hello, I have this exercice for practicing distance vectors:
it asks for me the point or points at the straight that are at two of distance from P(-2,2)
this is what the exercice tells me: r: (x - 1) / 2 = (y + 1) / -1, that the distance between the point P and the other points must be 2 (d = 2) and that the point P is P(-2,2)
What I though is to use the formula d = |Ax1 + By1 + C| / sqrt(a^2 + b^2), so first I got the general equation: x + 2y + 1 = 0, moreover, A = 1, B = 2, C = 1
(in what I was trying I supposed there is just 1 point at the line, but I don't know if it is good) the point that is in the line I called it O and then the width betweeen O and P is 2, so | OP | = 2, | OP | = P - O = (Px, Py) - (Ox, Py) = (-2,2) - (Ox, Oy) (this I am not pretty sure if it is right)
and I think aswell that | OP | = sqrt((Px2 - Ox1)^2 + (Py2 - Oy1)^2)
and then I am stucked here, because I don't know how to find the x1 and y1 from the formula distance between a point and a line, should I not use this formula?
how could I continue?