1

Find the locus of a point such that the square of its distance to the point of intersection of two perpendicular lines is equal to the sum of its distances to those lines.

Assume $P(x,y)$ is any point of the locus, then:

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

Am I doing it right? (intersection point at $(0,0)$, points A and B at $(0,2)$ and $(2,0)$.)

mobzopi
  • 1,335

1 Answers1

1

Take the x-axis and the y-axis as the two perpendicular lines, so that the intersection is the origin. The square of the distance to the origin is $x^2+y^2$. The sum of the distances to the axes is $|x|+|y|$. So the locus is

$$x^2+y^2=|x|+|y|$$

Note that this function is symmetric for both $x$ and $y$. So, to draw it, it is sufficient to solve it in the first quadrant and then draw its symmetric curves in the other quadrants. In the first quadrant, it corresponds to a semicircle with center in $(0.5,0.5)$ and radius $\displaystyle \frac{\sqrt{2}}{2}$ that intersects the axes in the points $(0,1)$ and $(1,0)$. Tracing the symmetric curves in the other quadrants, we get that the whole locus has the shape of a four-leaved clover with semicircular leaves.

Anatoly
  • 17,079
  • how did you graph it? I understand how you got the equation but I can't graph it – mobzopi Sep 17 '14 at 05:24
  • Draw the semicircumference with center in $(0.5,0.5)$ in the first quadrant, from point $(0,1)$ to point $(1,0)$. Draw three other semicircles that are symmetric to it in the other quadrants. The centers are in points $(-0.5,0.5)$, $(0.5,-0.5)$, and $(-0.5,-0.5)$. – Anatoly Sep 17 '14 at 05:59