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I will refer to the point of projection as as $(X_p,Y_p)$. Using the same observation, that two orthogonal slopes multiplied together make -1, the slope of the projection line is -1/m and it is also the rise over run for the arbitrary point (X,Y) and the point of projection $(X_p,Y_p)$ .
By first applying coordinate transformations a reduced algebra solution is possible. Given Circle (x1,y1,R) and Circle (x2,y2,P) find the two intersection points of the circles. Define d=distance(C1,C2). There are multiple conditions for Zero and One intersection points. Here we assume two points thus dR, and d-P>-R.
find the equations of the lines and equate the distance from the point P TO THE GIVEN TWO Pairs of lines .if the point is inside the rectangle,the distances to the pairs will be less than the distances between the line in the pair
Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
If, from our point we were translate in twice the direction of unit normal scaled up by the distance from line to point, we would reach the image point. Hence, the image coordinates $(I)$ are given by the vector:
Every point in space will have a unique such vector, so there's no ambiguity if I said something like "the vector for the point P". But it works the other way, too, in that if I take any arrow that starts at the origin, it is associated with a unique point in space (the tip of the arrow). So there's no ambiguity if I say, "the point for the vector V". Since there's a one-to-one dictionary ...
Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Intuitively, you want the distance between the point A and the point on the line BC that is closest to A. And the point on the line that you are looking for is exactly the projection of A on the line. The projection can be computed using the dot product (which is sometimes referred to as "projection product").