We are given y=3-x passes through (1,2) and that line L, y = (2-a) + ax, also does, for any value "a".
We are then told to:
i) find the equation of the line perpendicular to line L which passes through (1,2)
ii) Find equations of lines which pass through the point (1,2) and have perpendicular distance 1 from origin
So i) is not a problem.. y = -1/a(x-1) + 2b
but for ii) i thought of using the equation L to be a tangent to the circle x^2 + y^2 = 1
so substitute y = (2-a) + ax into the equation of the circle, this gives a = 3/4
so the equations of the lines are x=1 and 4y = 3x + 5
Is this correct? Are there any other methods?
Also, do we have to consider the case for the equation we got in i) to be a tangent to the circle? Substituting y = -1/a(x-1) + 2b into the circle gives me a quartic equation, with complicated roots.