how to find point lies counter clockwise of line

Question

After solving quadratic equation, and with real root condition. How do i find which solution or point lies counter clockwise direction of a line AB as shown below. Here AB is rotating with origin A with small angle. Here A=(x0,y0), B=(y1,y1) C and C' are the solutions of quadratic equation . How to find that solution C(x3,y3) lies at counter clock wise? not c'(x3',y3'). Here A and B coordinates are known. length of AB, BC, CA is also known. and yes i want to find direction (either left or right) in relation to line AB.

Answer 1

Do you mean (counter)clockwise around point $A$ or around an origin $(0,0)$?

Let $(a,b)$ be the origin of the clock. The could be a fixed point or moving point.

Let $(x_0,y_0)$ be the "anchor point"; the point that you want to compare a second point to.

Let $(w,v)$ be the point that you want to determine is clockwise or not.

Method 1:

Let $y=m(\frac {y_0-b}{x_0-a})x + (b-\frac {y_0-b}{x_0-a}a)$ or $y = mx + d$ where $m =\frac {y_0-b}{x_0-a}$ is the slope and $d = b-\frac {y_0-b}{x_0-a}a$ is the $y$-intercept.

Case 0: $x_0 = a$. Then the line is vertical.

Case 0a: $y_0 > b$. Then $(w,v)$ is counterclockwise/clockwise/colinear to $(x_0, y_0)$ if $w <|>|= a$.

Case 0b: $y_0 < b$. Then, the exact opposite of Case 0b, $(w,v)$ is counterclockwise/clockwise/colinear to $(x_0, y_0)$ if $w >|<|= a$

Case 1: $x_0 > a$. Then the line is "moving" left to right. $(w,v)$ is counter clockwise if it is above $(x_0,y_0)$ so $(w,v)$ is counterclockwise/clockwise/colinear to $(x_0, y_0)$ if $v >|<|= mw + d$.

Case 2: $x_0 < a$. The the line is "moving" right to left and the exact opposite of Case 1, so $(w,v)$ is counterclockwise/clockwise/colinear to $(x_0, y_0)$ if $v <|>|= mw + d$

Method 2:

Shift every thing to the the origin.

Let $(x_2, y_2) = (x_0-a,y_0 - b)$ and $(u,z) = (w-a, v-b)$.

Check the signs to find out which quadrants $(x_2,y_2)$ and $(u,z)$ are in.

If $c \ge 0; d\ge 0$ then $(c,d)$ is in quadrant I. If $c < 0; d\ge 0$ then $(c,d)$ is in quadrant II. If $c< 0$ and $d< 0$ then $(c,d)$ is in quadrant III, and If $c \ge 0, d< 0$ then $(c,d)$ is in quadrant IV$.

In counter clockwise direction the quadrants go: I, II, III, IV, with I opposite of III and II opposite of IV. (Opposite but neither clockwise nor counterclockwise.) If $(x_2,y_2)$ and $(u,z)$ are in different quadrants then $(w, v)$ is clockwise, counterclockwise to $(x_0,y_0)$ if $(u,z) $ is in a clockwise, counter clockwise quadrant to $(x_2,y_2)$.

If they are in the same quadrant or opposite quadrants then take $\theta_0 = \arctan {y_0}{x_0}$ and $\theta_1 = \arctan zu$.

(Notice: if the points are in opposite quadrants, the returned angles will be "normalized". The clockwise most point will be the one with the larger angle.)

If $\theta_1 > \theta_2$ then $(w,v)$ is counterclockwise to $(x_0, y_0)$. If $\theta 1 < \theta_2$ is is clockwise. If $\theta_1 = \theta_2$ they are colinear.

Answer 2

Look at $x_0y_1 - x_1y_0 +(y_0 - y_1)x_3 + (x_1 - x_0)y_3$ (twice the signed area of the triangle ABC). This is positive if A,B,C are in counter clockwise order, negative if clockwise.