Equations of a line

The gradient measures the steepness and direction of a line — how much y changes for every 1 unit increase in x. Positive gradient → rises left to right. Negative gradient → falls left to right. Zero gradient → horizontal line.

Gradient = rise ÷ run = 8 ÷ 2 = 4. The line goes up by 4 for every 1 unit to the right. This is a positive, fairly steep gradient.

The line falls steeply — for every 1 unit moved right, y drops by 5. Steepness = |−5| = 5 (compare using absolute value). The negative sign means it slopes downward from left to right.

Wrong — steepness uses absolute value: |−4| = 4 > |3| = 3. The line with gradient −4 is steeper. Never compare signed gradient values to decide steepness — always compare |m₁| and |m₂|.

m = (y₂ − y₁) / (x₂ − x₁) The y-change (rise) goes on top. The x-change (run) goes on the bottom. Use the same pair order for both: subtract in the same direction.

m = (9 − 1) / (7 − 3) = 8 / 4 = 2. y increased and x increased → positive gradient makes sense. ✓

m = (−1 − 5) / (4 − (−2)) = −6 / 6 = −1. Key step: 4 − (−2) = 4 + 2 = 6. Subtracting a negative flips the sign.

They have swapped Δy and Δx. The gradient formula is m = Δy/Δx, not Δx/Δy. Fix: always write the formula first — m = (y₂ − y₁)/(x₂ − x₁) — before substituting numbers.

The y-intercept is the point where the line crosses the y-axis — the value of y when x = 0. In y = mx + c, the y-intercept is c, the constant term. Example: y = 4x − 7 has y-intercept = −7, so it crosses at (0, −7).

m is the gradient — it is the coefficient of x. c is the y-intercept — it is the constant term. Example: y = −2x + 9 → gradient = −2, y-intercept = 9.

Gradient m = −3. y-intercept c = 7. Coordinates of y-intercept: (0, 7).

The equation is not in y = mx + c order. Rewrite: y = −3x + 5. Gradient m = −3, y-intercept c = 5. Always rearrange into y = mx + c form before reading off m and c.

Horizontal line: gradient = 0 (no rise — Δy = 0). Vertical line: gradient is undefined — Δx = 0, so we would divide by zero.

Compare the absolute values of their gradients. The line with the larger |m| is steeper. Example: |−5| = 5 > |2| = 2, so y = −5x is steeper than y = 2x.

y-intercepts: A → c = 1, B → c = −5. Line A crosses higher. Steepness: |−3| = 3 vs |4| = 4. Line B is steeper. Two different comparisons — do them separately.

They dropped the negative sign. The gradient is m = −1/3 (negative, because it is − times 1/3). The y-intercept is 9. Read the coefficient of x including its sign.

y = mx + c m = gradient (slope), c = y-intercept. This form directly shows both key features of the line.

Substitute directly into y = mx + c: y = 5x − 3. The gradient goes with x; the y-intercept is the constant.

Gradient m = −1/2. y-intercept c = 6. The line starts high on the y-axis and falls gently — it goes down 1 for every 2 units to the right.

IB always requires a full equation, not just the values of m and c. Write: y = 3x + 7. The equation must start with "y =" and show both m and c in the correct form.

1. Write y = mx + c with the known gradient m. 2. Substitute the coordinates of the given point for x and y. 3. Solve for c. 4. Write the full equation with both m and c.

y = 3x + c. Substitute (2, 8): 8 = 3(2) + c → 8 = 6 + c → c = 2. Equation: y = 3x + 2.

y = −2x + c. Substitute (−1, 5): 5 = −2(−1) + c → 5 = 2 + c → c = 3. Equation: y = −2x + 3. Check: plug in x = −1: y = −2(−1) + 3 = 5 ✓

They left m as a letter instead of replacing it with the actual gradient value. If gradient = 2 and c = 4, the equation must be: y = 2x + 4. Always replace m with its value in the final answer.

Step 1: Calculate the gradient using m = (y₂ − y₁)/(x₂ − x₁). Step 2: Use one point and the gradient to find c (substitute into y = mx + c).

m = (10 − 4)/(3 − 1) = 6/2 = 3. y = 3x + c. Use (1, 4): 4 = 3(1) + c → c = 1. Equation: y = 3x + 1.

m = (5 − (−3))/(4 − 0) = 8/4 = 2. y-intercept: when x = 0, y = −3, so c = −3 directly. Equation: y = 2x − 3. Shortcut: if one point is the y-intercept (x = 0), c = that y-value immediately.

They must use one of the given points to substitute into y = 4x + c and solve for c. The equation y = 4x only works if the line passes through the origin — that must be verified, not assumed.

ax + by + d = 0 (sometimes written ax + by = c). All terms are moved to one side, leaving zero on the other. IB accepts both y = mx + c and general form unless the question specifies which.

Move all terms to the left: 3x − y − 5 = 0. Or equivalently: −3x + y + 5 = 0 (both are valid; IB usually wants positive leading coefficient).

Rearrange: y = 2x + 8. Gradient m = 2, y-intercept c = 8.

IB requires the specific form asked for. Leaving it as y = 2x − 4 does not match ax + by + d = 0. Correct: 2x − y − 4 = 0. Always re-read what form the question requires before writing the final answer.

Two lines are parallel if and only if they have the same gradient. They never intersect (unless they are the same line). Example: y = 3x + 2 and y = 3x − 7 are parallel — both have m = 3.

Any line parallel to L₁ also has gradient m. The gradient is the same — only the y-intercept (c) can differ.

Yes — both have gradient m = −2. They are different lines (different y-intercepts: 5 and −3), so they are parallel, not the same line.

No — gradients are +2 and −2. These are different gradients, so the lines are not parallel. They intersect at (0, 1). Similar equations do not mean parallel lines — the gradient values must match exactly.

Two lines are perpendicular if the product of their gradients equals −1: m₁ × m₂ = −1. This means the gradients are negative reciprocals of each other.

The perpendicular gradient is −1/m (flip the fraction and change the sign). Examples: m = 3 → m⊥ = −1/3 m = −2/5 → m⊥ = 5/2 m = 4 → m⊥ = −1/4

m⊥ = −1 / (−3/4) = 4/3. Rule: flip the fraction (4/3) and change the sign. Starting negative → perpendicular is positive. Check: (−3/4) × (4/3) = −12/12 = −1 ✓

They only changed the sign but did not take the reciprocal. The perpendicular gradient is −1/5 (flip to 1/5, then negate). "Negative reciprocal" means both steps: flip AND change sign.

1. Identify the gradient: m = 4 (same as the original line — parallel). 2. Substitute into y = 4x + c using (3, 7): 7 = 4(3) + c → c = −5. 3. Equation: y = 4x − 5.

m⊥ = −1/2. y = −(1/2)x + c. Use (4, 1): 1 = −(1/2)(4) + c → 1 = −2 + c → c = 3. Equation: y = −(1/2)x + 3.

m⊥ = 1/3 (negative reciprocal of −3). The line passes through (0, 5), so c = 5 directly (it is the y-intercept). Equation of L₂: y = (1/3)x + 5.

Their answer will be a parallel line, not a perpendicular one — a completely different type of answer. Always find m⊥ = −1/m first, before substituting the given point to find c.

The perpendicular bisector is a line that: 1. Passes through the midpoint of AB. 2. Is perpendicular to AB (i.e. meets AB at a right angle). Every point on the perpendicular bisector is equidistant from A and B.

1. The midpoint of AB — the perpendicular bisector passes through this point. 2. The perpendicular gradient — find the gradient of AB first, then take the negative reciprocal.

Midpoint M = ((2+6)/2, (4+8)/2) = (4, 6). Gradient of AB: m = (8−4)/(6−2) = 4/4 = 1. So m⊥ = −1. y = −x + c. Use (4, 6): 6 = −4 + c → c = 10. Perpendicular bisector: y = −x + 10.

The line will pass through the wrong point — it will be perpendicular to AB but not at the midpoint. The perpendicular bisector must pass through the midpoint, not through A or B. Always substitute the midpoint to find c.

A linear model describes a situation where the output increases or decreases at a constant rate as the input changes. It has the form y = mx + c. Use it when: the rate of change is constant (e.g. fixed cost per unit, steady temperature drop).

C = 2.5d + 4. Gradient m = 2.50 (cost per km). y-intercept c = 4 (fixed booking fee — the cost when d = 0).

Model: C = 0.15t + 10. When t = 40: C = 0.15(40) + 10 = 6 + 10 = $16.

They have swapped m and c. m (gradient) = the rate — the amount added per unit (per km, per hour, etc.). c (y-intercept) = the fixed starting value — the value when the variable equals 0.

The gradient is the rate of change — how much y changes for each 1-unit increase in x. Examples: • m = 3 km/h → speed of 3 km per hour. • m = −50 → value decreases by 50 per unit. Always state the units when interpreting.

The y-intercept is the initial value — the value of y when x = 0. Examples: • c = 200 → 200 items in stock at the start. • c = 15 → the temperature was 15°C at time 0. It is the starting point before any change occurs.

Gradient m = 8: the cost increases by $8 per hour. y-intercept c = 25: the initial cost (before any time passes) is $25 — a fixed/setup fee.

IB requires contextual interpretation — the gradient must be described in terms of the variables in the problem. For example: "The cost increases by $50 per kilogram." Just stating the number "50" earns no credit for an interpretation question.

1. The rate of change (→ this becomes m). 2. An initial value or a specific data point (→ this lets you find c). If two data points are given, find m first using the gradient formula, then find c.

V = −60t + 800. m = −60 (rate of decrease — negative because draining). c = 800 (starting volume at t = 0).

m = (300 − 180)/(7 − 3) = 120/4 = 30. C = 30d + c. Use (3, 180): 180 = 30(3) + c → c = 90. Model: C = 30d + 90 (daily rate $30, fixed fee $90).

A decrease means a negative gradient: m = −3. When a quantity is falling or decreasing, the gradient must be negative. Always check the direction of change before assigning the sign to m.

Substitute the given input value for x into the model equation and calculate y. Example: If C = 12t + 30 and t = 4, then C = 12(4) + 30 = 78.

Interpolation: predicting within the range of the original data — generally reliable. Extrapolation: predicting outside the range of the original data — less reliable; the model may not hold. IB questions often award 1 mark for commenting on reliability.

Set P = 0: 0 = −3t + 120 → t = 40 years. This is extrapolation (t = 40 is beyond the data range of 0–20) — the prediction is less reliable.

No — IB always requires a reason for reliability judgements. A correct answer gives: (a) whether it is interpolation or extrapolation, and (b) a reason (e.g. "within the data range" or "outside the data range — the trend may not continue").