Find the point on a line that is of equal distance from two points.
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Find the point on a line that is of equal distance from two points.

[From: ] [author: ] [Date: 11-09-13] [Hit: ]
The perpendicular slope is 1.Thus we have derived this equation.(x=-7/2,And that is the point on the line defined by your equation that is equidistant from your two points. Understand?Ive looked over Ferozesos solution,......
Find the point on the line y=3 x + 4 that is of equal distance from the points (0, 6) and (9, -3).

I truly though I had a good grasp on distance problems until this. No clue what to do!

Any help would be great, thanks!

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The points that are equally distant from two points are on a line that bisects that line segment.

(midpoint 0 6 9 -3)
Point is 9/2, 3/2

So you have a line going through those points that is perpendicular through the line represented by (line-from-points 0 6 9 -3)
y=-1x+6

The perpendicular slope is 1.

So we get the y intercept by
3/2 = 1(9/2) + b -> b= -3

Thus we have derived this equation.

y = x - 3

Solve that in a system with the equation you got and you get the following:

(x=-7/2, y=-13/2)

And that is the point on the line defined by your equation that is equidistant from your two points. Understand?

I've looked over Ferozeso's solution, I do not believe it is correct at all.

(x−0)² + (3x+4−6)² = (x−9)² + (3x+4+3)² solves to x = -7/2, though, not to what he says it does (I just corrected it to solve((x-0)^2+(3x+4-6)^2=(x-9)^2+(3x+4+3… x) and the answer is x=-7/2 or -3.5 so x would be -6.5. His method would work if he could do the arithmetic :-)

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Any point on that line can be expressed as (x, 3x+4).
(x−0)² + (3x+4−6)² = (x−9)² + (3x+4+3)²
x² + (3x−2)² = (x−9)² + (3x+7)²
(3x−2)² − (3x+7)² = (x−9)² − x²
(−9)(6x+5) = (−9)(2x−9)
6x+5 = 2x−9
4x = −14
x = −3½
Hence the point is (−3½, −6½).
Careless mistake rectified. It is faster to use difference of two squares than to multiply everything out.
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