Straight Lines

We have written two types of n-tuples of numbers: the n-tuple $(x_1,x_2,\dots ,x_n)$ is a point in n-dimensional space, while the n-tuple $A=[a_1,a_2,\dots ,a_n]$ is a free vector in n-space. We now drop the notation of the square brackets and write $A=(a_1,a_2,\dots ,a_n)$. The apparent confusion resulting between points and free vectors is eliminated as follows. Given the point $(x_1,x_2,\dots ,x_n)$, consider the free vector determined by the located vector with initial point $(0,0,\dots ,0)$ and end point $(x_1,x_2,\dots ,x_n)$. In our new notation, this vector is also $(x_1,x_2,\dots ,x_n)$. Thus one may consider the n-tuple $(x_1,x_2,\dots ,x_n)$ as representing a point in n-space, or as representing the vector with $(0,0,\dots ,0)$ as initial point and this point as end point. By this identification, we may use vector addition to add objects which are ostensibly points, and then identify the resulting vector with a point.

Now we shall define a line in n-space. This is an object of a different category, neither point nor vector nor number. It is of the same category as the whole of n-dimensional space, in that it is a set, or collection of points. Thus to determine a line, we must give a rule which will enable us to decide whether a given point is on that line, or as we say, is a member of that set of points.

It is a familiar property of plane geometry that a line is determined by giving a point on it and a direction (say another line, or a vector) parallel to it. It is of course also determined by giving two distinct points on it, but this property is less convenient for our purposes. Let $P=(p_1,p_2)$ be a point in 2-space, and let $A=(a_1,a_2)$ be a vector, $A\ne 0$. The question we must then answer is: Given a point $X=(x_1,x_2)$, when is it on the line through $P$ and parallel to $A$ ? It is clear that the line segment $\overline{PX}$, or, rather, the vector with initial point $P$ and end point $X$, must be parallel to $A$.

It is evident from the diagram that this vector, when added to the vector $P$ (using our identification of points and vectors), gives the vector $X$. Hence the vector with initial point $P$ and end point $X$ is the vector $X-P$ (meaning $X+(-P)$; see the exercises on vector addition). Thus we must have $X-P=tA$, where $t$ is some non-zero number. Of course, if we set $t=0$, then $X=P$ is on the line. Thus $X=P+tA$, where $t$ is some number. Moreover, if $t$ is any number, the point $X=P+tA$ has the property that either $X=P$ or $X-P$ is parallel to $A$, i.e., $X$ lies on the desired line.

In 2-space, this takes the form $(x_1,x_2)=(p_1+t{a}_1,p_2+t{a}_2)$. Since $A\ne 0$, either $a_1\ne 0$ or $a_2\ne 0$. Let us assume $a_1\ne 0$. Then $$x_1=p_1+t{a}_1,\text{ or }t=\frac{x_1-p_1}{a_1}$$ and $$x_2=p_2+t{a}_2=p_2+\left(\frac{x_1-p_1}{a_1}\right)a_2=\frac{a_2}{a_1}{x}_1+(p_2-p_1\cdot \frac{a_2}{a_1})$$

If we change our notation and write $x$ for $x_1,y$ for $x_2,m$ for $\frac{a_2}{a_1}$ and $b$ for $p_2-p_1\frac{a_2}{a_1}$, we obtain $$y=mx+b$$ which is the ordinary “slope-intercept” equation of a line. It means that the point ( $x,y$ ) is on the line if and only if $y=mx+b$, where $m$ and $b$ are the numbers defined above. The form $P+tA$ for the equation of a line is customarily called the parametric form. It is used in this course because of its complete generality for all dimensions and for all vectors $A\ne 0$.

Now let $P=(p_1,p_2,\dots ,p_n)$ and $Q=(q_1,q_2,\dots ,q_n)$ be two points. What is the line through $P$ and $Q$ ? We have a point, say $P$, on the line, and a vector parallel to the line, namely the vector determined by the initial point $P$ and end point $Q$. But we have seen that this vector is just $Q-P$. Thus the desired line is the set of points $$X=P+t(Q-P)$$ where $t$ ranges through all numbers, or the set of points $$X=(1-t)P+tQ$$ where $t$ ranges through all numbers.