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

In this explainer, we will learn how to find the equation of a straight line in parametric form using a point on the line and the vector direction of the line.

Recall that the vector form of a straight line passing through the point 𝐴 and parallel to the direction vector 𝑑 is 𝑟=𝑂𝐴+𝑡𝑑.

Recall that the position vector of a point is the vector starting from the origin and ending at the point. The vector form of the equation of a line describes each point on the line as its position vector 𝑟. Each value of the parameter 𝑡 gives the position vector of one point on the line.

In the parametric form of the equation of a straight line, each coordinate of a point on the line is given by a function of 𝑡, called the parametric equation. Let us consider how the parametric form of the equation of a line may be derived from a given vector form of the equation of a line.

Consider a line passing through the point 𝐴(𝑥,𝑦) and parallel to the direction vector 𝑑=(𝑎,𝑏). Then, the vector form of the equation of the line is given by 𝑟=(𝑥,𝑦)+𝑡(𝑎,𝑏).

We can reduce the right-hand side of the vector equation into a single vector: 𝑟=(𝑥,𝑦)+𝑡(𝑎,𝑏)=(𝑥,𝑦)+(𝑎𝑡,𝑏𝑡)=(𝑥+𝑎𝑡,𝑦+𝑏𝑡).

The 𝑥- and 𝑦-coordinates of points on the line are given by its position vector 𝑟, which is the point (𝑥+𝑎𝑡,𝑦+𝑏𝑡). In other words, the 𝑥-coordinate of the point is given by 𝑥+𝑎𝑡, while the 𝑦-coordinate is given by 𝑦+𝑏𝑡 for some 𝑡. Hence, the parametric form of the equation of the line is obtained as written below.

Definition: The Parametric Form of the Equation of a Straight Line

The parametric form of the equation of a line passing through the point 𝐴(𝑥,𝑦) and parallel to the direction vector 𝑑=(𝑎,𝑏) is 𝑥=𝑥+𝑎𝑡,𝑦=𝑦+𝑏𝑡.

Let us examine how the parametric form of the equation of a line can be obtained from the vector form in our first example.

Example 1: Finding the Parametric Equations of a Line given a Point on the Line and Its Direction Vector

Straight line 𝐿 passes through the point 𝑁(3,4) and has a direction vector 𝑢=(2,5). Then, the parametric equations of line 𝐿 are .

Answer

The parametric form of the equation of a line passing through the point (𝑥,𝑦) and parallel to the direction vector (𝑎,𝑏) is 𝑥=𝑥+𝑎𝑘,𝑦=𝑦+𝑏𝑘.

We are given that our line has a direction vector 𝑢=(2,5) and passes through the point 𝑁(3,4), so we have (𝑥,𝑦)=(3,4),(𝑎,𝑏)=(2,5).

Using these values, we can write the parametric form 𝑥=3+2𝑘,𝑦=45𝑘.

Let us consider another example to familiarize ourselves with the process of converting the vector form to the parametric form of the equation of a straight line.

Example 2: Identifying the Parametric Equation of a Straight Line Given in Vector Form

The vector equation of a straight line is given by 𝑟=(1,3)+𝑘(5,2). Which of the following pairs of parametric equations represents this straight line?

  1. 𝑥=2+5𝑘, 𝑦=3𝑘
  2. 𝑥=15𝑘, 𝑦=3+2𝑘
  3. 𝑥=2+3𝑘, 𝑦=5𝑘
  4. 𝑥=5𝑘, 𝑦=2+3𝑘
  5. 𝑥=3+2𝑘, 𝑦=15𝑘

Answer

Recall that the vector form of the equation of a line is 𝑟=(𝑥,𝑦)+𝑘(𝑎,𝑏), where (𝑥,𝑦) is the position vector of the point (𝑥,𝑦) on the line and (𝑎,𝑏) is a direction vector of the line. Comparing this to the given equation, we see that a direction vector for our line is (5,2). Also, by substituting 𝑘=0, we get the vector 𝑟=(1,3), which tells us that the point (1,3) lies on this line. Thus, our line passes through point (1,3) and is parallel to the direction vector (5,2).

We also recall that the parametric form of the equation of a line passing through the point 𝐴(𝑥,𝑦) and parallel to the direction vector 𝑑=(𝑎,𝑏) is 𝑥=𝑎𝑘+𝑥,𝑦=𝑏𝑘+𝑦.

Substituting the point (1,3) and direction vector (5,2) into the parametric form of the equation of the line, we get 𝑥=5𝑘1,𝑦=2𝑘+3.

So, the correct answer is B.

In our next example, we will apply the definition for the parametric form of the equation of the line to obtain the direction vector from the parametric form.

Example 3: Finding the Direction Vector of a Line given the Parametric Equations of the Line

The direction vector of the straight line whose parametric equations are 𝑋=2 and 𝑌=2𝑘+4 is .

Answer

The parametric form of the equation of a line passing through the point 𝐴(𝑥,𝑦) and parallel to the direction vector 𝑑=(𝑎,𝑏) is 𝑥=𝑎𝑘+𝑥,𝑦=𝑏𝑘+𝑦.

We are given the parametric equations: 𝑋=2 and 𝑌=2𝑘+4; so, comparing terms, we obtain the values 𝑎=0,𝑏=2,𝑥=2,𝑦=4.

Hence, our line passes through the point (2,4) and is parallel to the direction vector (0,2).

Hence, the direction vector is (0,2).

In the example above, we obtained the direction vector from the given parametric form of the equation of the line. This shows that we also can relate a given parametric form to the vector equation.

Consider the parametric equations 𝑥=𝑎𝑡+𝑥 and 𝑦=𝑏𝑡+𝑦. Applying our definition for the parametric form of the equation of a line, we know that this line passes through the point (𝑥,𝑦) and is parallel to the direction vector (𝑎,𝑏). Hence, the vector form of the equation of this line is 𝑟=(𝑥,𝑦)+𝑡(𝑎,𝑏).

Using this method, the vector form of the equation of the line from our previous example is 𝑟=(2,4)+𝑘(0,2).

As we have seen, instead of giving the direction vector of a line directly, a problem may provide this information indirectly. In fact, the direction vector of a line may be indirectly given by

  • providing two points lying on the line,
  • providing the angle between the line and the positive 𝑥-axis, or
  • providing the slope of the line.

The next example indirectly provides the direction vector of the line by means of an angle.

Example 4: Finding Parametric Equations of Straight Lines

Find the parametric equations of the straight line that makes an angle of 135 with the positive 𝑥-axis and passes through the point (1,15).

  1. 𝑥=1+𝐾, 𝑦=15𝐾
  2. 𝑥=1+𝐾, 𝑦=115𝐾
  3. 𝑥=15𝐾, 𝑦=1+𝐾
  4. 𝑥=1, 𝑦=15𝐾

Answer

In this example, we are given the angle 𝜃 between the line and the positive 𝑥-axis. In order to obtain the parametric equations of a straight line, we need to obtain the direction vector of the line.

Method 1

Recall that the slope of the line that makes angle 𝜃 with the positive 𝑥-axis is given by tan𝜃. Since 𝜃=135, the slope of the line is tan135=1.

Hence, the slope of the line is 1. We know that the slope of a line is also given by riserun, so 1=11=,=1,=1.riserunwhichleadstoriserun

This leads to the direction vector of the line (1,1). We also know that this line passes through the point (1,15). We recall that the parametric form of the equation of a line passing through the point 𝐴(𝑥,𝑦) and parallel to the direction vector 𝑑=(𝑎,𝑏) is 𝑥=𝑥+𝑎𝐾,𝑦=𝑦+𝑏𝐾.

This leads to the parametric equations 𝑥=1+𝐾,𝑦=15𝐾.

This corresponds to the option A.

Method 2

We are given that the line passes through the point (1,15) and its angle with the positive 𝑥-axis is 135. Based on this angle, we need to determine a direction vector of the line. We recall the trigonometric ratios on the unit circle: the 𝑥- and 𝑦-coordinates of the point on the unit circle corresponding with angle 𝜃 are 𝑥=𝜃,𝑦=𝜃.cossin

Consider the following picture of the unit circle describing a direction vector 𝑣 at 135.

The components of this direction vector are 𝑥=135=22,𝑦=135=22.cossin

Therefore, a direction vector is given by 22,22. Substituting the direction vector 22,22 and the point (1,15) into the parametric form, we obtain 𝑥=122𝐾,𝑦=15+22𝐾.

We can see, however, that none of the listed choices contain direction vectors with components ±22. So, we know that this vector expression will not lead to any of the listed options.

To find the parametric equations, we must, therefore, search for alternate direction vectors that are parallel to 22,22 but have integer components as in the given options. Factoring 22 or 22 from each component of this vector, we can write 22,22=22(1,1)=22(1,1).

Recalling also that two vectors are parallel if one vector is a constant multiple of the other vector, we see that the vectors (1,1) and (1,1) are both parallel to the original direction vector. So, both (1,1) and (1,1) are direction vectors of the line.

Let us first write the parametric form of the equation of our line using the direction vector (1,1) and the given point (1,15). Substituting these values into the parametric form, we can obtain 𝑥=1𝐾,𝑦=15+𝐾.

Since this is not among the listed options, we try the other vector (1,1) as the direction vector. Then, the parametric form is given by 𝑥=1+𝐾,𝑦=15𝐾.

This corresponds to the option A.

Our next example indirectly provides the direction vector of the line by means of the slope of the line.

Example 5: Identifying the Parametric Equation of a Straight Line Given in a Point and the Slope

A straight line passes through the point (1,6) and has a slope of 12. Which of the following pairs of parametric equations represents this straight line?

  1. 𝑥=1+𝑘, 𝑦=6+2𝑘
  2. 𝑥=2+𝑘, 𝑦=1+6𝑘
  3. 𝑥=1+4𝑘, 𝑦=6+2𝑘
  4. 𝑥=1+6𝑘, 𝑦=2+𝑘
  5. 𝑥=1+4𝑘, 𝑦=6+𝑘

Answer

Recall that the parametric form of the equation of a line passing through the point 𝐴(𝑥,𝑦) and parallel to the direction vector 𝑑=(𝑎,𝑏) is 𝑥=𝑎𝑘+𝑥,𝑦=𝑏𝑘+𝑦.

We are given that the line passes through the point (1,6) and has a slope of 12. We first need to determine a direction vector of the line from the slope. Once we have found a direction vector, we can use this, together with the given point, to form the vector equation of the line. From this, we can then establish the parametric equations for 𝑥 and 𝑦.

We know that the slope of a line is also given by riserun. Since we know that the slope of this line is 12, we have 12=,=1,=2.riserunwhichleadstoriserun

This leads to the direction vector of the line (2,1).

Using the direction vector (2,1) and the given point (1,6), we can write the parametric form: 𝑥=2𝑘+1,𝑦=𝑘+6.

We note that this parametric form does not match any of the given options. We must, therefore, choose an alternate direction vector that is parallel to the vector (2,1).

Looking through the choices, we can identify the direction vector used for each option by recalling that the 𝑥- and 𝑦-components of the direction vector can be obtained from the coefficients of 𝑡 in the respective coordinate equations. Hence,

  1. 𝑥=1+𝑘, 𝑦=6+2𝑘𝑑=(1,2),
  2. 𝑥=2+𝑘, 𝑦=1+6𝑘𝑑=(1,6),
  3. 𝑥=1+4𝑘, 𝑦=6+2𝑘𝑑=(4,2),
  4. 𝑥=1+6𝑘, 𝑦=2+𝑘𝑑=(6,1),
  5. 𝑥=1+4𝑘, 𝑦=6+𝑘𝑑=(4,1).

Of the listed vectors, only (4,2) is parallel to our direction vector (2,1) since (4,2)=2(2,1). So, let us use (4,2) as the direction vector along with the point (1,6) to write the parametric form of the equation of our line: 𝑥=1+4𝑘,𝑦=6+2𝑘.

Thus, the correct answer is C.

In our final example, we will find the parametric equations of a line passing through a given point and the midpoint of two other points.

Example 6: Identifying the Parametric Equation of a Straight Line

Find the parametric equation of the line that passes through the midpoint of 𝐴𝐵, where 𝐴=(2,1) and 𝐵=(4,3), and point (2,3).

  1. 𝑥=3+𝑘, 𝑦=14𝑘
  2. 𝑥=3𝑘, 𝑦=14𝑘
  3. 𝑥=3𝑘, 𝑦=14𝑘
  4. 𝑥=14𝑘, 𝑦=3𝑘
  5. 𝑥=3𝑘, 𝑦=1+4𝑘

Answer

Recall that the parametric form of the equation of a line passing through the point (𝑥,𝑦) and parallel to the direction vector 𝑑=(𝑎,𝑏) is 𝑥=𝑥+𝑎𝑘,𝑦=𝑦+𝑏𝑘.

We are given that our line passes through the point (2,3), so we can use 𝑥=2 and 𝑦=3. However, we can see that none of the listed options use these values, so we should look for a different point to use for the parametric equations.

We know that our line also passes through the midpoint of 𝐴𝐵, so let us find the coordinates of this point. We recall that the coordinates of the midpoint is the averages of the coordinates of the endpoints. In other words, given two points 𝐴(𝑎,𝑎) and 𝐵(𝑏,𝑏), the 𝑥- and 𝑦-coordinates of the midpoint of 𝐴𝐵 is given by 𝑥=𝑎+𝑏2,𝑦=𝑎+𝑏2.

Hence, the 𝑥-coordinate of the midpoint is 2+42=62=3, and the 𝑦-coordinate of the midpoint is 1+32=22=1.

This leads to the coordinates (3,1). We notice that options A, C, and E use the values 𝑥=3 and 𝑦=1 for the parametric equations.

Let us identify the direction vector. We know that the line passes through points (2,3) and (3,1); hence, the direction vector of the line must be parallel to the vector connecting these two points. Subtracting the position vectors of these points, we obtain (2,3)(3,1)=(23,31)=(1,4).

Hence, a possible direction vector of this line is (1,4). We know also that this vector may be replaced by any constant multiple of the vector. Let us see if this direction vector is used in a listed option. Using the point (3,1) and the direction vector (1,4), we obtain the parametric equations 𝑥=3𝑘,𝑦=14𝑘.

This is option C.

We complete this explainer by noting some of the key points concerning the parametric equations of a straight line.

Key Points

  • The parametric form of a straight line gives 𝑥- and 𝑦-coordinates of each point on the line as a function of the parameter.
  • The parametric form of the equation of a line passing through the point 𝐴(𝑥,𝑦) and parallel to the direction vector 𝑑=(𝑎,𝑏) is 𝑥=𝑎𝑡+𝑥,𝑦=𝑏𝑡+𝑦.
  • Any point on a line may be used to obtain the parametric equations of the line. Also, the direction vector may be replaced by any constant multiple of the vector.

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