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


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 $x_C=0$

  • $\begingroup$ @ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.