<![CDATA[Math Is Fun Forum / Correct]]> 2016-12-18T00:33:19Z FluxBB https://www.mathisfunforum.com/viewtopic.php?id=4975 <![CDATA[Re: Correct]]> No problem.

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https://www.mathisfunforum.com/profile.php?id=33790 2016-12-18T00:33:19Z https://www.mathisfunforum.com/viewtopic.php?pid=391672#p391672
<![CDATA[Re: Correct]]> Well, I answered post #3, since he seemed to not understand implicit answers.

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https://www.mathisfunforum.com/profile.php?id=212031 2016-12-18T00:15:36Z https://www.mathisfunforum.com/viewtopic.php?pid=391670#p391670
<![CDATA[Re: Correct]]> Hi;

John gave an answer in post #2 also.

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https://www.mathisfunforum.com/profile.php?id=33790 2016-12-17T03:28:22Z https://www.mathisfunforum.com/viewtopic.php?pid=391554#p391554
<![CDATA[Re: Correct]]> Problem solved, after 10 years..

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https://www.mathisfunforum.com/profile.php?id=188456 2016-12-17T03:13:47Z https://www.mathisfunforum.com/viewtopic.php?pid=391550#p391550
<![CDATA[Re: Correct]]> No. The distance of a point and a line is 0 if the point is in the line. In your question, however, the coordinate (3,1) doesn't satisfy the equation y = x - 4 because 3 - 4 ≠ 1.

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https://www.mathisfunforum.com/profile.php?id=212031 2016-12-16T07:54:22Z https://www.mathisfunforum.com/viewtopic.php?pid=391528#p391528
<![CDATA[Re: Correct]]> so am i right?

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https://www.mathisfunforum.com/profile.php?id=5209 2006-11-08T19:42:26Z https://www.mathisfunforum.com/viewtopic.php?pid=48736#p48736
<![CDATA[Re: Correct]]> .
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.  / o
. /   o
./     o
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.        o
.         o
.
.     square root of 2
.     one down and one right.

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https://www.mathisfunforum.com/profile.php?id=1108 2006-11-08T18:10:07Z https://www.mathisfunforum.com/viewtopic.php?pid=48731#p48731
<![CDATA[Correct]]> Find the distance between the point (3,1) and the line y = x - 4

i did like this :

y = -2x + b < equation
1 = -2(2) + b < use (3,1)
1 = -6 + b < add 6
7 = b
y = -2x + 7 < completed equation
THAN
x - 4 = -2x + 7 < substitution
x = -2 + 11
3x = -2 + 11
3x = 9
x = 9/3 < solved for x
THAN
y = -2 (9/3) + 7
y = -6 + 7
y = 1 < solved for y
THAN distance formula
D = sqrt( ( (x2 - x1)^2 + (y2 - y1)^2 )
(3,1) and (3,1)
D = sqrt ( (3 - 3)^2 + (1 - 1)^2 )
D = sqrt( 0 + 0)
D = 0 < solved
correct? 