Parametric equations for given line

Question

How would you find the parametric equations for:

1) a line through $(3,1)$ and $(-5,4)$.

2) a segment joining $(1,1)$ and $(2,3)$.

Can anyone show me the steps of doing it cause the way my textbook shows it, it has gotten me in a loop.

Answer 1

A line will be parameterized by $y\colon x\mapsto ax +b$. For instance, to solve 1), you want $$ \begin{align*} 1 &= a\cdot3 +b \\ 4 &= a\cdot(-5) +b \end{align*} $$ since $(3,1)$ and $(-5,4)$ must be of the form $(x,y(x))$ for the desired line. Solve this system to get the coefficients $a$ and $b$.

Answer 2

The mechanical way to find the parametric equation for any lines going through two points $A$ and $B$ is to write it in vector form, for some constant $\mu\in\mathbb{R}$:

$$\vec{l}=\overrightarrow{OA}+\mu\overrightarrow{AB}$$

For instance, in your case, for the first line, we have: $A(3,1)$ and $B(-5,4)$, therefore the equation of our line is:

$$\vec{l_{1}}=\begin{pmatrix}3 \\ 1\end{pmatrix}+\mu\begin{pmatrix}-5-3 \\ 4-1\end{pmatrix}=\begin{pmatrix}3 \\ 1\end{pmatrix}+\mu\begin{pmatrix}-8 \\ 3\end{pmatrix}$$

If you want the non-vector parametric equations you simply take the $x$ and $y$ components separately, i.e.:

$$\begin{align}x&=3-8\mu \\ y&= 1+3\mu\end{align}$$

Answer 3

The equation of the first line $$\frac{y-1}{x-3}=\frac{1-4}{3-(-5)}\iff \frac{y-1}{-3}=\frac{x-3}8=a\text{(say)}$$ where $a$ is any real number

$$\implies y=1-3a,x=3+8a$$

Answer 4

Let $p,q \in \mathbb{R}^2$ be points on the plane. The vector $v$ with starting point $p$ and ending point $q$ can be understood as a transport of points, in other words, the vector $v$ will be the subtraction of the points to give a kind of machine that when applied via the sum to the starting point gives the endinging point. Multiplying a vector by a number $k$ is equivalent to get a transport of point in the same direction if $k>0$ that is $k$ times the original transport. It'll reverse orientation if $k <0$.

Now, what's a line? A line can be understood as a curve that never changes it's direction. In other words, you take a point and transport it forever in the same direction forward and backward and you'll get a line in euclidean plane. Now one can describe a line by the following function $\alpha : \mathbb{R} \to \mathbb{R}$

$$\alpha(t) = p + tv$$

So this function says that you take any point on the line and transport it always on the same direction (that of $v$) all the way forever (this aims to allowing $t$ to run all the reals). Of course $p$ can be any point on the line and $v$ can be any transport from one point on the line to any other point on the line. A segment of $\alpha$ is obtained by restricting the domain to some closed subinterval of the reals (so that the transport gets bounded).

Now with this theory let's tackle the exercise. Since we want a line through $(3,1)$ and $(-5,4)$ we know that both points are on the line. So $(-5, 4) - (3,1) = (-8, 3)$ is the vector that transports one point of the line to the other one. So by the preceding discussion the line is parametrized by $\alpha : \mathbb{R} \to \mathbb{R}^2$ given by the following:

$$\alpha(t) = (-5,4)+t(-8,3)=(-5-8t, 4-3t)$$

Now the segment joining $(1,1)$ and $(2,3)$ can be obtained in analogous way. However the domain must be a closed subinterval. The natural choice is $[0,1]$, so that for $0$ we get $(1,1)$ and for $1$ we get $(2,3)$. Again the vector responsible for transporting the initial to the final point is $(2,3) - (1,1) = (1,2)$. So the parametrization is $\alpha : [0,1] \to \mathbb{R}^2$ given by:

$$\alpha(t) = (1,1) + t(1,2) = (1+t, 1+2t)$$

Answer 5

1) The point-slope form for the equation of a line: $y-y_0=m(x-x_0)$. So an equation for the line is $y=\frac{4-1}{-5-3}(x-3)+1$. A parametric equation here would be $$y=\frac{4-1}{-5-3}(t-3)+1$$$$x=t$$ 2) You could use the same technique and restrict $t$, or you could restrict $t$ using another method. First, an equation for the line is $y=2(x-1)+1$. We want $1\le x\le2$, so we need a bounded function that goes from $1$ at $-\infty$ to 2 at $\infty$. One function that is similar is the tanh (hyperbolic tangent: $\displaystyle\frac{e^x-e^{-x}}{e^{-x}+e^x}$. The tanh function goes from $-1$ at $-\infty$ to $1$ at $\infty$, so $\frac12\tanh x + \frac32$ goes from $1$ to $2$, so this is something we can use.

The unrestricted parametric plot is:$$y=2(t-1)+1$$$$x=t$$ After substituting $\frac12\tanh t + \frac32$ for $t$, we get what we want:

$$y = 2(\frac12\tanh t + \frac32-1)+1$$ $$x = \frac12\tanh t + \frac32$$