Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**david****Guest**

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!!!!!

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 24,254

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.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

Pages: **1**