What's the locus of the intersection of these variable lines?

What's the locus of the intersection of these variable lines?

Question

The line $x+y=t$ meets the line $y=tx$ at the point $P$. Find the coordinates of $P$ in terms of $t$. Hence, or otherwise, show that the equation of the locus of $P$ as $t$ varies is $x^2+xy-y=0$. At the point on the locus where $x=3$, find the value of $y$ and the value of $t$.

Solution

We’d like to find the intersection of the two lines, and we’d like expressions for the

$x$

-coordinate and

$y$

-coordinate in terms of

$t$

. We already have an expression for

$y$

in terms of

$t$

and

$x$

, so the least work for us is to replace the

$y$

in

$x+y=t$

with

$y=tx$

and proceed from there.

$$\begin{align*} &&x+tx&=t,&&\quad \\ \Longrightarrow\quad&& x(1+t)&=t\\ \Longrightarrow\quad&& x&=\frac{t}{1+t}. \end{align*}$$

Can you see why $t$ cannot be $-1$?

Since $y=tx$, we see that $y=\frac{t^2}{1+t}.$ Therefore, $P=\left(\frac{t}{1+t},\frac{t^2}{1+t}\right).$

Since we would like to end up with an expression in $x$ and $y$, let’s try to eliminate $t$ from $y=tx$ and $x+y=t$. This gives us $y=(x+y)x,$ which when we rearrange, is the equation $x^2+xy-y=0.$

Alternatively, we could substitute our coordinates for

$P$

into the given equation:

$$\begin{align*} x^2+xy-y&=\left(\frac{t\vphantom{t^2}}{1+t}\right)^2+\left(\frac{t\vphantom{t^2}}{1+t}\right)\left(\frac{t^2}{1+t}\right)-\left(\frac{t^2}{1+t}\right)\\ &=\frac{t^2}{(1+t)^2}+\frac{t^3}{(1+t)^2}-\frac{t^2(1+t)}{(1+t)^2}\\ &=\frac{t^2+t^3-(t^2+t^3)}{(1+t)^2}\\ &=0. \end{align*}$$

To find $y$, let’s use $x^2+xy-y=0$. When $x=3$, this becomes $9+3y-y=0,$ which means that $y=-\frac{9}{2}.$ Since $x+y=t$, we can use $x=3$ and $y=-\frac{9}{2}$ to find that $t=-\frac{3}{2}$.

It’s always worth checking your answer if you have more than one equation that your variables should satisfy. For example, we could check that our values of $x$, $y$ and $t$ satisfy $y=tx$. We could also have used this equation to determine $t$, and then used $x+y=t$ as a check.

The following GeoGebra applet helps us to see what is happening. The blue line is $y=tx$, the red line is $x+y=t$. What happens when $t=-1$?

Download GeoGebra file