Suppose you graphed every single point of the form (2t + 3, 3-3t).

Question

Suppose you graphed every single point of the form $(2t + 3, 3-3t)$. For example, when $t=2$, we have $2t + 3 = 7$ and $3-3t = -3$, so $(7,-3)$ is on the graph. Explain why the graph is a line, and find an equation whose graph is this line.

Answer 1

$$x=2t+3, y=3-3t$$ $$(x-3) \frac{3}{2}=3t$$ $$y=3-3t=3-\frac {3}{2} x +\frac{9}{2}$$ $$y=-\frac {3}{2} x+ \frac{15}{2}$$ Tip: "Parametric equations of a curve express the coordinates of the points of the curve as functions of a variable, called a parameter. For example, $x= \cos t$ and $y= \sin t \ $are parametric equations for the unit circle, where t is the parameter."

Answer 2

The equations for the points $(x(t), y(t))$ $$ x = 2t+3 \\ y = 3−3t $$ can be solved for $t$ and then gives $$ t = (x-3)/2 = (3 - y)/3 \iff \\ (1/2) x + (1/3) y = 1 +3/2 = 5/2 \iff \\ 3 x + 2 y = 15 \quad (*) $$

Interpretation 1:

We solve equation $(*)$ for $y$ and get $$ y = -(3/2) x + (15/2) $$ which is of the form $y = m x + n$, which is a linear function.

Interpretation 2:

Equation $(*)$ can as well be interpreted as the equation of an affine hyperplane $$ n \cdot u = 15 \quad (**) $$ for a normal vector $n = (3,2)$ applied to the general vector $u=(x,y)$.

Affine means it must not necessarily include the origin, hyperplane means a space of dimension one less than the considered space. Here we have two dimensions, so a hyperplane has dimension one, in other words it is a line.

Equation $(**)$ can be used to read the (oriented) distance of the line from the origin as well. It is $d = 15 / \sqrt{3^2 + 2^2} = 15 / \sqrt{13}$.