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I have the following relation:

$M =${$ (x,y), x =$$ {1}\over{t+1}$, $y =$$ {5t + 8}\over{t + 1}$,$t\in\mathbb R$}

The task is to sketch the points of M into a coordinate system! But my opinion is that this would mean that i would have to paint the whole area?

I also transformed it to:

$$ y = 5 + 3x$$

This would be a graph! But i think that with the points of M another thing is asked! What would you say? Thanks

mathlove
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John Smith
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2 Answers2

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Yes, the teachers means the line.

When asked to draw the points of $M$, you need to color only the points $(x,y)$ in the plane that satisfy the existence of a $t$ such that $x =$$ {1}\over{t+1}$, $y =$$ {5t + 8}\over{t + 1}$. The important point that you might be missing is that for most random $(x,y)$ in the plane, there does not exist such a $t$!

You showed actually that if for any fixed $(x,y)$, the existence of such a $t$ is equivalent to the relation $y = 5 + 3x$ to hold. So drawing all points of $M$ is the same as drawing all points $(x,y)$ for which there exists such a $t$, which again is the same as drawing all $(x,y)$ that satisfy your nice linear relation.

So, indeed, you draw the line.

Does this help? Was this indeed the step you were missing?

Edit: There was an unnoted error. It's about the equivalence between the two statements: \begin{align*} &(i) \text{ for (x,y) in the plane the relation y = 5 + 3x holds}\\ &(ii) \text{ for (x,y) in the plane there is a $t\in \mathbb{R}$ such that $x = \frac{1}{t+1 }$ and $y= \frac{5t+8}{t+1}$} \end{align*}

The statements cannot be equivalent, since in (ii), the value for $x$ can never be zero, but in (i) the point $(0,5)$ satisfies! Take a really close look at how you prove that (ii) implies (i), probably you assumed $x\neq 0$ somewhere. The correct equivalence is:

\begin{align*} &(i) \text{ for (x,y) in the plane the two relations y = 5 + 3x and $x \neq 0$ hold}\\ &(ii) \text{ for (x,y) in the plane there is a $t\in \mathbb{R}$ such that $x = \frac{1}{t+1 }$ and $y= \frac{5t+8}{t+1}$} \end{align*}

Now if we draw the points of $M$, this is the line you already drew, except that you take out the point $(0,5)$. That point is not on $M$.

Joachim
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1

I'm not quite sure about what you want.

Your equation represents a line. And a line consists of points. Also, drawing a whole line is actually impossible for us because a line is infinitely long.

mathlove
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