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

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

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 \(t=-\frac{3}{2}\).

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 \(t=-1\)?