Review question

# What's the locus of the intersection of these variable lines? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5255

## Solution

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

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, $P=\left(\frac{t}{1+t},\frac{t^2}{1+t}\right).$

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 $t=-1$?