geometry
posted by xavier .
Find the equation of a line parallel to 5x +6y = 0 at distance sqrt 61 from (3,7)
I need help please. I don't have any idea

y = (5/6)x for the first line, so the slope of your second (parallel) line is also 5/6. Its equation is
y' = (5/6)x + C
C is the yintercept of the second line.
You have to choose C so that y and y' are sqrt61 apart.
C is the hypotenuse of a right triangle with sides sqrt61 and (5/6)*sqrt61. Its value is sqrt[61 + (25/36)*61]
= sqrt[(61/36)*61] = 61/6
The line you want is
y' = (5/6)x + 61/6 
There is a second line that is also a solution, below the first line
y' = (5/6)x  61/6