Here's an animation showing the shrinking triangle:
As is required by the statement of the problem, the sides of the triangle are everywhere tangent to the blue curves.
The blue curves are logarithmic spirals. The curve on the right has endpoints $(0,0)$ and $(1,0)$, and is defined by the equation
$$
r \;=\; \exp\left(-\sqrt{3}\;\theta\right).
$$
Note that the triangle actually undergoes an infinite number of rotations as it shrinks towards the origin.
The rate at which the sides of the triangle shrink is equal to 3/2 of the speed at which the points move. (This follows from the fact that one endpoint of each edge has velocity tangent to the edge, while the other has a velocity component of $\sin(30^\circ)=1/2$ in the direction of the edge.) As a consequence, the length of each blue curve is $2/3$ of the side length of the large triangle.
Edit: Here is the Mathematica code for the animation above:
PolarToRectangular[{r_, theta_}] := {r*Cos[theta], r*Sin[theta]}
tmax = 2/Sqrt[3];
PolarCurve[t_] := {1 - t/tmax, -Log[1 - t/tmax]/Sqrt[3]}
f1[t_] := PolarToRectangular[PolarCurve[t]]
f2[t_] := PolarToRectangular[PolarCurve[t] + {0, 2 Pi/3}]
f3[t_] := PolarToRectangular[PolarCurve[t] + {0, 4 Pi/3}]
spirals = ParametricPlot[{f1[t], f2[t], f3[t]}, {t, 0, tmax},
Axes -> None, ImageSize -> 300, PlotStyle -> Darker[Blue],
PlotRange -> {{-0.7, 1.1}, {-0.9, 0.9}}];
triangle[t_] := Graphics[{Opacity[0], EdgeForm[Black],
Polygon[{f1[t], f2[t], f3[t]}]}]
points[t_] := Graphics[{PointSize[Large], Point/@{f1[t],f2[t],f3[t]} }]
dt = tmax/75;
myframes = Table[Show[spirals, triangle[t], points[t]], {t, 0, 75*dt, dt}];
Export["ShrinkingTriangle.gif", myframes,
"DisplayDurations" -> {1}~Join~ConstantArray[0.04, 74]~Join~{1}]
This code exports the animation as a GIF. If you want to see the animation from within Mathematica, the last command would be ListAnimate[myframes] instead.
A line in $\mathbb R^3$ (3d real space) can't be represented by a single equation. The reason for this is that a line is one-dimensional whereas space is 3-dimensional. The goal of writing a line in three-dimensional space is to eliminate two of these dimensions. To do this we need two equations: one to eliminate each extra dimension.
A little bit less abstractly, a line in $\mathbb R^3$ is just the intersection of two planes. This should be fairly intuitive, just imagine two infinite non-parallel planes and look at where they cross. So if we want to represent a line algebraically, all we need is a system of two plane equations. The equation of a plane is
$$ax+by+cz = r$$
where $a,b,c,r$ are constants. Now we can get our representation of a line by taking the solution set a system of two of these:
$$\begin{cases}a_1x+b_1y+c_1z=r_1\\a_2x+b_2y+c_2z=r_2\end{cases}$$
The solutions to this will form a line. As a concrete example, the intersection of the simple planes $x = 0$ and $y = 0$ will form the line that is the $z$ axis.
Best Answer
Both:
You can use slope to determine how steep a line is; if a line is steep, it's slope will be larger than a line that is less steep, and the steeper the line, the larger its slope.
So they are mutually correlated: steepness increases as slope increases (directly and positively proportional): (each gives information about the other...
But steep, as a description itself, is relative to some orientation. Usually we mean that steepness is a measure of the absolute value of the slope: the larger the magnitude of the slope, the closer a line with that slope is to the y-axis.
The sign of the slope tells us in what in what direction the line is tilted, if it is "tilted" whether y is increasing from "left to right" (positive), whether $y$ is decreasing from left to right, or neither(0 = horizontal, or slope is not defined = vertical.)