Find every point whose distance from each of the two coordinate axes equals its distance from the point $(4, 2)$

Question

Here is my attempt:

Let $(x, y)$ be any such point.

Then the distance of $(x, y)$ from the $x$-axis is $\lvert y \rvert$, whereas the distance of that point from the $y$-axis is $\lvert x \rvert$. Thus we have the equalities $$ \lvert x \rvert = \lvert y \rvert = \sqrt{ (x-4)^2+(y-2)^2}. $$ From $\lvert x \rvert = \lvert y \rvert$, we obtain $y = \pm x$.

Thus we have the equations $$ \sqrt{ (x-4)^2 + (\pm x -2)^2} = \lvert x \rvert, $$ which implies $$ (x-4)^2 + ( \pm x - 2)^2 = x^2, $$ which simplifies to $$ x^2 - 2(4 \pm 2)x + 20 = 0. $$

Thus we have the following two quadratic equations $$ x^2 - 12x + 20 = 0 \qquad \mbox{ and } \qquad x^2 -4x + 20 = 0, $$ and the solutions of these quadratic equations are $$ x = \frac{12 \pm 8 }{ 2 } \qquad \mbox{ and } \qquad x = \frac{ 4 \pm 8 \iota }{ 2 }, $$ that is, $$ x = 10, 2 \qquad \mbox{ and } \qquad x = 2 \pm 4 \iota. $$ We will of course only need the real values for our $x$.

Thus there are eight possible points satisfying the condition given in the problem, namely $$ (10, 10), (10, -10), (-10, 10), (-10, -10), (2, 2), (2, -2), (-2, 2), (-2, -2). $$

Is my solution correct in terms of the technique employed as well as the answers obtained? Or, are there any mistakes?

Answer 1

If you had actually checked each candidate point for equality of distance from the coordinate axes and $(4,2)$, you would have seen that only $(2,2)$ and $(10,10)$ satisfy the conditions of the original question.

Answer 2

All solutions must lie in the same quadrant as $(4,2)$. If any solution is in a different quadrant, it's distance from both axes is less than distance from $(4,2)$. Can you see why?

The vector joining any point in a different quadrant cuts the axes before reaching the point $(4,2)$. The length of perpendicular dropped on the closest axis from the solution point is less than the length of this vector (along hypotenuse).

Answer 3

Point lies on the bisector $(t,t)$ $$(t-4)^2+(t-2)^2=t^2$$

$$t^2-12 t+20=0\to t=2;\;t=10$$

Points are $(2,2);\;(10,10)$

The other possibility, $A$ on the other bisector $(t,-t)$, leads to the equation: $$(-t-2)^2+(t-4)^2 =t^2$$ which has no real solutions. $$$$ $$ $$

Answer 4

$$\dfrac{x-4}{\cos t}=\dfrac{y-2}{\sin t}=r\ge0$$

So, the point $P(4+r\cos t,2+r\sin t)$$

We need $$(4+r\cos t)^2=(2+r\sin t)^2=r^2$$

$$r(1+\cos t)=4$$ as $r(1-\cos t)\ge0$

Similarly, $r(\sin t+1)=2$

Using https://en.m.wikipedia.org/wiki/Weierstrass_substitution#The_substitution, $$\dfrac42=\dfrac2{(1+p)^2}$$

where $p=\tan\dfrac t2$

$\implies(1+p)^2=1\implies p=0,-2$

If $p=0,r=2$

If $p=-2,r=?$

Answer 5

(-4,2) (-4,-2) (4,-2)

The original point is 4 units from the y axis and two units from the x axis. i.e distance from each axis. The question is asking to find every point which is 4 units from the y axis AND 2 units from the x axis. There is one such point in each quadrant, which are the three additional points I listed. If you graph it, it should make sense. Those three points are the only ones which will satisfy the condition of distance of 4 from the y axis and distance of 2 from the x axis. It is basically a question about symmetry around each axis.

These represent the remaining three corners of an 8x4 rectangle, centered on the origin. One point in each quadrant. I believe most people are overthinking what the problem actually asks for.