Review linear functions and system solving methods. Grade 11 Math 20-1 Alberta curriculum.
Lesson 2.1 of Linear Functions & Systems of Equations in Math 20-1 — Alberta curriculum lessons.
Definition: Slope. The slope (m) of a line measures how steep it is. There are two ways to find it.
Method 1 — From a graph: count the rise over the run between two points.
m = (rise)/(run)
Method 2 — From two points (x_1, y_1) and (x_2, y_2):
m = (y_2 - y_1)/(x_2 - x_1)
There are six important slope situations to recognize:
Parallel Lines. Parallel lines have equal slopes.
If m_1 = (2)/(3), then m_2 = (2)/(3) as well.
Perpendicular Lines. Perpendicular lines have negative reciprocal slopes.
If m_1 = (2)/(3), then m_2 = -(3)/(2).
A useful test: m_1 × m_2 = -1 for any pair of perpendicular lines (when neither is vertical).
How to Find a Negative Reciprocal.
Example 1 — Slope of AB and a Perpendicular Line. Given A(3, -5) and B(0, 12), find the slope of line AB. If line CD is perpendicular to AB, find its slope.
Step 1: Apply the slope formula.
m_AB = (y_2 - y_1)/(x_2 - x_1) = (12 - (-5))/(0 - 3) = (17)/(-3) = -(17)/(3)
Step 2: Find the negative reciprocal.
Reciprocal of -(17)/(3) is -(3)/(17). Change the sign: (3)/(17).
Result: m_CD = (3)/(17)
Quick check: m_AB × m_CD = (-(17)/(3))((3)/(17)) = -1 ✓
Any linear equation can be expressed in three standard forms. Choose the form that fits the information you are given.
Example 2 — Find Slope and y-Intercept from General Form. Determine the slope and y-intercept of the line 3x + 9y + 18 = 0.
Step 1: Subtract 3x and 18 from both sides.
9y = -3x - 18
Step 2: Divide every term by 9.
y = -(1)/(3)x - 2
Slope: m = -(1)/(3)
y-intercept: b = -2
Example 3 — Equation of a Parallel Line. Determine the equation of the line through (-2, 1) that is parallel to 3x + y = 5. Write in general form.
Step 1: Rearrange the given line to slope-intercept form.
3x + y = 5 → y = -3x + 5, so slope = -3.
Step 2: The parallel line has the same slope: m = -3.
Step 3: Use slope-point form with (-2, 1).
y - 1 = -3(x - (-2))
y - 1 = -3(x + 2)
Step 4: Distribute and simplify.
y - 1 = -3x - 6
y = -3x - 5
Step 5: Move everything to one side (positive x coefficient).
Answer: 3x + y + 5 = 0
Quick check: 3(-2) + 1 + 5 = -6 + 1 + 5 = 0 ✓
Example 4 — Equation Through Two Points. Determine the equation of the line through (1, -4) and (3, 6) in general form.
Step 1: Find the slope.
m = (6 - (-4))/(3 - 1) = (10)/(2) = 5
Step 2: Use slope-point form with (1, -4).
y - (-4) = 5(x - 1)
y + 4 = 5x - 5
Step 3: Rearrange to general form.
y = 5x - 9
Answer: 5x - y - 9 = 0
Quick check: 5(3) - 6 - 9 = 15 - 6 - 9 = 0 ✓
Example 5 — Find a Missing Coordinate. Two points A(2, -k) and B(3, 4) have a slope of (6)/(5). Find the value of k.
Step 1: Substitute into the slope formula.
(6)/(5) = (4 - (-k))/(3 - 2) = (4 + k)/(1) = 4 + k
Step 2: Solve for k.
k = (6)/(5) - 4 = (6)/(5) - (20)/(5) = -(14)/(5)
Answer: k = -(14)/(5)
The System. Solve: x - y = 6 and 3x + 2y = 8
Method a) Graphically. Step 1: Rearrange each equation to slope-intercept form.
x - y = 6 → y = x - 6
3x + 2y = 8 → y = -(3)/(2)x + 4
Step 2: Graph both lines and find the intersection visually.
Calculator Steps for Finding the Intersection.
Method b) Elimination. Step 1: Write both equations.
x - y = 6 ·s (1)
3x + 2y = 8 ·s (2)
Step 2: Multiply equation (1) by 2 so the y terms cancel when added.
2x - 2y = 12 ·s (1')
Step 3: Add (1') and (2).
(2x - 2y) + (3x + 2y) = 12 + 8
5x = 20 → x = 4
Step 4: Substitute x = 4 back into equation (1).
4 - y = 6 → y = -2
Solution: (4, -2)
Method c) Substitution. Step 1: Solve equation (1) for y.
x - y = 6 → y = x - 6
Step 2: Substitute y = x - 6 into equation (2).
3x + 2(x - 6) = 8
Step 3: Distribute and solve for x.
3x + 2x - 12 = 8
5x = 20 → x = 4
Step 4: Plug x = 4 into y = x - 6.
y = 4 - 6 = -2
Solution: (4, -2)
Choosing the Best Method.
Create a free Studyio account to take the full lesson with voice-over teaching, practice questions and instant feedback.