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I seem to be missing something obvious here.

I am given a line r: $x + y -2 = 0$ and two points $A = (1,2)$ and $B = (3,4)$

So I decided that I needed to find a point P in which d(AP) = d(BP)

I thus equaled the equations of $d(AP)$ and $d(BP)$, eventually reaching that $2y + 6x - 20 = 0$

Taking $y = -3x + 10$, I equaled this equation to the equation of my line, which I took to be $y = -x + 2$

This yielded the point $P = (6,-4)$ as a result, which is a point on r. However, when I proceed to calculate d(AP) and d(BP), they are not equal. Where did I go wrong?

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  • $\begingroup$ What were your “equations of $d(AP)$ and $d(BP)$?” $\endgroup$ – amd Apr 20 '18 at 20:51
  • $\begingroup$ This is an exercise in a linear geometry class, can you think of any way to resolve this problem that makes use of techniques involving dot product, vectors, etc? $\endgroup$ – Jake Apr 21 '18 at 13:39
  • $\begingroup$ Sorry, I meant linear algebra $\endgroup$ – Jake Apr 21 '18 at 14:43
  • $\begingroup$ Your equation of the bisector is incorrect. $\endgroup$ – amd Apr 23 '18 at 22:30
  • $\begingroup$ $ d(AP)= d(BP)$ gives you the radical axis or straight line when $x^2,y^2 $ terms cancel out. I got $x+y=5, $ please check. Find where this line cuts $ x+y=2 $. They are parallel ! $\endgroup$ – Narasimham Apr 28 '18 at 7:25
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By eyeballing, the bisector (=locus of equal distances) of $(1,2)$ and $(3,4)$ should be $$ x-y+1=0.$$ Check again how you arrived at $2y+6x-20=0$ instead.

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  • $\begingroup$ This is an exercise in a linear geometry class, can you think of any way to resolve this problem that makes use of techniques involving dot product, vectors, etc? $\endgroup$ – Jake Apr 21 '18 at 13:38
  • $\begingroup$ Apologies, I meant a linear algebra class $\endgroup$ – Jake Apr 21 '18 at 14:43
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Bisector should be $$x+y=5$$ and not $2y+6x-20=0$.

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The parametric equations of the line r are

$$x=t $$ $$y=2-t $$

$$d ^2(AP)=(1-t)^2+(2-2+t)^2$$ $$d^2 (BP)=(3-t)^2+(4-2+t)^2$$

thus $$d^2 (AP)=2t^2-2t+1$$ $$=d^2 (BP)=2t^2-2t+13$$

The point P does not exist.

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  • $\begingroup$ Does that means that the line in between the two points and the given line are parallels? $\endgroup$ – Max Ft Apr 20 '18 at 20:32
  • $\begingroup$ @MaxFt yes {}{}{}{}{}{}{}{}{}{} $r:y=2-x$ and $y=5-x $ are parallel. $\endgroup$ – hamam_Abdallah Apr 20 '18 at 20:35
  • $\begingroup$ This is an exercise in a linear geometry class, can you think of any way to resolve this problem that makes use of techniques involving dot product, vectors, etc? $\endgroup$ – Jake Apr 21 '18 at 13:39
  • $\begingroup$ Apologies, I meant linear algebra $\endgroup$ – Jake Apr 21 '18 at 14:44
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The line containing $A$ and $B$ is $x - y +1=0$

This line is perpendicular to $x+y+2 = 0.$ Which means that the only way that a point on the line $x+y+2=0$ is the same distance to $A$ as to $B$ is if the line perpendicularly bisects $AB.$ (which it doesn't) And if it did, then every point on the line would be equal distance from $A$ as it is to $B.$

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