Line and distance equation

Question

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

Answer 1

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.

Answer 2

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.