points equidistant from two lines

elifast · September 16th, 2012, 06:51 PM

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!

MarkFL · September 16th, 2012, 06:56 PM

Is your line $y=x+2$ ?

elifast · September 16th, 2012, 07:17 PM

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)

MarkFL · September 16th, 2012, 07:34 PM

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$

HallsofIvy · September 17th, 2012, 02:00 PM

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.

skipjack · September 17th, 2012, 03:35 PM

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.

elifast · September 18th, 2012, 06:35 PM

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!!!

September 16th, 2012, 06:51 PM	#1
elifast 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!
$elifast is offline$

September 16th, 2012, 06:56 PM	#2
MarkFL Global Moderator Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 11,808 Thanks: 268 Math Focus: The calculus	Re: Find the point on the line that is equidistant from 2 po Is your line $y=x+2$ ?
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September 16th, 2012, 07:17 PM	#3
elifast 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)
$elifast is offline$

September 16th, 2012, 07:34 PM	#4
MarkFL Global Moderator Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 11,808 Thanks: 268 Math Focus: The calculus	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$
$MarkFL is offline$

September 17th, 2012, 03:35 PM	#6
skipjack Global Moderator Joined: Dec 2006 Posts: 11,462 Thanks: 300	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.
$skipjack is offline$

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