Find the locus of a point P? I don't understand how to go on with this question:

"As the point $R$ moves on the line $x+y=1$ from $(1,0)$ to $(0,1)$, the point $P$ moves such that it has the same $x$-coordinate as $R$, and its $y$-coordinate is equal to the square root of that of $R$. Describe the locus of $P$ and draw the loci of both $R$ and $P$ on the same set of axes."

Any help would be appreciated,
Thanks
First, just graph the line first on the Cartesian plane. And then,
determine the relationship between $R$ and $P$. Any point $R$ would
be of the form $\left(t,1-t\right)$, where $0\leq t\leq1$.

This means that $P$ will have coordinates $\left(t,\sqrt{1-t}\right)$.
So actually what you have is
\begin{eqnarray*}
f:R &amp; \rightarrow &amp; P\\
\left(t,1-t\right) &amp; \mapsto &amp; \left(t,\sqrt{1-t}\right).
\end{eqnarray*}
After some changes in the coordinates, what you have is $\left(1-s,s\right)\mapsto\left(1-s,\sqrt{s}\right)$,
where $0\leq s\leq1$. Just run through all the points and you'll
eventually see a graph, which is actually $y=\sqrt{1-x}$ on $\left[0,1\right]$.
First draw the line hence it will be between $y = 1$ and $x = 1$ and then use the $1/x$ method and it will increase hence the curve will go above the line and create a parabola shape.