How can I graph $x + y + z = 1$ without using graphing devices?
I equal $z = 0$ to find the graph on the xy plane. So I got a line, $y = 1-x$ But when I equal 0 for either the $x$ or the $y,$ I get $z = 1-y$ or $z = 1-x$ , and those are two different lines from different angles. Different graphing websites were telling me different answers...
Please don't show some crazy and complicated methods to graph this. I just want simple steps just as plugging $x,y,z$ as zeros and etc.
I hope you don't mind if I post here a silence solution. I think it is telling you everything, however, other nice answers give you the theoretical points. :-)
Sketching 3D graphs can be super difficult to do by hand. When it comes to sketching simple planes, what I like to do is sketch the part of the plane that is only in a single octant (for example, in this case, we can consider only the first octant where $x,y,z \ge 0$). To do this, we find where the plane intersects each of the 3 axes.
- For the $x$-intercept, set $y=z=0$ to obtain $x+0+0=1$, which yields the point $(1,0,0)$.
- For the $y$-intercept, set $x=z=0$ to obtain $0+y+0=1$, which yields the point $(0,1,0)$.
- For the $z$-intercept, set $x=y=0$ to obtain $0+0+z=1$, which yields the point $(0,0,1)$.
Finally, draw the given octant that contains these 3 intercepts and label each of these 3 points. Connect these points to form a triangle that represents the portion of the plane inside this octant.
In 2D, a graph $$\frac xa +\frac yb=1$$ is a straight line connecting $(a,0)$ and $(0, b)$. This can be confirmed easily by substitution.
Similarly, in 3D, $$\frac xa +\frac yb+\frac zc=1$$ is a plane which goes through $(a,0,0), (0, b, 0), (0,0,c)$.
The question above refers to the case where $a=b=c=1$.
$x+y+z=1$ forms a plane. This plane contains all three lines you've found (and lots more). But what you've done should be enough - if you know a few points on a plane (in this case, for example, you know (1,0,0), (0,1,0) and (0,0,1), and more), you can sketch the whole plane.
