# Given a right angled triangle and side lengths , finding locus of one of the point?

Q) a right angled triangle ABC having a right angle at C , $CA=b$ and $CB=a$ move such that the angular points A and B slide along x-axis and y-axis axis respectively. Find locus of C ?

Let C=(h,k) and A lies on x-axis and B lies on y-axis , i assumed an angle $\theta$ between x-axis and side $CA$ then i got coords of A, B as

$$A=(h+b\cos \theta ,0)$$ $$B=(0,a\sin \theta -k)$$

Then i wrote equations of equations of CA and CB . Then i found out their slopes and i used $m_{CA}.m_{CB}=-1$. Then i got a huge second degree differential equations , and i stopped there .

Should i resume my method or how else should i begin a approach ?

Let $C=(h,k)$, $A=(p,0)$, $B=(0,q)$.

Slope of AC=$\frac{k}{h-p}$

Slope of BC=$\frac{k-q}{h}$

Using distance formula, $$a=\sqrt{(h-p)^2+k^2}$$ $$h-p=\sqrt{a^2-k^2}$$ Similarly, $$k-q=\sqrt{b^2-h^2}$$

Substituting the above values on the slope equation,

Slope of AC=$\frac{k}{\sqrt{a^2-k^2}}$

Slope of BC=$\frac{\sqrt{b^2-h^2}}{h}$

Since AB is perpendicular to BC,$$\frac{k}{\sqrt{a^2-k^2}}\cdot \frac{\sqrt{b^2-h^2}}{h}=-1$$

Squaring both sides you get $$\frac{k^2}{a^2-k^2}\cdot \frac{b^2-h^2}{h^2}=1$$

$$(b^2-h^2)k^2=(a^2-k^2)h^2$$ $$bk \pm ah=0$$

Hence locus of Point C is $$by \pm ax=0$$

No matter where $A$ and $B$ are located on the $x$- and $y$- axis, the quadrilateral $ACBO$ is cyclic, due to $\widehat{AOB}=\widehat{ACB}=\frac{\pi}{2}$, hence $\widehat{AOC}=\widehat{ABC}$ is constant and $C$ lies on a line through the origin.

• I understood what your trying to explain but what if a case like B lies on +Y axis and A lies on +X axis and C lies in fourth quadrant ? – Sujith Sizon Oct 30 '15 at 15:08
• @SujithSizon: in such a case $ACBO$ is still a cyclic quadrilateral, hence $C$ still lies on a line through the origin, just another one. – Jack D'Aurizio Oct 30 '15 at 15:15
• I'm able to understand the cyclic quadrilateral part but could you please elaborate how you analysed that c lies on a straight line ? – Sujith Sizon Oct 30 '15 at 15:24
• @SujithSizon: if $A$ always lies on the $x$-axis and the angle $\widehat{AOC}$ is constant... – Jack D'Aurizio Oct 30 '15 at 16:08
• @SujithSizon: try drawing the graph. – Aditya Dev Oct 30 '15 at 16:21