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It is a simpler problem, as uses both compass & ruler; but still need help.

Need draw perpendicular bisector of a line segment with the help of desmos, as here, as if with compass with a slightly greater radius (taken $1.3$ times the radius, as by value of parameter $a=1.3$) than the half of length of the line segment.

The issue is what dimension of $y$ to take in #$7$ function to enable full length of the segment between the two intersection points of the two circles, for all possible values of $(h_1, k_1), (h_2, k_2)$.

Also, I hope that taking ruler is logically same as being able to put a formula for calculating $y$ in function #$7$. Please correct me in this part, if wrong.

Edit -- The below ones are the additional & linked questions that arose in the same context:

There are other errors also, that concern with extra segment of arc appearing elsewhere, and hence logic is wrong & needs correction. Also, wrong logic is showing up for some negative values of $h_1, h_2, k_1, k_2$. Although for the few values used by me, obtained by changing slider for $h_1$ in the given graph always gives the bisector, but the issue is the two arcs, and line segment joining them in function #$7$.

Want to find the formula for generating label for the two intersection points.

jiten
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  • This question is about maths or about making diagrams with Desmos?? – jonsno Apr 25 '18 at 12:14
  • @samjoe It is about maths, as formula for function $7$ is not giving complete segment. Also, the arc is appearing elsewhere too. Also negative coordinates do not work. Also, it is about logic, as what does the 'ruler' imply in this problem. – jiten Apr 25 '18 at 12:16
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    There are far far too many "also"s here. You should ask just one question at a time. One possible problem, after a quick read: there may not be a way to specify the circles in advance so that they work for all values of the parameters. – Ethan Bolker Apr 25 '18 at 12:33
  • @EthanBolker Kindly elaborate your answer that states :"there may not be a way to specify the circles in advance so that they work for all values of the parameters". – jiten Apr 25 '18 at 12:37
  • Sorry - I won't take the time to elaborate. I just posted an idea that might work, for you to think about. If you narrow your question down to just one problem that you've tried to solve perhaps someone here will help you. – Ethan Bolker Apr 25 '18 at 12:46
  • @EthanBolker Done the same, by restricting to one question about maths, & one question about the 'meaning' of ruler usage. Rest are shown as 'extra' questions. – jiten Apr 25 '18 at 12:49

1 Answers1

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Guide:

Suppose your two coordinates are $(h_1, k_1)$ for $A$ and $(h_2, k_2)$ for $B$. Let the midpoint is $(h_3,k_3)$ and the gradient of the line connecting $(h_1,k_1)$ and $(h_2, k_2)$ be $-\frac1m$ where $m\ne 0$ (handle the case when $m=0$ separately).

Then the slope of the perpendicular bisector would be $m$.

Let the distance from $(h_1, k_1)$ to $(h_3, k_3)$ be $d$ and the radius of the circles be $r$, then by Pythagoras theorem, the distance from the intersection point to the midpoint would be $\sqrt{r^2-d^2}.$

Hence the coordinate of the intersection point would be

$$\left(h_3+\frac{\sqrt{r^2-d^2}}{\sqrt{1+m^2}}, k_3+\frac{m\sqrt{r^2-d^2}}{\sqrt{1+m^2}}\right)$$ and

$$\left(h_3-\frac{\sqrt{r^2-d^2}}{\sqrt{1+m^2}}, k_3-\frac{m\sqrt{r^2-d^2}}{\sqrt{1+m^2}}\right).$$

As for your second question, function $7$ is mentioned, are you implying that you are working with a fixed coordinate system, and you can read off any coordinate of any point any connect lines between them. In that case, writing out the equation would of course construct a line and given a line, we can write its equation as well. However, in the classical setting, I don't think we have the abiity to read off the coordinate system. We do not have a ruler, what we have is a straightedge.

Siong Thye Goh
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  • My view of why a quesiton per post: if a user only knows one of the answer... posting it would not be a good answer, hence a potential answerer might choose to completely ignore you, hence reducing the probability that you get any progress. – Siong Thye Goh Apr 27 '18 at 08:29
  • I hope that would get an answer to my earlier post at https://math.stackexchange.com/a/2747005/424260 soon. Thanks for your kind answer. – jiten Apr 27 '18 at 12:03
  • I am also stuck at https://math.stackexchange.com/q/2755914/424260, as have derived an equality $C$ based on difference $A-B$, & seemingly sign in book ($D$) should be opposite with $(y_2-y_1)(y_3-y_1)(y_3-y_2)$. It seems the sign change / error (in book) is 'in-consequential'? Also, there is a confusion: book states to use 'factor theorem' on beginning of pg 4, to derive in page 3. Usage of factor theorem is hopefully reference to reverse usage of product of 3 terms $(y_2-y_1)\cdots$ to derive alternate expression for equality of differences: $C = A-B$. – jiten Apr 27 '18 at 12:50