<![CDATA[Math Is Fun Forum / LINe]]> 2006-12-26T21:54:14Z FluxBB https://www.mathisfunforum.com/viewtopic.php?id=5525 <![CDATA[Re: LINe]]> the way to do this would be to see that this line would be a perpendicular bisector of the line joining the points that passes through the middle.

imagining this line between the two points, its gradient would be

so the gradient of the line we want would be

the midpoint of this line would be

so our line is

you can ofcourse just solve algebraicly,

youre errors came in in changing the right side to 0 and in the expansion of the squared terms

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https://www.mathisfunforum.com/profile.php?id=3758 2006-12-26T21:54:14Z https://www.mathisfunforum.com/viewtopic.php?pid=54234#p54234
<![CDATA[LINe]]> Write the equation of the line that is equidistant from the points (4, 2) and (-4, 3).

distance (x, y) to (-4, 3) = distance (x, y) to (4, 2)

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sqrt( [(x) - (-4)]^2 + [(3) - (y)]^2 ) = sqrt( [(4) - (x)]^2 + [(2) - (y)]^2 )

(x + 4)^2 + (3 - y)^2 = (4 - x)^2 + (2 - y)^2

x^2 + 16x + 16 + 9 - 6y + y^2 = 16 - 8x + x^2 + 4 - 4y + y^2

16x + 16 + 9 - 6y = 16 - 8x + 4 - 4y

24x + 10y - 5 = 0

y = - 12/5x + 5/10

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https://www.mathisfunforum.com/profile.php?id=5209 2006-12-26T21:33:18Z https://www.mathisfunforum.com/viewtopic.php?pid=54232#p54232