Given the points P(-1,1) and S(2,2), give the equation of the line passing through P and perpendicular to PS.

First, we can find the slope of the line PS, with the help of the formula m = (y2- y1)/(x2- x1) or if we substitute for the given points mPS = (2 - 1)/(2 - (-1)) = (-1)/(3)= -1/3. Since the line l passing through P is perpendicular to PS, for l's slope ml and PS's slope mPS we know that mPCml=-1. From 1/3 * ml = -1 we can conclude that ml = -3. Any line's equation can be represented in the form y = mx + b - so l's equation is y = -3x + b, and b = y + 3x. If we substitute (x,y) for the coordinates of P(-1,1), we will get b = 1 - 3 = -2. Finally, l's equation can be written as y = -3x -2 or 3x + y + 2 = 0.

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Answered by Kalina S. Maths tutor

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