In 2d space, given P1, and P2, find a perpendicular point (270 degrees) to the left of the line between P1, and P2

Question

Find a perpendicular point, P3, given a vector P1P2, and the distance between P2 and P3. Assuming the direction along the line P1 and P2 is 0 degrees, find the point 270 degrees to the left of P2. This is in 2d space.

Answer 1

Take coordinates of $P_1$, $P_2$ and $P_3$ as $(x_1,y_1)$, $(x_2,y_2)$ and $(x,y)$ respectively. Then we have:$$P_1P_2=(x_2-x_1,y_2-y_1)\\P_2P_3=(x-x_2,y-y_2)\\P_1P_2.P_2P_3=0\to (x_2-x_1)(x-x_2)+(y_2-y_1)(y-y_2)=0$$ also $$(x-x_2)^2+(y-y_2)^2=b^2\to {x=x_2+\frac{(y_2-y_1)b}{P_1P_2}\\y=y_2-\frac{(x_2-x_1)b}{P_1P_2}}$$