Find the equations of the lines which pass through (1,2) and have perpendicular distance 1 from the origin.?

Question

There's the obvious vertical one, y=2, and the other one you can say has equation:
y-2=m(x-1)
Where m is the gradient.
The line perpendicular to this is y= -x/m , so the way i did it was to say that you know that where these two lines meet, x^2 + y^2 = 1, by Pythagoras's Theorem.
So then you can set up simultaneous equations and work through some really complicated algebra and the answer does eventually come out, but i was just wondering if there is a better way than that? Can anyone help?

Answer 1

If you sketch it you'll see that there are only two more lines that behave that way. They form two right triangles that share a hypothenuse from (0,0) to (1,2). The bases, short side, are the perpendicular lines of length 1.

So, now you know two sides of both triangles. You can proceed in several ways but I would locate the coordinates of the corner points, where the right angle is, and use these points and (1,2) to find the equations of both lines.

Just draw a sketch and it will become clear.

Answer 2

DO YOUR OWN HOMEWORK