The line \(x+y=t\) meets the line \(y=tx\) at the point \(P\). Find the coordinates of \(P\) in terms of \(t\).

Can you see why \(t\) cannot be \(-1\)?

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

Hence, or otherwise, show that the equation of the locus of \(P\) as \(t\) varies is \(x^2+xy-y=0\).

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: \[\begin{align*} x^2+xy-y&=\left(\frac{t\vphantom{t^2}}{1+t}\right)^2+\left(\frac{t\vphantom{t^2}}{1+t}\right)\left(\frac{t^2}{1+t}\right)-\left(\frac{t^2}{1+t}\right)\\ &=\frac{t^2}{(1+t)^2}+\frac{t^3}{(1+t)^2}-\frac{t^2(1+t)}{(1+t)^2}\\ &=\frac{t^2+t^3-(t^2+t^3)}{(1+t)^2}\\ &=0. \end{align*}\]At the point on the locus where \(x=3\), find the value of \(y\) and the value of \(t\).

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 \(t=-1\)?