What is the perpendicular bisector equation of a line segment with endpoints of p q and 7p 3q?

What is the perpendicular bisector equation of a line segment with endpoints of p q and 7p 3q?
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A line with slope m has a perpendicular with slope m' such that:mm' = -1→ m' = -1/mThe line segment with endpoints (p, q) and (7p, 3q) has slope:slope = change in y / change in x→ m = (3q - q)/(7p - p) = 2q/6p = q/3p→ m' = -1/m = -1/(q/3p) = -3p/qThe perpendicular bisector goes through the midpoint of the line segment which is at the mean average of the endpoints:midpoint = ((p + 7p)/2, (q + 3q)/2) = (8p/2, 4q/2) = (4p, 2q)A line through a point (X, Y) with slope M has equation:y - Y = M(x - x)→ perpendicular bisector of line segment (p, q) to (7p, 3q) has equation:y - 2q = -3p/q(x - 4p)→ y = -3px/q + 12p² + 2q→ qy = 12p²q + 2q² - 3pxAnother Answer: qy =-3px +12p^2 +2q^2
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