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C. Geometry and Algebra of Vectors · Lines and Planes

Connecting Scalar, Vector, Parametric and Symmetric Equations of a Line

C4.1 C3.1About 70 minutes
Learning goals
  • I can convert between the scalar, vector, parametric and symmetric equations of a line in two-space.
  • I can use point-normal form n · (r − r₀) = 0.
  • I can decide whether lines are parallel or perpendicular from their direction or normal vectors.
  • I can find the intersection of two lines given in different forms.

Words to know

point-normal form
n · (r − r₀) = 0, or A(x − x₀) + B(y − y₀) = 0: the line through P₀ with normal n (two-space only).
symmetric equation
(x − x₀)/m₁ = (y − y₀)/m₂, a form of the line found by solving each parametric equation for t. It needs m₁ ≠ 0 and m₂ ≠ 0.
eliminate the parameter
Solve one parametric equation for t and substitute into the other to get an equation in x and y only.
perpendicular bisector
The line through the midpoint of a segment, perpendicular to the segment.

One line, many outfits

A line on a hiking map is the same line whether a park ranger describes it as "the trail that heads 2 km north for every 1 km east from the lodge" or a surveyor writes "2x − y − 3 = 0". Each description is useful for a different job:

  • The scalar form Ax + By + C = 0 is best for testing points and reading off a normal.
  • The vector and parametric forms are best for generating points and for motion.
  • The symmetric form is a compact way to write the parametric form.

In this lesson you will switch between them confidently, and use the switch to find intersections quickly.

ScalarAx + By + C = 0Vectorr = r₀ + tmParametricx = x₀ + tm₁, y = y₀ + tm₂Symmetric(x − x₀)/m₁ = (y − y₀)/m₂m = [B, −A], a pointn ⟂ mcomponentssolve for tcross-multiply
One line, four ways to write it. Read a normal n = [A, B] from the scalar form, swap to get a direction m, split the vector form into components, and solve each parametric equation for t to get the symmetric form.

From vector (or parametric) to scalar

Method 1: use a normal. If m = [m₁, m₂] is a direction vector, then n = [m₂, −m₁] is perpendicular to it (check: m₁m₂ − m₂m₁ = 0). So the scalar equation is m₂x − m₁y + C = 0. Find C by substituting the known point.

Method 2: eliminate t. Solve one parametric equation for t and substitute into the other.

Point-normal form. For a point P₀ with position vector r₀ and a normal n = [A, B], a point P with position vector r is on the line exactly when n · (r − r₀) = 0. In coordinates: A(x − x₀) + B(y − y₀) = 0. This is a third vector equation of a line, and it only works in two-space (you will see why in the lesson Normals and the Scalar Equation of a Plane).

P₀P(x, y)n = [A, B]r − r₀n · (r − r₀) = 0A(x − x₀) + B(y − y₀) = 0
Point-normal form: P(x, y) is on the line exactly when the vector from P₀ to P is perpendicular to the normal n, that is, when n · (r − r₀) = 0. Expanding gives the scalar equation.

From scalar to vector (or parametric)

Start with Ax + By + C = 0.

  1. Find a point. Set x = 0 or y = 0 (an intercept is usually easiest) and solve.
  2. Find a direction. Use m = [B, −A], which is perpendicular to the normal [A, B].
  3. Write it. r = r₀ + tm.

Slope connection. A line with direction [m₁, m₂] (m₁ ≠ 0) has slope m₂/m₁. A line y = kx + b has direction [1, k] and passes through (0, b).

The symmetric equation

If m₁ ≠ 0 and m₂ ≠ 0, solve each parametric equation for t and set them equal:

(x − x₀)/m₁ = (y − y₀)/m₂.

For example, the line through (3, −1) with direction [2, 5] has symmetric equation (x − 3)/2 = (y + 1)/5. If a component of m is zero, there is no symmetric form; write that coordinate as a constant instead (for example, direction [0, 3] through (4, 1) is just x = 4).

Parallel and perpendicular, in vector language

Two lines are parallel when their direction vectors (or their normals) are scalar multiples. They are perpendicular when their direction vectors have dot product 0, which is the same as their normals having dot product 0. You never need slopes, so vertical lines cause no trouble.

Worked example

Example 1: Vector to scalar

Find a scalar equation of the line r = [3, −1] + t[2, 5].

  1. Direction m = [2, 5], so a normal is n = [5, −2].
  2. The scalar equation has the form 5x − 2y + C = 0.
  3. Substitute (3, −1): 15 + 2 + C = 0, so C = −17.
  4. Check by eliminating t: t = (x − 3)/2, so y = −1 + 5(x − 3)/2. Multiply by 2: 2y = −2 + 5x − 15, which rearranges to 5x − 2y − 17 = 0 ✓.

5x − 2y − 17 = 0

Worked example

Example 2: Scalar to vector and parametric

Write vector and parametric equations for 3x + 4y − 12 = 0.

  1. A point: y = 0 gives x = 4, so (4, 0) is on the line.
  2. A direction: m = [B, −A] = [4, −3].
  3. Vector equation: r = [4, 0] + t[4, −3]. Parametric: x = 4 + 4t, y = −3t.
  4. Check: 3(4 + 4t) + 4(−3t) − 12 = 12 + 12t − 12t − 12 = 0 for every t ✓.

r = [4, 0] + t[4, −3]; x = 4 + 4t, y = −3t.

Worked example

Example 3: Intersection, mixing forms

Find where the line 2x − y − 4 = 0 meets the line x = 1 + 3t, y = 2 + t.

  1. Substitute the parametric expressions into the scalar equation: 2(1 + 3t) − (2 + t) − 4 = 0.
  2. Simplify: 2 + 6t − 2 − t − 4 = 0, so 5t = 4 and t = 0.8.
  3. Point: x = 1 + 3(0.8) = 3.4, y = 2 + 0.8 = 2.8.
  4. Check in the scalar equation: 2(3.4) − 2.8 − 4 = 6.8 − 6.8 = 0 ✓.

They meet at (3.4, 2.8). Substituting parametric into scalar gives one equation in one unknown: the fastest way to intersect two lines.

Worked example

Example 4: A perpendicular line

Find vector and scalar equations of the line through (−2, 5) perpendicular to 4x − y + 7 = 0.

  1. The given line has normal [4, −1]. A line perpendicular to it runs along that normal, so use direction m = [4, −1].
  2. Vector equation: r = [−2, 5] + t[4, −1].
  3. For the scalar form, a normal to the new line is [1, 4]: x + 4y + C = 0.
  4. Substitute (−2, 5): −2 + 20 + C = 0, so C = −18.

r = [−2, 5] + t[4, −1] and x + 4y − 18 = 0.

Just for fun

Scalar, vector, parametric, symmetric: a line has more outfits than most people's closets. Luckily it is the same line in every one, so it never has to decide what to wear.

Hands-on activity

Matching cards

You need: six index cards (or slips of paper) and a pencil.

Write one line on each card:

  • 2x − y − 3 = 0
  • r = [1, −1] + t[1, 2]
  • x = 3 + 2t, y = 1 − t
  • x + 2y − 5 = 0
  • x/1 = (y − 4)/(−3)
  • 3x + y − 4 = 0

Shuffle them, then sort them into three pairs that describe the same line. For each pair, show your evidence: a point from one form that satisfies the other, and a direction vector that is perpendicular to the normal.

Answer: 2x − y − 3 = 0 pairs with r = [1, −1] + t[1, 2] (the point (1, −1) works and [2, −1] · [1, 2] = 0). x + 2y − 5 = 0 pairs with x = 3 + 2t, y = 1 − t ((3, 1) works and [1, 2] · [2, −1] = 0). 3x + y − 4 = 0 pairs with x/1 = (y − 4)/(−3) ((0, 4) works and [3, 1] · [1, −3] = 0).

Your turn

1

Find a scalar equation of r = [1, 2] + t[3, −4].

2

Write a vector equation of 2x − 5y + 10 = 0.

3

Find the slope of the line x = 2 + 3t, y = 1 − 6t.

4

Write the symmetric equation of the line through (4, −3) with direction [2, 7].

5

Write y = 3x − 2 as a vector equation.

6

Are the lines 6x + 2y = 5 and r = [0, 1] + t[−1, 3] parallel, perpendicular or neither? If parallel, are they distinct?

7

Find the intersection of x − y = 1 with x = 2t, y = 3 + t.

More practice: warm-up

1

Give a normal vector of 7x − y + 2 = 0.

2

Give a direction vector of 7x − y + 2 = 0.

3

Give a normal vector of the line r = t[2, 3].

4

What is the slope of a line with direction vector [4, 2]?

5

Write a scalar equation of the line x = t, y = t.

More practice: on your own

1

Convert x = −1 + 4t, y = 3 − 2t to a scalar equation by eliminating t.

2

Find vector and scalar equations of the line through (2, −3) parallel to 5x + 2y = 1.

3

Use point-normal form to find the scalar equation of the line through P₀(1, 4) with normal [3, −2].

4

Find the intersection of r = [1, 1] + t[2, 1] and r = [4, −2] + s[1, 3].

5

Show that r = [3, 0] + t[3, 2] is a vector equation of 2x − 3y = 6, and use it to find the y-intercept.

Challenge

1

Explain why n · (r − r₀) = 0 describes a line in two-space but would not describe a line in three-space.

2

Find a scalar equation of the perpendicular bisector of A(−1, 2) and B(5, 6).

3

Use a vector equation to find the foot of the perpendicular from P(7, 1) to the line x + 2y − 4 = 0, and the distance from P to the line.

Check your understanding

Quiz

1.

Which is a scalar equation of the line r = [2, 1] + t[1, 3]?

2.

Which vector is a direction vector of 4x − 3y + 1 = 0?

3.

Which is the symmetric equation of the line through (1, −2) with direction [3, 5]?

4.

Which line is perpendicular to r = [0, 2] + t[1, −2]?

5.

Where does x + y = 10 meet the line x = 2t, y = 3t?

6.

In two-space, the equation n · (r − r₀) = 0 describes:

For parents and teachers

Big idea (C4.1): the same line can be written in scalar, vector, parametric or symmetric form. The bridge between them is that the normal [A, B] and the direction [B, −A] are perpendicular.

How to help: Give your teen one line and ask for all four forms. Then pick a point from one form and check it in another. If a point from the vector form fails the scalar form, there is an arithmetic slip.

Common difficulties:

  • Using the direction vector as the normal (or the other way round). Ask: "Is this vector along the line or across it?"
  • Sign errors in the symmetric form: the point (1, −2) gives (y + 2), not (y − 2).
  • Trying to write a symmetric form when a direction component is 0. Then that coordinate is simply constant.
  • Solving for an intersection with two parameters when one line is already in scalar form; substituting is quicker.