SOLUTION: The equation of the perpendicular bisector of the line segment joining the points P
(7,4) and Q (a, b) is 4x + 3y - 15 = 0. The point Q is:
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Question 1199457: The equation of the perpendicular bisector of the line segment joining the points P
(7,4) and Q (a, b) is 4x + 3y - 15 = 0. The point Q is:
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
The equation of the perpendicular bisector of PQ is 4x+3y-15=0, or y=(-4/3)x+5.
The slope of the perpendicular bisector is -4/3, so the slope of PQ is 3/4.
I'll outline the process for solving the problem and let you fill in the details.
(1) Use the given point (7,4) and the slope of 3/4 to find the equation of line PQ.
(2) Solve the pair of equations representing PQ and the perpendicular bisector to find the point of intersection. (You should find it is (3,1); that point is the midpoint of PQ).
(3) Use the coordinates of P and the midpoint of PQ to find the coordinates of Q. (You should find it is Q(-1,-2)).
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