1
$\begingroup$

Let $l_1$ and $l_2$ be two lines in the plane. The locus of all points $P$, such that the sum of squares of the distances of $P$ to $l_1$ and $l_2$ is constant, is a circle. Prove that $l_1$ and $l_2$ are perpendicular.

Now I can prove the converse of this statement really easily, but I'm stuck on proving this. I've let the centre of the circle be (0,0); I'm not sure if this helps. Also, how can we assume that the two lines will intersect at the centre of the circle?

$\endgroup$
1
$\begingroup$

Ignoring the case of parallel (or coincident) lines, suppose $\ell_1$ and $\ell_2$ meet at the unique point $O$. Any circle about $O$ meets the lines at the vertices of a rectangle $\square ABCD$; it also meets the bisectors of the angles formed by those lines at the vertices of a square $\square WXYZ$ (because the bisectors are necessarily perpendicular).

enter image description here

Each vertex of $\square ABCD$ is at distance $0$ from either $\ell_1$ or $\ell_2$, and is at some common distance (say $k$) from the other of $\ell_1$ or $\ell_2$. Thus, "the sum of the squares of the distances to the lines" is a constant (namely, $k^2$) across all four points, which implies that $\bigcirc O$ must be one of the "locus-circles" determined by the lines. That sum must then also be constant across the vertices of $\square WXYZ$, as those vertices lie on that locus-circle; in particular, the sums for $W$ and $X$ alone should match. However, because $W$ is on an angle bisector, the sum of the squares of the distances from $W$ to both lines is just double the square of the distance to either line; likewise for $X$. We conclude that the distances from each of $W$ and $X$ to each of $\ell_1$ and $\ell_2$ all match, making one of the lines a parallel to segment $\overline{WX}$ and the other line the perpendicular bisector of that segment. $\square$

$\endgroup$
0
$\begingroup$

Let $l_1$ be given by $a_1x+b_1y+c_1=0$ and $l_1$ be given by $a_2x+b_2y+c_2=0$.

Let $P$ be the point $(h,k)$ and $d_j$ be the distance of the point $P$ from line $l_j$. Then, $$d_1=\frac{|a_1h+b_1k+c_1|}{\sqrt{a_1^2+b_1^2}} \quad \text{ and } \quad d_2=\frac{|a_2h+b_2k+c_2|}{\sqrt{a_2^2+b_2^2}}$$

We are given that the locus of the points for which $d_1^2+d_2^2=s$ (where $s$ is some constant) is a circle. Let $\lambda_1=a_1^2+b_1^2$ and $\lambda_2=a_2^2+b_2^2$. Observe that $$d_1^2+d_2^2=\left(\frac{\lambda_2a_1^2+\lambda_1a_2^2}{\lambda_1 \lambda_2}\right)h^2+\left(\frac{\lambda_2b_1^2+\lambda_1b_2^2}{\lambda_1 \lambda_2}\right)k^2+\left(\frac{2a_1b_1}{\lambda_1}+\frac{2a_2b_2}{\lambda_2}\right)hk+\dotsb$$ For this to be a circle,

  1. coefficient of $hk$ must be $0$, and
  2. coefficients of $h$ and $k$ should be equal.

Thus we have \begin{align*} \lambda_2a_1^2+\lambda_1a_2^2 & = \lambda_2b_1^2+\lambda_1b_2^2\\ a_1b_1\lambda_2+a_2b_2\lambda_1 & = 0 \end{align*}

Upon solving this, we get $$(a_1a_2)^2 = (b_1b_2)^2 \implies a_1a_2 = \pm b_1b_2.$$

Try to see why $a_1a_2=b_1b_2$ will not occur. Then the only thing left is $a_1a_2+b_1b_2=0$ which is the condition for perpendicularity of $l_1$ and $l_2$.

$\endgroup$
  • $\begingroup$ Let $P = (h, k)$ and $L_1$ be $ax + by + c = 0$. It could be easier if we let $L_2$ be $y = 0$. Then, $d_2 = k$. Following the same logic, we end up with $b = 0$ and $L_1$ is in the form $ax + c = 0$ which is perpendicular to $L_2$. $\endgroup$ – Mick Nov 4 '15 at 16:25
  • $\begingroup$ @Mick you are absolutely right. I did think of that approach but was not sure if that would have been clear to OP because he said something about center being (0,0) and I didn't quite get as to what he wanted to try. So I ended up choosing this approach without making any more subtle choices. $\endgroup$ – Anurag A Nov 4 '15 at 18:32
  • $\begingroup$ The important thing is your logic is sound. $\endgroup$ – Mick Nov 5 '15 at 4:19

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.