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Suppose there is given two point $A=(-4;-2)$ and $B=(3,2)$ we have to find such $C$ point on $OY$ axis, such that

a) $C$ is equidistant from $A$ and $B$

b) $ACB$ spline must be minimum.

As I know for solving part (a), we should write equation of line, which is perpendicular of $AB$ segment and goes through it's midpoint as well, in a) case coordinates of C is $(0,y)$

midpoint of $AB=(-0.5,0)$,so equation would be $y=k(x+0.5)$ where $k$ is equal to $-7/4$,because the slope of segment $AB$ is $4/7$ and we know that for perpendicular lines,$\text{slope}_1*\text{slope}_2=-1$, so we get that $y=-7/4*(x+0.5)$. If $x=0$ then $y=-0.5*7/4=-7/8$, so I have got that for a $C$ coordinates are $(0,-7/8)$. Am i correct for part (a)?

As for (b) I think that I can take symmetry point of $B'$ related to $B$ along so that $B'=(-3,2)$, equation of $AB'$ would be $y+2=4(x+4)$ or $y=4*x+14$ coordinated of $C$ is if we put $x=0$ we get $(0,14)$. Am i right? please help me to check my work

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I thing your method is corretc.A short method to solve a) is : if $I$ is the middle of $[AB]$ then $C$ is determined by : $\overrightarrow{CI}.\overrightarrow{AB}=0$ and $x_C=0$

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  • $\begingroup$ @ dato: I want help you about the question b) but I don't know what is 'spline' definition . If you can define this I try to help you. $\endgroup$ – Mohamed Jun 18 '12 at 11:48
  • $\begingroup$ please check at internet if you are interested $\endgroup$ – dato datuashvili Jun 18 '12 at 12:02
  • $\begingroup$ I'm not talking about the general definition of the word but the context of your exercises I still know the meaning of the word in general $\endgroup$ – Mohamed Jun 18 '12 at 12:41
  • $\begingroup$ spline in my question refers geometry figure,which can be represented by points connected or not connected,sorry for such explanation,but is like a not connected ordering of points $\endgroup$ – dato datuashvili Jun 18 '12 at 13:14
  • $\begingroup$ it is example of spline ka.wikipedia.org/wiki/… here it is connected $\endgroup$ – dato datuashvili Jun 18 '12 at 13:19

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