3
$\begingroup$

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.

$\endgroup$

2 Answers 2

3
$\begingroup$

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

enter image description here

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

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.