In this explainer, we will learn how to find the equation of a straight line in vector form.

There are many different ways to represent a straight line in the plane. In fact, each of these different representations can be useful for different situations. For example, we recall that if we know the -intercept, , of the line, and its slope, , then we can write the equation of the line in the slopeβintercept form: .

However, this form for the equation of a straight line assumes that we know both the slope and -intercept of our line and it is difficult to sketch a line given in this form. It also assumes that the line is not vertical (since the slope of a vertical line is undefined). If we were instead given a point on the line together with the slope, we could use this information to find the -intercept, and, hence, the equation of the line in slopeβintercept form. However, it is easier to represent this line in the pointβslope form, which is defined as follows.

### Recap: PointβSlope Form of the Equation of a Line in Two Dimensions

A line passing through the point with slope has the equation

This is called the pointβslope equation of the line.

This is known as the pointβslope form for the equation of a line because we only need to know a single point on the line and its slope to find the equation of the line in this form.

There are still some problems with this form, however. First, since the pointβslope form requires the slope of the line to exist, it assumes that the line is not vertical. Second, we will not always be given the slope of the line, which means we would need to calculate the slope to find its equation. Lastly, it is not easy to sketch a line in this form, since we do not know either of its intercepts.

There is another form for the equation of a straight line, known as the standard form, which is easier to sketch than the previous two forms.

### Recap: Standard Form of the Equation of a Line in Two Dimensions

If , , and are integers where is nonnegative, then is called the standard form for the equation of a line.

At most one of or is allowed to be equal to zero.

The standard form for the equation of a line has a few advantages over the other forms noted above. First, if we take , then we can construct the equation of any vertical line.

For example, if , , and , we have the line , that is, the vertical line . Second, if both and are nonzero, we can easily find the - and -intercepts of the line in this form, by substituting and respectively.

We find the -intercept by substituting : and the -intercept by substituting :

Since we can easily find both intercepts, we can use this form to sketch the line by plotting both intercepts and sketching the line that passes through these two points.

This form of the equation of a line also has some drawbacks, however. First, it is not always possible to find integer values for , , and ; this means we cannot write every line in the standard form. Second, we cannot easily see the slope of the line in standard form. Finally, the standard form is difficult to find and usually involves manipulating one of the other forms of the equation of a line or being given both intercepts.

This leads us to the vector form for the equation of a line. As we have seen in the pointβslope form, we can think of a line as a point on the line and a slope representing the direction of the line. The problem with using the slope is that it assumes the line is not vertical. However, recalling that direction can also be represented using vectors, we can overcome this problem. Letβs first explore how to find the vector equation of any line and then consider how this applies in the case of a vertical line.

We can find the position vector of any point on a straight line by using a known point, , on the line that has a position vector, , together with any nonzero vector, , that is parallel to the line. Vector is called the direction vector of the line.

If the point lies on the line which is parallel to the nonzero vector , then we can find the position vector of any point on the line by adding a scalar multiple of to the position vector of : . This gives us the following equation for the line called the vector form of the line.

### Definition: Vector Form of the Equation of a Line in Two Dimensions

The position vector, , of any point on a line containing the point , with position vector , is given by where is the direction vector of the line and is any scalar value.

Since is dependent on the value of the scalar , we often write this equation as

For example, we can sketch the line . When , we have . This is the position vector of a point on the line. Hence, this line passes through the point . Next, the direction vector of this line is . Remember, the first component tells us the horizontal displacement and the second component tells us the vertical displacement.

Therefore, the vector represents one unit of movement to the right for two units of moving upwards. This gives us the following.

We move along the line from the point by scalar multiples of the direction vector .

For example, moving from the point with direction vector , a scalar multiple of 1, we reach the point .

If the scalar multiple was instead 2, then we move from the point reaching the point .

We can also use negative values for our scalar; if the scalar had been , then we would move from the point reaching the point .

In fact, any nonzero vector parallel to the line is an equivalent direction vector of the line since we are allowed to take any scalar multiples of the direction vector.

We can use the vector form of the equation of a line to find the - and -intercepts of the line.

### How To: Finding the π₯- and π¦-Intercepts from the Vector Form of the Equation of a Line

To find the - and - intercepts, we set each component of the vector equation equal to zero and then solve for . For example, in the line , to find the -intercept, we set the first component equal to zero:

Each component on the left and right side of this equation must be equal. This gives us two equations:

For the first equation to be true, we must have . We can then substitute this value into the second equation:

Hence, the -intercept is at . Following the same process with the second component, we find that the line has an -intercept at .

As we have seen, finding the intercepts involves finding the value of the scalar and then substituting this value into an equation. This seems to be more work than the other forms of a straight line. However, as we noted earlier, there is one big advantage to using the vector form of an equation of a line. That is, we can find the vector form of the equation of any line, including vertical lines. Letβs look at an example.

### How To: Writing the Equation of a Vertical Straight Line in Vector Form

The line passes through the point and its direction is vertical. This means it has no horizontal direction at all. Hence, the first component of the direction vector will be 0. Since is any scalar value, we can choose any nonzero constant for the second component. We will choose the direction vector but it is important to remember that we can choose any nonzero scalar multiple of this vector.

This gives us the equation which is illustrated in the following sketch.

In our next example, we will form the vector equation of a line given the position vector of a point on the line and its direction vector.

### Example 1: Finding the Vector Equation of a Straight Line That Passes through a Point with a Direction Vector

Write the vector equation of the straight line that passes through the point with direction vector .

### Answer

We recall that the vector equation of a line is where is the position vector of any point on the line and is the direction vector of the line. We are told that the line passes through the point with position vector and that the line has direction vector . Substituting these vectors into the vector equation of the line gives us

Finally, the value of in this equation represents any scalar multiple; to match the options, we will call this .

Hence, the vector equation of the straight line that passes through the point with direction vector is

In our next example, to find the vector form of the equation of a line, we will see how to calculate the direction vector of a line given its slope.

### Example 2: Finding the Vector Equation of a Line given the Slope of the Line and a Point on the Line

Find the vector equation of the straight line whose slope is and passes through the point .

### Answer

We recall that the vector equation of a line is where is the position vector of any point on the line, is the direction vector of the line, and is any scalar value.

We are told that the line passes through the point which has the position vector , and, in the vector equation of the line, this is our point with position vector .

Therefore, to find the vector form of the equation of this line, we only need to find its direction vector .

We can do this by recalling what is meant by the slope of a line. The slope of a line is the change in divided by the change in . Therefore, if the slope of a line is , this means that for every three units we move horizontally, we must move 8 units vertically. There are two equivalent ways of writing this as a vector.

We can think of this as moving 3 units right and 8 units down or as 3 units left and then 8 units up. These are the two vectors, and , respectively. In fact, these are equivalent; they are both direction vectors of the line. Only the vector appears in the options, so we will choose this as our direction vector.

Hence, the vector equation of the line is , which is option C.

This example demonstrates a useful property for finding the direction vector of a line given its slope. Since the slope of a line is the change in over the change in , we can always find the direction vector of a nonvertical line by taking the change in to be 1 and then the change in to be the slope. Therefore, the line will have the direction vector , where is the slope of the line.

### Definition: Direction Vector of a Line Given Its Slope

If a line has slope , then the line will have the direction vector .

If is a rational number, we can multiply the direction vector by the denominator, , to get a direction vector in terms of integers. Hence, its direction vector is .

In our next example, to determine the vector equation of a line, we will need to find the direction vector, given two distinct points on the line.

### Example 3: Finding the Vector Equation of a Line given Two Points on the Line

Find the vector equation of the straight line passing through the points and .

### Answer

We recall that the vector equation of a line is where is the position vector of any point on the line, is the direction vector of the line, and is any scalar. We are given two points, both of which lie on the line, and we can choose either of these to determine a vector equation for this line. We will choose the point that has the position vector . This will be our vector .

To find the vector form for the equation of this line, all we need now is the direction vector of the line. We can find this by recalling that the vector between two points and is given by

If and are distinct points of the line, then will be a direction vector of this line.

The order of the points is not important, so, for example, choosing and , we have

However, this is not the direction vector of any of the given options, but we can find an equivalent vector to by multiplying by the scalar , which will be equal to ; this gives

Therefore, substituting this direction vector and the position vector into the vector form of the equation of a line gives us , which is option A.

It is worth reiterating that the vector form of the equation of a line is not unique. We can take any point on our line to be the position vector and any direction vector of the line to be the direction vector .

So, in the above example, all of the following would be valid vector equations of the same line:

The above example gives us a useful result about finding the direction vector of a line when we are given two distinct points on the line.

### Definition: Vector Form of the Equation of a Line in Two Dimensions

If and are distinct points on a line, then one vector form of the equation of the line through and is given by

We can use this to help us determine if three or more points in the -plane are collinear, as we will see in our next example.

### Example 4: Identifying Whether Three Points Are Collinear Using the Vector Form of the Equation of a Straight Line

Using the vector form of the equation of a straight line, identify whether the points , , and are collinear.

### Answer

Recall that a list of points is said to be collinear if all of the points lie on the same straight line. There are a few ways of checking whether the three given points are collinear, one of which is to find the equation between one pair of points and then check if the third point satisfies the equation. Letβs do this by finding the vector form of the equation of the line between and .

We use the fact that a vector form of the equation of a line between two distinct points, and , is given by where is any scalar.

Substituting in the coordinates of the two points gives us which can be written as

If the three points are colinear, our third point, , will lie on this line. Letβs check whether this is the case or not. This point has the position vector and substituting its position vector into the vector equation of the line for the vector gives

For the point to lie on the line there must be a value of that satisfies the equation.

Equating the first component of each vector gives us which we can then solve for :

Now, substituting this value of into the vector equation gives us which is the position vector of the point , so it also lies on this line containing the points and . Hence, the points are collinear.

We could check if these points are colinear by finding the vectors between each pair of points. If three points are colinear, all vectors between any pair of points must be parallel.

We can calculate these vectors:

We then see that

Hence, all three vectors are parallel and we know all three points must be colinear.

In our final example, we will determine the vector equation of a line given in slopeβintercept form.

### Example 5: Converting an Equation from SlopeβIntercept Form to Vector Form

The equation of a straight line is given by . Which of the following vector equations represents the same line?

### Answer

We recall that the vector equation of a line is given by where is the position vector of any point on the line, is the direction vector of the line, and is any scalar. Therefore, the point with position vector must lie on the line and the vector must be parallel to the line.

The line given to us in the question is in slopeβintercept form, that is, the form , where is the slope and is the -intercept. Hence, the line has slope 7 and -intercept . The slope of the line being 7 means that for every 1 unit the line moves in the positive horizontal direction it moves 7 units in the positive vertical direction and we can represent this as the vector , which is one possible direction vector of the line.

Any nonzero scalar multiple of this vector will also be parallel to the line; this means must be a scalar multiple of . Of the options given, we can see that only options A, D, and E have direction vectors which are scalar multiples of :

Since the direction vectors in options B and C are not scalar multiples of , the lines in options B and C are not possible vector equations of the line .

In the vector form of an equation of a line, the vector must be the position vector of a point on the line. We can check the three remaining options, A, D, and E, to see if the points, in each case, representing the position vectors lie on the line.

Comparing with our general vector equation of a line, we see that in option A, , so to check if is on the line , we substitute :

Since in option A the corresponding -coordinate when is , and not , we can conclude that the point is not on the line , and option A does not represent this line. In fact, since the direction vectors of these lines are parallel and the lines are distinct, we have shown that this is a parallel line.

In option D, to check if lies on the line, we substitute into the line , which gives

Therefore, the point does not lie on the line and hence option D does not represent this line. Once again, since these two lines are distinct and have parallel direction vectors, these lines must be parallel.

Finally, considering option E, we want to see if the point lies on the line.

Substituting into the slopeβintercept equation of the line, we have

Therefore, lies on the line and this is a valid representation of this line.

Hence, of the options listed, only option E, , is a representation of the straight line .

Letβs finish by recapping some of the key points of this explainer.

### Key Points

- The position vector, , of any point on a line containing the point with position vector is given by where is the direction vector of the line and is any scalar value.

This is called the vector equation of the line. - If we have two distinct points and on the line, we can find the direction vector by calculating the vector :
- A line of slope has the direction vector .
- The vector equation of a line is not unique; we can choose any point on the line for the position vector and any nonzero vector parallel to the line for the direction vector .
- Any two direction vectors and are equivalent; they are nonzero scalar multiples of each other.
- Any three or more points are collinear if the direction vectors between each pair of points are equivalent.