Let a given line $L_1$ intersect the $x$- and $y$ axis at $P$ and $Q$ respectively. Let another line $L_2$ perpendicular to $L_1$, cut the $x$ and $y$ axis at $R$ and $S$ respectively. Show that the locus of the point of intersection of the lines $PS$ and $QR$ is a circle passing through the origin.

Any help would be appreciated, I am new to coordinate geometry.


2 Answers 2


Presumably, we leave $L_1$ fixed and seek the locus as $L_2$ varies.

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Let $T$ be the point where $PS$ meets $QR$.

We're given that $RS \perp PQ$. Also, $RP\perp QS$ (since the $\overleftrightarrow{RP}$ is the $x$-axis, and $\overleftrightarrow{QS}$ is the $y$-axis). This says that $R$ lies on two altitudes of $\triangle PQS$; since $R$ must then also lie on the third altitude, we have $RQ\perp PS$: therefore, $\angle PTQ$ is a right angle.

The locus of points $T$ that make a right angle with $P$ and $Q$ is a circle with diameter $PQ$. (This is an aspect of Thales' Theorem.) Since $O$ is such a point, it lies on that circle.


Let us write the equation of line $L_1$ as $y= x \tan {\theta} +j$. Using this notation, $\theta$ represents the angle identified at the crossing of the line with the $x$-axis.

Similarly, considering that $L_2$ is perpendicular to $L1$ and then its slope is the negative inverse of that of $L_1$, we can write $L_2$ as $y= -x \cot {\theta} +k$.

The coordinates of $P,Q,R,S$ can be easily obtained setting $x=0$ and $y=0$ in the equations above. So we get that they are $(-j \cot{\theta}, 0)$, $(0,j)$, $(k \tan{\theta},0)$, and $(0,k)$, respectively.

The equation of the line $PS$ is then

$$y=\frac{k}{j \cot{\theta}}x+k$$

and that of the line $QR$ is

$$y=-\frac{j}{k \tan{\theta}}x+j$$

The locus of the intersection point between $PS$ and $QR$ is obtained by considering the set of all possible lines $L_2$ perpendicular to $L_1$ (that is to say, moving $L_2$ by changing its intercept $k$ and keeping its slope constant). Therefore, it can be determined by solving the two equations for $k$ and then equalizing them, so that $k$ is canceled out. This leads to

$$k=\frac{jy \cot{\theta}}{(j\cot{\theta} + x)}$$

$$k=\frac {jx}{ \tan{\theta} (j-y)} $$

and then

$$ \frac{jy \cot{\theta}}{(j\cot{\theta} + x)}= \frac {jx}{ \tan{\theta}(j-y)}$$

$$ y \cot{\theta} (j \tan{\theta}-y \tan{\theta})=x (j \cot{\theta} + x)$$

which reduces to

$$ -y^2 +jy=xj\cot{\theta}+ x^2$$

Setting $m=j \cot{\theta}$ we finally get

$$ x^2+y^2 +mx-jy=0$$

which is the equation of a circle passing through the origin.


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