Given a point P in 2-D plane and equation of a line, the task is to find the foot of the perpendicular from P to the line.
Note: Equation of line is in form ax+by+c=0.
Input : P=(1, 0), a = -1, b = 1, c = 0 Output : Q = (0.5, 0.5) The foot of perpendicular from point (1, 0) to line -x + y = 0 is (0.5, 0.5) Input : P=(3, 3), a = 0, b = 1, c = -2 Output : Q = (3, 2) The foot of perpendicular from point (3, 3) to line y-2 = 0 is (3, 2)
Since equation of the line is given to be of the form ax + by + c = 0. Equation of line passing through P and is perpendicular to line. Therefore equation of line passing through P and Q becomes ay – bx + d = 0. Also P passes through line passing through P and Q, so we put coordinate of P in the above equation:
ay1 - bx1 + d = 0 or, d = bx1 - ay1
Also, Q is the intersection of the given line and the line passing through P and Q. So we can find the solution of:
ax + by + c = 0 and, ay - bx + (bx1-ay1) = 0
Since a, b, c, d all are known we can find x and y here as:
Below is the implementataion of the above approach:
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