0
$\begingroup$

I've studied line geometry especially one that has to do with distance formula and question. But I just don't know how to approach this question

Find the equation for the set of all points

  • equidistant from the line y = 1 and the point (-1,-1)

  • equidistant from the line x = -2 and the point (2,3)

  • equidistant from the points (-2,3) and (4,1)

These three questions, I know, have the same similar approach but I just need a clue to proceed. Any help will be really appreciated

$\endgroup$
  • $\begingroup$ The question stated equidistant from a line to a point $\endgroup$ – Michael Umande Oct 15 '18 at 21:47
0
$\begingroup$

Hint:

The points equidistant from the line $y=0$ and the point $(0,1)$ fulfill the equation

$$y=\sqrt{x^2+(y-1)^2}$$ or

$$x^2-2y+1=0$$ which describes a parabola.

The points equidistant from the points $(1,0)$ and $(-1,0)$ fulfill the equation

$$\sqrt{(x+1)^2+y^2}=\sqrt{(x-1)^2+y^2}$$ or

$$x=0$$ which describes the $y$ axis.

Now, the specified constraints are metric and remain invariant under a similarity transform. So you can find the similarity transforms that turn the given points/lines to the reduced ones that I used.

$\endgroup$
  • $\begingroup$ Okay I grab the analogy used by you $\endgroup$ – Michael Umande Oct 15 '18 at 21:45
0
$\begingroup$

You have to think about what it means to be equidistant to a line and a point. Consider any point $(x, y)$ in the plane. The distance of the point $(x, y)$ to the point $p: (-1, -1)$ for example is: $\sqrt{(x- (-1))^2 + (y-(-1))^2}$. The distance from the point $(x, y)$ to the line $l: y = 1$ is $|{y - 1}|$ (The distance to the line is taken perpendicular to the line). Now the point $(x, y)$ is in the set of all points which are equidistant to $p$ and $l$ if and only if $\sqrt{(x- (-1))^2 + (y-(-1))^2} = |y - 1|$. From here on you should be able to solve it. Hope this helps.

$\endgroup$
  • $\begingroup$ Of course it does. I really appreciate- gracias $\endgroup$ – Michael Umande Oct 15 '18 at 21:49
  • $\begingroup$ Sorry I had to comment again. Now, looking at the rhs of the equation, I noticed that you took the absolute value of the line equation. $\endgroup$ – Michael Umande Oct 15 '18 at 22:11
  • $\begingroup$ Does it add significant to the overall equation? $\endgroup$ – Michael Umande Oct 15 '18 at 22:12
  • $\begingroup$ When you solve the equation for y, you square the whole thing anyway. But i added the absolute value because that's what a distance should be. It would not make sense to speak about a distance and then allow negative values in my formula. $\endgroup$ – araomis Oct 16 '18 at 4:39
  • $\begingroup$ If you think about it, we took the absolute value for the lefthand side too. Taking the square root always means that you get two solutions (negative and positive), but we're only considering the positive one. (Because we're thinking about distances in this exercise) $\endgroup$ – araomis Oct 16 '18 at 4:58

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.