## Find the point on the line that is equidistant from 2 points

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### 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!
elifast
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### Re: Find the point on the line that is equidistant from 2 po

Is your line $y=x+2$ ?
Living in the pools, They soon forget about the sea...— Rush, "Natural Science" (1980)

MarkFL
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### 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)
elifast
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### 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$
Living in the pools, They soon forget about the sea...— Rush, "Natural Science" (1980)

MarkFL
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### Re: Find the point on the line that is equidistant from 2 po

elifast wrote: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.
HallsofIvy
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Joined: Tue Sep 11, 2007 9:14 am

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.
skipjack
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### Re:

skipjack wrote: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!!!
elifast
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