Webposting #6
New Expressions For Lines: Part #1
Vector Equation of a Line in Â2
We're used to the equation of a line in two dimensions (Â2), being of the form y = m x + b,
but that's not the only way we can express the structure of a line. Now that we understand
vectors, we can use them as the basis of our description.
What if I wish to represent any point on a line by a general combination of vectors? Well,
first we could jump to the line, by moving from the origin, our to an arbitrary point on our line,
p =< p1, p2 >. From here, every other point lays either forward or backwards along the same
direction. We can represent this by a vector v pointing along the line. Any point on the line
then can be reached by adding some multiple of v to our specified point p. That is:
r= p+t v ⇔
< x, y > = < p1 , p2 > + t < v1 , v2 >
One Choice of p and v
Alternate Choice of p and v
So for instance, if we have a line passing through two points, say A = (2,3) and B = (-1, 1),
we can construct a vector pointing along our line by subtracting these, getting:
v = < 2,3 > - < -1,1 > = < 3,2 >
Note that order doesn't matter, as either direction will still produce a vector pointing along
our line.
Now that we have our v, we can pick either A or B to be our p. Here lets choose A, so
p = < 2, 3 >
Putting this together, we get:
r = < x, y > = < 2, 3 > + t < 3, 2 >
Unfortunately, this is a highly non-unique result. That is to say there is an infinite number
of alternate solutions generated by working with any choice of points on the line, which
produce a different looking equation describing the same line
Just for example, I could have used B as my starting point, and the opposite direction
choice of v.
r = < − 1, 1 > + t < − 3, − 2 >
Alternately, notice the points < -4, -1 > and < 5, 5 > are also on this line ( t = 1, t = -2
respectively from the above equation). If instead I had been given these points with which to
construct my line, I could of taken the difference v = < 5, 5 > - < -4, -1 > = < 9, 6 >, and
written:
r = < 5, 5 > + t < 9, 6 >
All of these functions then have describe the same line. They all can “reach” the same
family of points by selecting all possible t values. The only difference is that the same point
will require different t-values in each representation.
Also notice, in each version of the equation, the v vectors are always scalar multiples of
each other. They always have the same direction (or exactly the reverse) so they can always
point along our line.
Converting From the Scalar Equation to a Vector Equation:
If we have a vector form of the equation of the line, converting back into the traditional
“scalar” form of y = mx + b, is quite straightforward.
First we pick any point on our line. Remember, the vector form of a line is not unique: any
initial point on the line produces an equivalent result. So pick something easy. Since our
equation is being expressed in the form:
y = mx + b
the point (0,b) is a convenient option. Although any convenient choice will do. Again for v,
we can choose any vector pointing the same direction as our line, length (magnitude) is not
important.
There are two common ways to choose v:
Method #1:
In the most common method, we select two points on our line, and take the difference.
That difference must inevitably be a multiple of v.
For example, if we have the equation y = 2x + 1, we can see that < 0, 1 >, and < 2, 5 > lie
on our line. Then I can construct v = < 2, 5 > - < 0,1 > = < 2, 4 >. Then our vector equation
becomes:
r = < 0, 1 > + t < 2, 4 >
Method #2:
We know from the slope, that m =y/x = rise/run. So we could write the v-vector as
< x,y >, or better yet, (rescaling our vector by x), we could write v =< 1, m >.
For example, if once again we use y = 2x + 1, and our point on the line, < 0, 1 >, we can
construct v = <1,m > = < 1, 2 >. Then our vector equation is:
r = < 0, 1 > + t < 1, 2 >
Notice, both v's are clearly scalar multiples of each other, so they generate the same
direction of line.
Converting From Vector to Scalar equations:
Since slope is rise over run, we can actually read the slope off of our direction vector.
As we travel along the line, we go up v2 for each change of v1, so our slope is m = v2/v1.
Combine that with our reference point/offset vector <p1,p2>, and we have the information to
use the point slope form of the line:
v 
y − p2 =  2  ( x − p1 )
 v1 
So, for example if we have the vector equation of a line:
r = < 3, 1 > + t < 1, − 2>
Then we can convert this to our standard scalar equation of the line.
Note that m = -2/1 = -2, so we can construct our equation:
y − 1 = ( − 2 ) ( x − 3) ⇒ y = − 2 x + 6 + 1 ⇒ y = − 2 x + 7
Normal form of the line in Â2:
An alternate, and interesting method of generating a scalar equation of a line in two
dimensional space occurs if we look at any normal to the line.
The normal direction to the line is the direction perpendicular to the line. A normal vector,
then, is any vector orthogonal (ie perpendicular) to the line, and thus orthogonal to our v.
Two Representatives of all Possible n's
for a given Line
Out of all the possibilities in two dimensions, probably the simplest way to construct a
normal is to flip the entries in v, and make one of them negative. That is, we can let:
n = < n1 , n2 > = < v2 , − v1 >
We can directly confirm that this n is, in fact, normal, Remember, perpendicular vectors have
zero dot product, and:
n • v = < n1 , n2 > • < v1 ,v2 > = < v2 , − v1 > • < v1 ,v2 > = v2 v1 − v1 ,v2 = 0
Once we have this normal (or any other for that matter), we can construct the scalar
equation of the line by noting that the vector between any two points on the line is a scalar
multiple of v, and thus orthogonal to n. Since the resulting dot product will be zero, we can
say, for any x = < x, y > on the line, that:
n • ( x − p) = 0 ⇒ < n1 , n2 > • ( < x, y > − < p1 , p2 > ) = 0 ⇒ n1 x + n2 y = < n1 , n2 > • < p1 , p2 >
⇒ n1 x + n2 y = c,
where c = < n1 , n2 > • < p1 , p2 >
For example if we look at our previously discussed vector equation of the line:
r = < 3, 1 > + t < 1, − 2>
We can see a possible n would be < 2, 1 >, and our equation, unsurprisingly, becomes:
2x + y = < 2, 1 > • < 3, 1 > = 6 + 1 ⇒ 2x + y = 7
Sample Questions:
1. What is the slope of the line r = < 2, 1 > + t < 3, -2 > ?
2. What is the slope of the line r = < 0, 3 > + t < -2, -2 > ?
3. Which of the following points are on the line r = < 0, 3 > + t < -2, -2 > ?
a) < 2. 5 >
b) <-1, 6 >
c) <-7,-4 >
d) < 0, 0 >
4. Which of the following points are on the line r = < 3, 1 > + t < 1, 6 > ?
a) < 2.5, -2 >
b) <6,19>
c) < 2,-5 >
d) < 5,13 >
5. Write scalar equations for the following lines:
a) r = < 0, 3 > + t < -2, -2 >
b) r = < 3, 4 > + t < 1, -3 >
c) The line through <1,1> in the direction given by the vector < 2,-1 >
6. Write vector equations for the lines:
a) y = 2x + 1
b) y = -x - 2
c) The line through <0,0> in the direction <-1,1>
d) the line through <-1, 1 > and <-1, 0 >
7. Which of the following vectors is a normal to the line r = < 0, 3 > + t < -2, -2 > ?
a) < 1.5, -1.5 >
b) <6,19>
c) < 2,-5 >
d) < -13,13 >
Solutions to Sample Questions:
1. m =-2/3
2. m = 1
3. c) and a)
4. all! (For a), t = -1/2, for b), t = 3, for c) t = -1, and for d), t = 2)
5. a) y = x + 3
b) y = -3x + 13
c) y = (-1/2)x + (3/2)
7. a) and d)
6. {Remember, vector equations are not unique
There are many other correct answers}
a) r = < 0, 1 > + t < 1, 2 >
b) r = < 0, -2 > + t < -1, 1 >
c) r = t < -1, 1 >
d) r = < -1, 1 > + t < 0, 1 >