Find the locus of a point P?

Question

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

Answer 1

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.

Answer 2

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 & \rightarrow & P\\ \left(t,1-t\right) & \mapsto & \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 $\left[0,1\right]$.

Answer 3

Hint: The question can be interpreted as such:

Plot the line $(x,\sqrt{1-x})$ from $x=1$ to $0$.