C. Geometry and Algebra of Vectors · Lines and PlanesLines and Linear Systems in Two-Space
C3.1About 65 minutes
Learning goals- I can explain why the solution points of Ax + By + C = 0 form a line.
- I can read a normal vector and a direction vector from a scalar equation of a line.
- I can decide algebraically whether two lines meet in one point, are parallel and distinct, or are coincident.
- I can describe infinitely many solutions using a parameter.
Words to know
scalar equation of a line
An equation Ax + By + C = 0 (A and B not both 0) whose solution points form a line in two-space.
solution set
All the points that make an equation, or every equation of a system, true.
normal vector
A non-zero vector perpendicular to a line or plane. For Ax + By + C = 0, one normal is [A, B].
direction vector
A non-zero vector parallel to a line. For Ax + By + C = 0, one direction vector is [B, −A].
coincident lines
Two equations that describe the same line; one equation is a non-zero multiple of the other.
parameter
A variable, such as t, that can take any real value and is used to list infinitely many solutions.
Streets on a map
Picture two streets in your town drawn on a map. If they cross, they share exactly one intersection. If they run side by side, like two parallel avenues, they never meet. And if two street names turn out to label the same road, the "two" streets share every point.
That is the whole story of two lines in a plane. In this unit you will tell the same story in three dimensions, where lines and planes can meet in more surprising ways. We start where you already feel at home: two-space.
Remember from Grade 10: in Principles of Mathematics (MPM2D, Analytic Geometry) you solved linear systems by graphing, substitution and elimination, and you predicted whether a system had one, zero or infinitely many solutions. You also found the distance from a point to a line. This lesson revisits those ideas with a Grade 12 lens: we describe each case algebraically and connect it to vectors.
A linear equation in two variables is a line
An equation of the form
Ax + By + C = 0 (with A and B not both zero)
is called a scalar equation of a line. Its solution set is the set of all points (x, y) that make the equation true. That set is always a straight line:
- If B ≠ 0, solve for y: y = −(A/B)x − C/B. This is slope-intercept form, with slope −A/B. Every such equation graphs as a line.
- If B = 0, the equation becomes Ax + C = 0, so x = −C/A. This is a vertical line.
So "a linear equation in two variables" and "a line in two-space" are two names for the same thing. You will often see the same equation written as Ax + By = C instead; just move the constant across.
To check whether a point lies on a line, substitute it. The point (4, −1) is on 3x + 5y − 7 = 0 because 3(4) + 5(−1) − 7 = 0.
The normal vector hiding in the coefficients
Take two points P₁(x₁, y₁) and P₂(x₂, y₂) on the line Ax + By + C = 0. Subtract their equations:
A(x₂ − x₁) + B(y₂ − y₁) = 0.
The left side is the dot product of n = [A, B] with the vector P₁P₂ = [x₂ − x₁, y₂ − y₁]. A zero dot product means the vectors are perpendicular (you met this in the unit Dot and Cross Products). Since P₁P₂ points along the line, n = [A, B] is perpendicular to the whole line. We call it a normal vector.
A vector that points along the line is a direction vector. Swapping the coefficients and changing one sign gives one: m = [B, −A] (or [−B, A]). Check: [A, B] · [B, −A] = AB − BA = 0.
Two equations: one point, no point, or every point
A system of two linear equations asks for the points that lie on both lines.
| Case | Picture | What elimination gives | Number of solutions |
|---|
| Normals not parallel | lines cross | a single value for x and for y | one point |
| Normals parallel, equations not multiples | parallel, distinct lines | a false statement such as 0 = 5 | none |
| One equation is a multiple of the other | coincident lines | a true statement 0 = 0 | infinitely many |
Here is the algebraic test. Write the system as
A₁x + B₁y = C₁ and A₂x + B₂y = C₂.
- If A₁B₂ − A₂B₁ ≠ 0, the normals [A₁, B₁] and [A₂, B₂] are not parallel, so the lines cross at exactly one point.
- If A₁B₂ − A₂B₁ = 0, the normals are parallel: [A₂, B₂] = k[A₁, B₁] for some k. Then the lines are parallel. They are coincident if also C₂ = kC₁, and distinct (no solution) if C₂ ≠ kC₁.
This answers the curriculum's question: when does a system of two linear equations in two variables fail to determine a point? Exactly when the two equations have proportional left sides, that is, when their normal vectors are parallel.
Why the number A₁B₂ − A₂B₁ matters
It is the determinant of the coefficients. You will see it again in first-year university linear algebra, where it decides whether a system has a unique solution in any number of dimensions. In this unit, the three-dimensional version is the triple scalar product n₁ · (n₂ × n₃), which you met in Dot and Cross Products and will use in the lesson Three Planes: Geometric Configurations.
Worked exampleExample 1: Intercepts and a point check
For the line 4x − 3y − 24 = 0, find the intercepts, a normal vector and a direction vector. Is (3, −4) on the line?
- x-intercept: set y = 0. Then 4x = 24, so x = 6.
- y-intercept: set x = 0. Then −3y = 24, so y = −8.
- A normal vector is the coefficient vector n = [4, −3].
- A direction vector is m = [3, 4], since [4, −3] · [3, 4] = 12 − 12 = 0.
- Test (3, −4): 4(3) − 3(−4) − 24 = 12 + 12 − 24 = 0. It is on the line.
Intercepts (6, 0) and (0, −8); n = [4, −3]; m = [3, 4]; yes, (3, −4) is on the line.
Worked exampleExample 2: A system with one solution
Solve 2x + 3y = 12 and x − y = 1.
- Check first: A₁B₂ − A₂B₁ = 2(−1) − 1(3) = −5 ≠ 0, so there is exactly one solution.
- From the second equation, x = y + 1.
- Substitute: 2(y + 1) + 3y = 12, so 5y = 10 and y = 2.
- Then x = 2 + 1 = 3.
- Verify in both: 2(3) + 3(2) = 12 ✓ and 3 − 2 = 1 ✓.
The lines meet at the single point (3, 2).
Worked exampleExample 3: Coincident lines and a parameter
Classify the system 3x − 6y = 9 and −x + 2y = −3, and describe all its solutions.
- Multiply the second equation by −3: 3x − 6y = 9. That is exactly the first equation.
- So the lines are coincident: every point on 3x − 6y = 9 is a solution.
- To list them, let y = t, where t is any real number. Then 3x = 9 + 6t, so x = 3 + 2t.
- Check: 3(3 + 2t) − 6t = 9 + 6t − 6t = 9 ✓.
Infinitely many solutions: (x, y) = (3 + 2t, t), t ∈ ℝ. Writing the solutions with a parameter like this is the first step toward the parametric equations of the next lesson.
Worked exampleExample 4: Choosing a coefficient
For which value of k does the system kx + 2y = 5 and 3x + 6y = 1 have no solution? Can it ever have infinitely many?
- No unique solution happens when A₁B₂ − A₂B₁ = 0: k(6) − 3(2) = 0, so k = 1.
- With k = 1, multiply the first equation by 3: 3x + 6y = 15. The second is 3x + 6y = 1.
- Same left side, different right side: 15 ≠ 1. The lines are parallel and distinct.
- Infinitely many solutions would need the whole equation to be a multiple, but for k = 1 the constants do not match, and for k ≠ 1 there is one solution.
No solution when k = 1. The system never has infinitely many solutions; for every k ≠ 1 it has exactly one.
Just for funWhy are parallel lines so sad? They have so much in common, but they will never meet. (Coincident lines have the opposite problem: they never get any time apart.)
Hands-on activitySpaghetti systems
You need: a sheet of graph paper (or draw a 10 by 10 grid on plain paper with a ruler), a pencil, and two pieces of uncooked spaghetti (or two straight straws).
- Draw x- and y-axes on the grid from −1 to 8.
- Lay the two spaghetti strands so they cross at a grid point. Read two grid points on each strand and find the equation of each line in the form Ax + By = C. Solve the system by elimination and check that you get the crossing point.
- Now lay them parallel. Find both equations. What happens when you eliminate? (You should get a false statement such as 0 = 4.)
- Lay one strand exactly on top of the other. Write two equations that look different but describe this line (for example, multiply one by 2). What does elimination give now?
- For each case, compute A₁B₂ − A₂B₁.
What to look for: a non-zero value of A₁B₂ − A₂B₁ in step 2, and zero in steps 3 and 4. In step 3 the constants are not in the same ratio; in step 4 they are.
Your turn
1Is the point (4, −1) on the line 3x + 5y − 7 = 0?
Substitute: 3(4) + 5(−1) − 7 = 12 − 5 − 7 = 0. Yes, it is on the line.
2Find the x- and y-intercepts of 4x − 3y = 24.
y = 0 gives x = 6; x = 0 gives y = −8. Intercepts (6, 0) and (0, −8).
3Give a normal vector and a direction vector for 5x − 2y + 9 = 0.
Normal n = [5, −2]. Direction m = [2, 5] (check: 5(2) + (−2)(5) = 0). Any non-zero multiples also work.
4Solve x + 2y = 7 and 3x − y = 7.
From the second, y = 3x − 7. Substitute: x + 6x − 14 = 7, so x = 3 and y = 2. Point (3, 2). Check: 3 + 4 = 7 ✓, 9 − 2 = 7 ✓.
5Classify the system 4x − 6y = 10 and 6x − 9y = 15.
Multiply the first by 1.5: 6x − 9y = 15, the second equation exactly. The lines are coincident (infinitely many solutions).
6Classify the system 2x + y = 3 and 4x + 2y = 1.
Divide the second by 2: 2x + y = 0.5. Same left side, different constant (3 ≠ 0.5): parallel and distinct, no solution.
7Describe all solutions of the single equation x − 3y = 6 using a parameter t.
Let y = t. Then x = 6 + 3t. Solutions: (x, y) = (6 + 3t, t), t ∈ ℝ.
More practice: warm-up
1What is the slope of the line 2x + y = 5?
y = −2x + 5, so the slope is −2.
2Is (0, 0) on the line y = 3x?
3Write the vertical line x = 4 in the form Ax + By + C = 0.
x − 4 = 0 (A = 1, B = 0, C = −4).
4Write down a normal vector to 3x + 4y = 12.
[3, 4] (the coefficients of x and y).
5How many points do two distinct parallel lines share?
More practice: on your own
1Solve 5x + 2y = 1 and 3x − 4y = 24 by elimination.
Double the first: 10x + 4y = 2. Add to the second: 13x = 26, so x = 2. Then 2y = 1 − 10 = −9, y = −4.5. Point (2, −4.5). Check: 3(2) − 4(−4.5) = 6 + 18 = 24 ✓.
2Find k so that 6x + ky = 4 is parallel to 3x − 2y = 5. Are the lines distinct?
Normals must be parallel: [6, k] = 2[3, −2], so k = −4. Then 6x − 4y = 4, while doubling the other line gives 6x − 4y = 10. Since 4 ≠ 10 the lines are distinct parallel lines.
3Find k so that x + ky = 3 and 2x + 8y = 6 are coincident.
Halve the second: x + 4y = 3. The constants already match, so k = 4.
4The line 2x − 5y = 10 passes through (5, 0) and (0, −2). Show directly that n = [2, −5] is perpendicular to the segment joining these points.
The segment vector is [0 − 5, −2 − 0] = [−5, −2]. Dot product: 2(−5) + (−5)(−2) = −10 + 10 = 0, so they are perpendicular.
5Find the acute angle between the lines 2x + y = 4 and x − 3y = 1, using their normals.
Normals [2, 1] and [1, −3]. Dot product = 2 − 3 = −1. |n₁| = √5, |n₂| = √10. cos θ = −1/√50 gives θ ≈ 98.1°. The angle between the lines is the acute one: 180° − 98.1° ≈ 81.9°.
Challenge
1Show by elimination that the system a₁x + b₁y = c₁, a₂x + b₂y = c₂ gives (a₁b₂ − a₂b₁)x = c₁b₂ − c₂b₁. What does this say when a₁b₂ − a₂b₁ ≠ 0?
Multiply the first equation by b₂ and the second by b₁: a₁b₂x + b₁b₂y = c₁b₂ and a₂b₁x + b₁b₂y = c₂b₁. Subtract: (a₁b₂ − a₂b₁)x = c₁b₂ − c₂b₁. If a₁b₂ − a₂b₁ ≠ 0 we can divide, so x = (c₁b₂ − c₂b₁)/(a₁b₂ − a₂b₁) is a single value (and similarly for y): the system has exactly one solution.
2Find k so that the three lines x + y = 5, 2x − y = 1 and x − 2y = k all pass through one point.
Add the first two: 3x = 6, so x = 2 and y = 3. The third line must pass through (2, 3): k = 2 − 2(3) = −4.
3For which values of m does the system mx + y = 2, x + my = 2 have (a) exactly one solution, (b) no solution, (c) infinitely many solutions?
A₁B₂ − A₂B₁ = m·m − 1·1 = m² − 1. (a) m ≠ ±1: one solution (by symmetry x = y, and then (m + 1)x = 2 gives x = y = 2/(m + 1)). (b) m = −1: −x + y = 2 and x − y = 2, that is, −x + y = −2. Same left side, different constant: no solution. (c) m = 1: both equations are x + y = 2, coincident, infinitely many solutions.
Quiz
1. Which point lies on the line 2x − 3y + 6 = 0?
2. Which vector is a normal to the line 5x − 2y = 7?
3. The system 3x + 2y = 4 and 6x + 4y = 9 has:
4. Two linear equations in two variables fail to determine a single point exactly when:
5. What is the intersection of x + y = 6 and x − y = 2?
6. For which k are the lines kx + 3y = 1 and 2x + 6y = 5 parallel?
Big idea (C3.1): one linear equation in x and y is a line; two equations are two lines, and the system's solution is where they meet: one point, no point (parallel), or every point (coincident). The Grade 12 twist is reading the normal vector [A, B] straight from the equation and using it to classify the system.
How to help: Your teen has seen linear systems before (Grade 10, MPM2D). Ask them to predict the case before solving: "Are the normals parallel?" Then have them solve and check the answer in both equations.
Common difficulties:
- Stopping at 0 = 0 and writing "no solution". A true statement like 0 = 0 means infinitely many solutions; a false one like 0 = 5 means none.
- Mixing up normal and direction vectors. The normal uses the coefficients as they are: [A, B]. The direction vector swaps them and changes one sign.
- Reporting the obtuse angle between lines. The angle between two lines is the acute one (or 90°).
- Sign slips when moving C across: Ax + By + C = 0 is the same as Ax + By = −C.