GROUP 4 · MATH 1

The Straight Line — Lessons

Every formula from the presentation, worked step by step. Use this page to study before the quiz.

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Lesson 1

Slope of a line

Slope measures steepness: how far the line goes up (rise) for every step across (run).

m = (y₂ − y₁) / (x₂ − x₁)

Line through A (4,3) and B (8,5) with rise 2 and run 4 — A (4, 3) and B (8, 5): rise = 2, run = 4.

Example. Slope through (4, 3) and (8, 5): m = (5 − 3)/(8 − 4) = 2/4 = 1/2.

Example. Slope through (−2, 5) and (3, −5): m = (−5 − 5)/(3 − (−2)) = −10/5 = −2 (falling line).

The four faces of slope

Rising, falling, horizontal and vertical lines

Line	Slope	Why
Rising (↗)	m > 0	rise and run have the same sign
Falling (↘)	m < 0	rise and run have opposite signs
Horizontal	m = 0	rise = 0
Vertical	undefined	run = 0, and dividing by zero is impossible

Angle of inclination

θ is the angle from the +x axis to the line (0° ≤ θ < 180°), and m = tan θ. Acute θ → positive slope; obtuse θ → negative slope; θ = 90° → undefined.

Watch out. Keep the same order in the numerator and denominator: if y₂ comes first on top, x₂ must come first on the bottom.

Lesson 2

Parallel & perpendicular lines

Parallel: m₁ = m₂ (different intercepts)
Perpendicular: m₁ · m₂ = −1

Two parallel lines — y = ½x + 1 and y = ½x − 2: same slope, different intercepts.

Example (parallel). 2x + 3y = 6 → y = −(2/3)x + 2. And 4x + 6y = 10 → y = −(2/3)x + 5/3. Same slope, different intercepts → parallel.

To get a perpendicular slope: flip the fraction and change the sign. −2/3 → 3/2, and (−2/3)(3/2) = −1 ✓.

Example. Line through (1, 2) perpendicular to y = 2x + 3. The slope is −1/2, so y − 2 = −(1/2)(x − 1), i.e. y = −x/2 + 5/2, or x + 2y − 5 = 0. Check: 1 + 4 − 5 = 0 ✓.

Collinear points

Three points are on one line when the slope between the first two equals the slope between the last two.

Yes: (0, 1), (2, 2), (6, 4): slopes 1/2 and 2/4 = 1/2 ✓. No: A(−2, 8), B(0, 4), C(3, 2): slopes −2 and −2/3 — not equal.

Lesson 3

Distance from a point to a line

The shortest path from a point to a line is the perpendicular one.

d = |A x₀ + B y₀ + C| / √(A² + B²) for the line Ax + By + C = 0

Steps: (1) write the line in general form, (2) substitute the point, (3) take the absolute value, (4) divide by √(A² + B²).

Example. Line 5x − 12y + 3 = 0, so √(5² + 12²) = 13.

P (−2, 1): |−10 − 12 + 3| / 13 = 19/13 ≈ 1.46
Q (2, −1): |10 + 12 + 3| / 13 = 25/13 ≈ 1.92
R (5, 0): |25 − 0 + 3| / 13 = 28/13 ≈ 2.15

Another. Distance from (3, 4) to 3x + 4y − 5 = 0: |9 + 16 − 5| / 5 = 20/5 = 4.

Bonus — distance between parallel lines. For 3x + 4y + 5 = 0 and 3x + 4y − 10 = 0: |5 − (−10)| / 5 = 3.

Lesson 4

Forms of lines

One line with two points and both intercepts — The same line y = 2x − 1 can be written five ways.

Form	Equation	Use when you are given…
Two-point	y − y₁ = [(y₂ − y₁)/(x₂ − x₁)] (x − x₁)	two points
Point-slope	y − y₁ = m (x − x₁)	slope and one point
Slope-intercept	y = m x + b	slope and y-intercept
Intercept	x/a + y/b = 1	both intercepts
General	A x + B y + C = 0	the standard final answer

Examples.

Through (2, 3) and (5, 9): m = 6/3 = 2 → y − 3 = 2(x − 2) → y = 2x − 1
m = 2 through (3, 4): y − 4 = 2(x − 3) → y = 2x − 2
m = −2, b = 5: y = −2x + 5
2x + 5y = 20: divide by 20 → x/10 + y/4 = 1 (intercepts 10 and 4)
y = (2/3)x + 1: multiply by 3 → 3y = 2x + 3 → 2x − 3y + 3 = 0

Watch out. A line through the origin has an intercept of 0, so it cannot be written in intercept form (you would divide by zero).

Application

Area of a triangle from coordinates

No base, no height — only the three vertices. List them counterclockwise so the determinant comes out positive.

Triangle with vertices A (-4,-2), B (1,-1), C (3,4)

Area = ½ | x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂) |

The “weave” is the same computation drawn as a diagram: copy the first two columns to the right, add the ↘ diagonals, subtract the ↗ diagonals, then halve.

Example. A (−4, −2), B (1, −1), C (3, 4).

↘ (add): (−4)(−1) + (1)(4) + (3)(−2) = 4 + 4 − 6 = 2
↗ (subtract): (3)(−1) + (−4)(4) + (1)(−2) = −3 − 16 − 2 = −21
Area = ½ | 2 − (−21) | = 23/2 = 11.5 square units

Check with the formula: −4(−1 − 4) + 1(4 + 2) + 3(−2 + 1) = 20 + 6 − 3 = 23 → ½ · 23 ✓.

Collinearity test. If the area comes out 0, the points are collinear. (0, 1), (2, 2), (6, 4): 0(2 − 4) + 2(4 − 1) + 6(1 − 2) = 0 + 6 − 6 = 0 ✓.

Common slips. Forgetting the ½ at the end, or mixing up which diagonals are subtracted. Area is never negative — take the absolute value if the points were clockwise.

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