Lesson Explainer: Equation of a Straight Line: Vector Form Mathematics

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 𝑦-intercept, 𝑏, of the line, and its slope, π‘š, then we can write the equation of the line in the slope–intercept form: 𝑦=π‘šπ‘₯+𝑏.

However, this form for the equation of a straight line assumes that we know both the slope and 𝑦-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 𝑦-intercept, and, hence, the equation of the line in slope–intercept form. However, it is easier to represent this line in the point–slope form, which is defined as follows.

Recap: Point–Slope Form of the Equation of a Line in Two Dimensions

A line passing through the point 𝑃(π‘₯,𝑦) with slope π‘š has the equation π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯).

This is called the point–slope equation of the line.

This is known as the point–slope 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 point–slope 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 𝐴, 𝐡, and 𝐢 are integers where 𝐴 is nonnegative, then 𝐴π‘₯+𝐡𝑦=𝐢 is called the standard form for the equation of a line.

At most one of 𝐴 or 𝐡 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 𝐡=0, then we can construct the equation of any vertical line.

For example, if 𝐴=3, 𝐡=0, and 𝐢=6, we have the line 3π‘₯+0𝑦=6, that is, the vertical line π‘₯=2. Second, if both 𝐴 and 𝐡 are nonzero, we can easily find the π‘₯- and 𝑦-intercepts of the line in this form, by substituting 𝑦=0 and π‘₯=0 respectively.

We find the π‘₯-intercept by substituting 𝑦=0: 𝐴π‘₯+𝐡(0)=𝐢π‘₯=𝐢𝐴, and the 𝑦-intercept by substituting π‘₯=0: 𝐴(0)+𝐡𝑦=𝐢𝑦=𝐢𝐡.

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 𝐴, 𝐡, and 𝐢; 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 point–slope 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. Let’s 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, 𝑃(π‘₯,𝑦), on the line that has a position vector, βƒ‘π‘ŸοŠ¦, together with any nonzero vector, ⃑𝑑, that is parallel to the line. Vector ⃑𝑑 is called the direction vector of the line.

If the point 𝑃(π‘₯,𝑦) lies on the line which is parallel to the nonzero vector ⃑𝑑, then we can find the position vector of any point on the line by adding a scalar multiple of ⃑𝑑 to the position vector of 𝑃: βƒ‘π‘Ÿ=(π‘₯,𝑦). 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, βƒ‘π‘Ÿ, of any point on a line containing the point 𝑃(π‘₯,𝑦), with position vector βƒ‘π‘ŸοŠ¦, is given by βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+𝑑⃑𝑑, where ⃑𝑑 is the direction vector of the line and 𝑑 is any scalar value.

Since βƒ‘π‘Ÿ is dependent on the value of the scalar 𝑑 , we often write this equation as βƒ‘π‘Ÿ(𝑑)=βƒ‘π‘Ÿ+𝑑⃑𝑑.

For example, we can sketch the line βƒ‘π‘Ÿ=(1,1)+𝑑(1,2). When 𝑑=0, we have βƒ‘π‘Ÿ=(1,1). This is the position vector of a point on the line. Hence, this line passes through the point (1,1). Next, the direction vector of this line is (1,2). Remember, the first component tells us the horizontal displacement and the second component tells us the vertical displacement.

Therefore, the vector (1,2) 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 (1,1) by scalar multiples of the direction vector (1,2).

For example, moving from the point (1,1) with direction vector (1,2), a scalar multiple of 1, we reach the point (2,3).

If the scalar multiple was instead 2, then we move 2(1,2)=(2,4) from the point (1,1) reaching the point (3,5).

We can also use negative values for our scalar; if the scalar had been βˆ’1, then we would move βˆ’(1,2)=(βˆ’1,βˆ’2) from the point (1,1) reaching the point (0,βˆ’1).

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 π‘₯- and 𝑦-intercepts of the line.

How To: Finding the π‘₯- and 𝑦-Intercepts from the Vector Form of the Equation of a Line

To find the π‘₯- and 𝑦- intercepts, we set each component of the vector equation equal to zero and then solve for 𝑑. For example, in the line βƒ‘π‘Ÿ=(1,1)+𝑑(1,2), to find the 𝑦-intercept, we set the first component equal to zero: (0,𝑦)=(1,1)+𝑑(1,2),(0,𝑦)=(1,1)+(𝑑,2𝑑),(0,𝑦)=(1+𝑑,1+2𝑑).

Each component on the left and right side of this equation must be equal. This gives us two equations: 0=1+𝑑,𝑦=1+2𝑑.

For the first equation to be true, we must have 𝑑=βˆ’1. We can then substitute this value into the second equation: 𝑦=1+2(βˆ’1)=βˆ’1.

Hence, the 𝑦-intercept is at 𝑦=βˆ’1. Following the same process with the second component, we find that the line has an π‘₯-intercept at π‘₯=12.

As we have seen, finding the intercepts involves finding the value of the scalar 𝑑 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. Let’s look at an example.

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

The line π‘₯=βˆ’1 passes through the point (βˆ’1,0) 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 𝑑 is any scalar value, we can choose any nonzero constant for the second component. We will choose the direction vector (0,1) but it is important to remember that we can choose any nonzero scalar multiple of this vector.

This gives us the equation βƒ‘π‘Ÿ=(βˆ’1,0)+𝑑(0,1), 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 (6,βˆ’9) with direction vector (9,βˆ’2).

  1. βƒ‘π‘Ÿ=(9,βˆ’2)+𝐾(6,βˆ’9)
  2. βƒ‘π‘Ÿ=(6,βˆ’9)+𝐾(9,βˆ’2)
  3. 𝐾=(6,βˆ’9)+βƒ‘π‘Ÿ(9,βˆ’2)
  4. 𝐾=(9,βˆ’2)+βƒ‘π‘Ÿ(6,βˆ’9)

Answer

We recall that the vector equation of a line is βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+𝑑⃑𝑑, where βƒ‘π‘ŸοŠ¦ is the position vector of any point on the line and ⃑𝑑 is the direction vector of the line. We are told that the line passes through the point with position vector (6,βˆ’9) and that the line has direction vector (9,βˆ’2). Substituting these vectors into the vector equation of the line gives us βƒ‘π‘Ÿ=(6,βˆ’9)+𝑑(9,βˆ’2).

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

Hence, the vector equation of the straight line that passes through the point (6,βˆ’9) with direction vector (9,βˆ’2) is βƒ‘π‘Ÿ=(6,βˆ’9)+𝐾(9,βˆ’2).

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 βˆ’83 and passes through the point (4,βˆ’9).

  1. βƒ‘π‘Ÿ=(βˆ’9,4)+𝐾(3,βˆ’8)
  2. βƒ‘π‘Ÿ=(4,βˆ’9)+𝐾(8,βˆ’3)
  3. βƒ‘π‘Ÿ=(4,βˆ’9)+𝐾(3,βˆ’8)
  4. βƒ‘π‘Ÿ=(3,βˆ’8)+𝐾(4,βˆ’9)

Answer

We recall that the vector equation of a line is βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+𝐾⃑𝑑, where βƒ‘π‘ŸοŠ¦ is the position vector of any point on the line, ⃑𝑑 is the direction vector of the line, and 𝐾 is any scalar value.

We are told that the line passes through the point (4,βˆ’9) which has the position vector (4,βˆ’9), and, in the vector equation of the line, this is our point with position vector βƒ‘π‘ŸοŠ¦.

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

We can do this by recalling what is meant by the slope of a line. The slope of a line is the change in 𝑦 divided by the change in π‘₯. Therefore, if the slope of a line is βˆ’83, 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, (3,βˆ’8) and (βˆ’3,8), respectively. In fact, these are equivalent; they are both direction vectors of the line. Only the vector (3,βˆ’8) appears in the options, so we will choose this as our direction vector.

Hence, the vector equation of the line is βƒ‘π‘Ÿ=(4,βˆ’9)+𝐾(3,βˆ’8), 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 𝑦 over the change in π‘₯, we can always find the direction vector of a nonvertical line by taking the change in π‘₯ to be 1 and then the change in 𝑦 to be the slope. Therefore, the line will have the direction vector (1,π‘š), where π‘š is the slope of the line.

Definition: Direction Vector of a Line Given Its Slope

If a line has slope π‘š, then the line will have the direction vector (1,π‘š).

If π‘š=π‘π‘ž is a rational number, we can multiply the direction vector by the denominator, π‘ž, to get a direction vector in terms of integers. Hence, its direction vector is (π‘ž,𝑝).

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 (6,βˆ’7) and (βˆ’4,6).

  1. βƒ‘π‘Ÿ=(6,βˆ’7)+𝐾(10,βˆ’13)
  2. βƒ‘π‘Ÿ=(βˆ’4,6)+𝐾(βˆ’13,10)
  3. βƒ‘π‘Ÿ=(6,βˆ’4)+𝐾(βˆ’7,6)
  4. βƒ‘π‘Ÿ=(βˆ’4,6)+𝐾(10,13)

Answer

We recall that the vector equation of a line is βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+𝐾⃑𝑑, where βƒ‘π‘ŸοŠ¦ is the position vector of any point on the line, ⃑𝑑 is the direction vector of the line, and 𝐾 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 (6,βˆ’7) that has the position vector (6,βˆ’7). This will be our vector βƒ‘π‘ŸοŠ¦.

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 𝐴(π‘₯,𝑦) and 𝐡(π‘₯,𝑦) is given by 𝐴𝐡=(π‘₯,𝑦)βˆ’(π‘₯,𝑦)=(π‘₯βˆ’π‘₯,π‘¦βˆ’π‘¦).

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

The order of the points is not important, so, for example, choosing 𝐴(6,βˆ’7) and 𝐡(βˆ’4,6), we have ⃑𝑑=(βˆ’4,6)βˆ’(6,βˆ’7)=((βˆ’4)βˆ’6,6βˆ’(βˆ’7))=(βˆ’10,13).

However, this is not the direction vector of any of the given options, but we can find an equivalent vector to 𝐴𝐡 by multiplying by the scalar βˆ’1, which will be equal to 𝐡𝐴; this gives ⃑𝑑=βˆ’(βˆ’10,13)=(10,βˆ’13).

Therefore, substituting this direction vector and the position vector into the vector form of the equation of a line gives us βƒ‘π‘Ÿ=(6,βˆ’7)+𝐾(10,βˆ’13), 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 βƒ‘π‘ŸοŠ¦ and any direction vector of the line to be the direction vector ⃑𝑑.

So, in the above example, all of the following would be valid vector equations of the same line: βƒ‘π‘Ÿ=(6,βˆ’7)+𝐾(10,βˆ’13),βƒ‘π‘Ÿ=(6,βˆ’7)+𝐾(βˆ’10,13),βƒ‘π‘Ÿ=(βˆ’4,6)+𝐾(10,βˆ’13),βƒ‘π‘Ÿ=(βˆ’4,6)+𝐾(βˆ’10,13).

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 𝐴(π‘₯,𝑦) and 𝐡(π‘₯,𝑦) are distinct points on a line, then one vector form of the equation of the line through 𝐴 and 𝐡 is given by βƒ‘π‘Ÿ=(π‘₯,𝑦)+𝑑(π‘₯βˆ’π‘₯,π‘¦βˆ’π‘¦).

We can use this to help us determine if three or more points in the π‘₯𝑦-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 (βˆ’7,5), (βˆ’1,2), and (5,βˆ’1) 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. Let’s do this by finding the vector form of the equation of the line between (βˆ’7,5) and (βˆ’1,2).

We use the fact that a vector form of the equation of a line between two distinct points, 𝐴(π‘₯,𝑦) and 𝐡(π‘₯,𝑦), is given by βƒ‘π‘Ÿ=(π‘₯,𝑦)+𝑑(π‘₯βˆ’π‘₯,π‘¦βˆ’π‘¦), where 𝑑 is any scalar.

Substituting in the coordinates of the two points gives us βƒ‘π‘Ÿ=(βˆ’7,5)+𝑑((βˆ’1)βˆ’(βˆ’7),2βˆ’5)=(βˆ’7,5)+𝑑(6,βˆ’3), which can be written as (βˆ’7,5)+𝑑(6,βˆ’3)=(βˆ’7+6𝑑,5βˆ’3𝑑).

If the three points are colinear, our third point, (5,βˆ’1), will lie on this line. Let’s check whether this is the case or not. This point has the position vector (5,βˆ’1) and substituting its position vector into the vector equation of the line for the vector βƒ‘π‘Ÿ gives (5,βˆ’1)=(βˆ’7+6𝑑,5βˆ’3𝑑).

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

Equating the first component of each vector gives us 5=βˆ’7+6𝑑, which we can then solve for 𝑑:12=6𝑑𝑑=2.

Now, substituting this value of 𝑑 into the vector equation gives us (βˆ’7+6(2),5βˆ’3(2))=(βˆ’7+12,5βˆ’6)=(5,βˆ’1), which is the position vector of the point (5,βˆ’1), so it also lies on this line containing the points (βˆ’7,5) and (βˆ’1,2). 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: ⃑𝑑=(βˆ’7,5)βˆ’(βˆ’1,2)=(βˆ’6,3),⃑𝑑=(βˆ’7,5)βˆ’(5,βˆ’1)=(βˆ’12,6),⃑𝑑=(βˆ’1,2)βˆ’(5,βˆ’1)=(βˆ’6,3).

We then see that ⃑𝑑=2⃑𝑑=2⃑𝑑.

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 slope–intercept form.

Example 5: Converting an Equation from Slope–Intercept Form to Vector Form

The equation of a straight line is given by 𝑦=7π‘₯βˆ’3. Which of the following vector equations represents the same line?

  1. βƒ‘π‘Ÿ(𝑑)=(0,3)+𝐾(1,7)
  2. βƒ‘π‘Ÿ(𝑑)=(0,βˆ’3)+𝐾(2,7)
  3. βƒ‘π‘Ÿ(𝑑)=(3,18)+𝐾(7,3)
  4. βƒ‘π‘Ÿ(𝑑)=(3,7)+𝐾(2,14)
  5. βƒ‘π‘Ÿ(𝑑)=(βˆ’1,βˆ’10)+𝐾(3,21)

Answer

We recall that the vector equation of a line is given by βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+𝐾⃑𝑑, where βƒ‘π‘ŸοŠ¦ is the position vector of any point on the line, ⃑𝑑 is the direction vector of the line, and 𝐾 is any scalar. Therefore, the point with position vector βƒ‘π‘ŸοŠ¦ must lie on the line and the vector ⃑𝑑 must be parallel to the line.

The line given to us in the question is in slope–intercept form, that is, the form 𝑦=π‘šπ‘₯+𝑏, where π‘š is the slope and 𝑏 is the 𝑦-intercept. Hence, the line has slope 7 and 𝑦-intercept βˆ’3. 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 (1,7), 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 ⃑𝑑 must be a scalar multiple of (1,7). Of the options given, we can see that only options A, D, and E have direction vectors which are scalar multiples of (1,7): (1,7)=1(1,7),(2,14)=2(1,7),(3,21)=3(1,7).

Since the direction vectors in options B and C are not scalar multiples of (1,7), the lines in options B and C are not possible vector equations of the line 𝑦=7π‘₯βˆ’3.

In the vector form of an equation of a line, the vector βƒ‘π‘ŸοŠ¦ 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 βƒ‘π‘ŸοŠ¦ lie on the line.

Comparing with our general vector equation of a line, we see that in option A, βƒ‘π‘Ÿ=(0,3), so to check if (0,3) is on the line 𝑦=7π‘₯βˆ’3, we substitute π‘₯=0: 𝑦=7(0)βˆ’3,𝑦=βˆ’3.

Since in option A the corresponding 𝑦-coordinate when π‘₯=0 is 𝑦=3, and not βˆ’3, we can conclude that the point (0,3) is not on the line 𝑦=7π‘₯βˆ’3, 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 (3,7) lies on the line, we substitute π‘₯=3 into the line 𝑦=7π‘₯βˆ’3, which gives 𝑦=7(3)βˆ’3𝑦=18.

Therefore, the point (3,7) does not lie on the line 𝑦=7π‘₯βˆ’3 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 (βˆ’1,βˆ’10) lies on the line.

Substituting π‘₯=βˆ’1 into the slope–intercept equation of the line, we have 𝑦=7(βˆ’1)βˆ’3𝑦=βˆ’10.

Therefore, (βˆ’1,βˆ’10) lies on the line and this is a valid representation of this line.

Hence, of the options listed, only option E, βƒ‘π‘Ÿ(𝑑)=(βˆ’1,βˆ’10)+𝐾(3,21), is a representation of the straight line 𝑦=7π‘₯βˆ’3.

Let’s finish by recapping some of the key points of this explainer.

Key Points

  • The position vector, βƒ‘π‘Ÿ, of any point on a line containing the point 𝑃(π‘₯,𝑦) with position vector βƒ‘π‘ŸοŠ¦ is given by βƒ‘π‘Ÿ=βƒ‘π‘Ÿ+𝑑⃑𝑑, where ⃑𝑑 is the direction vector of the line and 𝑑 is any scalar value.
    This is called the vector equation of the line.
  • If we have two distinct points 𝐴(π‘₯,𝑦) and 𝐡(π‘₯,𝑦) on the line, we can find the direction vector ⃑𝑑 by calculating the vector 𝐴𝐡: 𝐴𝐡=(π‘₯βˆ’π‘₯,π‘¦βˆ’π‘¦).
  • A line of slope π‘š has the direction vector (1,π‘š).
  • The vector equation of a line is not unique; we can choose any point on the line for the position vector βƒ‘π‘ŸοŠ¦ and any nonzero vector parallel to the line for the direction vector ⃑𝑑.
  • Any two direction vectors βƒ‘π‘‘οŠ§ and βƒ‘π‘‘οŠ¨ 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.

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