1. ## Geometry/conics

I hope this question goes here.
Heres my question:
The line y-2x+u=0 is 2sqrt5 units from the point (1,-3). Find the possible values of u.

Here is my attempt:
y-2x+u
((-3)-2(1)+u)/√(2^2+1^2 )=2√5
(-5+u)/√5×√5/1=(2√5)/1×√5/1
-5+u=10
u=15 Or u=?

I hope that makes sense, sorry about the mess I dont know how to use the program here.
I can't find the other answer, but I know that the answer is u=15 or u=-5.
Thanks.

2. You forgot the absolute values in the perpendicular distance formula $d = \frac{|Am_1 + Bm_2 + C |}{\sqrt{A^2 + B^2}}$

So really, you need to solve $|u-5| = 10$, not just $u - 5 = 10$

3. ## Thanks

does that mean it turns into u+5=10.
Thanks for help.

4. The absolute value means u - 5 can be a positive or negative number of the same magnitude.

So aside from (u-5) = 10, you also have -(u-5) = 10

5. ## thanks

thanks, that really helped.
I hope people continue to help.
Thanks