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")

OMG