor equation solving (nothing too hard!)

NOTE 2: there may be a simpler geometrical solution to this!

(Solution not by me, but by "Astronomer")

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

#3 2005-05-19 09:07:32

Milos
Member

Offline

Re: Analytic geometry

Wow, a very long one - but correct. Yes, there is a shorter solution for this:
As we have distance which is cut by two straight lines on the one that we need to find equation for we will do this:
d(A,L2)-distance between L1: y1=2x1+5 and L2: y2=2x1+10, where A is point that belongs to L1-(we will pick a number for y, let it be 1, and then we find value for x, which is -2):
so A(-2,1)
d(A,L2)= |mx1-y2+n|/sqr m^2+1 ; x1= -2;y2= 1
d(A,L2)=sqr5
*Now triangle can be formed(right angled)- draw it. we have sqr10 and sqr5
c^2=a^2+b^2; c=sqr10 and a=sqr5
then b=sqr5
Now as a=b=sqr5 it means that triangle's angles are 90, 45, 45.(if this was not the case we would solve it using trigonometry)
As we have formula:
tanß(m2-m)/1+m*m2|  where ß=45 is angle between two lines; m2=2,  m- unknown
find m from here and you will get
m'= -3 ; L' : y'=-3x'
m''=1/3;  L'' : y''=1/3x''

#4 2005-05-31 21:05:31

Mr T
Super Member

Offline

Re: Analytic geometry

OMG

I come back stronger than a powered-up Pac-Man
I bought a large popcorn @ the cinema the other day, it was pretty big...some might even say it was "large
Fatboy Slim is a Legend

#5 2005-05-31 21:11:39

MathsIsFun

Offline

Re: Analytic geometry

Is that McCulkin McSkulkin around?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman