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 September 16th, 2012, 07:51 PM #1 Newbie   Joined: Sep 2012 Posts: 15 Thanks: 0 Find the point on the line that is equidistant from 2 points Find the point (x, y) on the line y=1 x + 2 that is equidistant from the points (?2, 2) and (?10, 4). I missed my last lecture and she explained to the class about these types of equations! I don't have a buddy in the math class yet so I couldn't turn to anyone for help! Could someone please answer this practice question so I can look at the steps and understand it? Would be greatly appreciated! Thank you!
 September 16th, 2012, 07:56 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Find the point on the line that is equidistant from 2 po Is your line $y=x+2$ ?
 September 16th, 2012, 08:17 PM #3 Newbie   Joined: Sep 2012 Posts: 15 Thanks: 0 Re: Find the point on the line that is equidistant from 2 po I think the question is messed up or something.... I have the points distance and midpoint, but when I plug it into that line it makes the question incorrect..... Distance: sqrt68 Mid point: (-6,4)
 September 16th, 2012, 08:34 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Find the point on the line that is equidistant from 2 po I'll show you the general method by deriving a formula for any non-vertical line and two arbitrary points. Let the line be $y=mx+b$ and the two arbitrary point be $$$x_1,y_1$$,\,$$x_2,y_2$$$ We let the point on the line be $$$x,y$$=$$x,mx+b$$$ and we equate the square of the distances from this point on the line to the two arbitrary points: $$$x-x_1$$^2+$$mx+b-y_1$$^2=$$x-x_2$$^2+$$mx+b-y_2$$^2$ Next, we expand the squared binomials: $x^2-2x_1x+x_1^2+m^2x^2+2bmx-2my_1x+b^2-2by_1+y_1^2=x^2-2x_2x+x_2^2+m^2x^2+2bmx-2my_2x+b^2-2by_2+y_2^2$ Remove terms common to both sides: $-2x_1x+x_1^2-2my_1x-2by_1+y_1^2=-2x_2x+x_2^2-2my_2x-2by_2+y_2^2$ Arrange with terms involving x on the left and the rest on the right: $2x_2x-2x_1x+2my_2x-2my_1x=x_2^2-x_1^2+2by_1-2by_2+y_2^2-y_1^2$ Factor 2x from the left side: $2x$$x_2-x_1+my_2-my_1$$=x_2^2-x_1^2+2by_1-2by_2+y_2^2-y_1^2$ Divide through by $2$$x_2-x_1+my_2-my_1$$$ $x=\frac{x_2^2-x_1^2+2by_1-2by_2+y_2^2-y_1^2}{2$$x_2-x_1+my_2-my_1$$}$ $y=m$$\frac{x_2^2-x_1^2+2by_1-2by_2+y_2^2-y_1^2}{2\(x_2-x_1+my_2-my_1$$}\)+b$
September 17th, 2012, 03:00 PM   #5
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Re: Find the point on the line that is equidistant from 2 po

Quote:
 Originally Posted by elifast I think the question is messed up or something.... I have the points distance and midpoint, but when I plug it into that line it makes the question incorrect..... Distance: sqrt68 Mid point: (-6,4)
You are assuming that the two points are on the given line and that is not true.

 September 17th, 2012, 04:35 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,288 Thanks: 1968 The mid-point of the line segment joining (-2, 2) and (-10, 4) is (-6, 3). The line y = 4x + 27 is the perpendicular bisector of that line segment, so every point on that line is equidistant from the points (-2, 2) and (-10, 4). Find where that line intersects the line given in the question.
September 18th, 2012, 07:35 PM   #7
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Re:

Quote:
 Originally Posted by skipjack The mid-point of the line segment joining (-2, 2) and (-10, 4) is (-6, 3). The line y = 4x + 27 is the perpendicular bisector of that line segment, so every point on that line is equidistant from the points (-2, 2) and (-10, 4). Find where that line intersects the line given in the question.
Thank you so much! You're honestly a life saver!

All of you!
Thank you so much for the help!!!

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