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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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Topic review (newest first)

ganesh
2005-10-05 14:57:53

Given two points, the equation of the line joining the points is
(y-y1)/(y2-y1) = (x-x1)/(x2-x1)
Substituting y1 = 5, y2 = -1, x1 = 3 and x2 = -3, the equation of the line is
x-y+2 = 0,
The slope of this line m=(y2-y1)/(x2-x1), i.e., m = 1.
The slope of a line perpendicular to this line would be -1
(Because, if two lines are perpendicular, the product of their slopes is -1).
The midpoint of U(3,5) and V(-3, -1) is given by
[(x1+x2)/2, (y1+y2)/2]
Therefore, the midpoint of the line UV is
(0,2).
The equation of a line of given slope passing through a given point is
(y-y1) = m(x-x1)
We know, m=-1 for the perpendicular line.
Therefore, the equation of the perpendicular bisector is
(y-2) = -1(x-0)
or
x+y-2 = 0
Substitute the x and y coordinates of point W(2,-1) in this equation.
We see that it does not satisfy the equation.
Therefore, the point does not lie on the perpendicular bisector.

david
2005-10-05 10:37:59

Determine whether the point W(2,-1) lies on the perpendicular bisector of line segment UV, endpoints U(3,5) and V(-3,-1). Explain and justify your answer.


I cant figure it out, if someone could, it would be a pleasure!!!!!

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