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**Milos****Member**- Registered: 2005-05-06
- Posts: 44

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.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,534

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

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**Milos****Member**- Registered: 2005-05-06
- Posts: 44

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''

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**Mr T****Member**- Registered: 2005-03-30
- Posts: 1,012

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

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,534

Is that McCulkin McSkulkin around?

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

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