General-form lines and contextual models
Введено| English | Значение |
|---|---|
| general form/ˈdʒenərəl fɔːm/ | общий вид |
Which part of a delivery charge changes with distance?
- A straight-line cost model has a fixed charge and a rate per kilometre. Its intercept and gradient have different meanings and units.
- This lesson studies general form 一般式: The form ax+by+c=0 for a straight line, with a and b not both zero.
The method
- For a nonvertical line through two points, m=(y₂−y₁)/(x₂−x₁) and y−y₁=m(x−x₁).
- Rearrange into ax+by+c=0.
- Parallel nonvertical lines have equal gradients; perpendicular finite gradients multiply to −1.
- Treat vertical and horizontal lines separately.
Why can a vertical line not be written as y=mx+c?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Gradient = Δy/Δx and a vertical line has Δx=0. It takes the form x=4 in the general family ax+by+c=0 with b=0 - a separate case from finite-gradient lines.
A worked case
Through A(−1,3) and B(5,−1), m=−4/6=−2/3. Thus y−3=−2(x+1)/3, giving 2x+3y−7=0. The perpendicular through A has gradient 3/2 and equation 3x−2y+9=0. A vertical line x=4 is perpendicular to a horizontal line y=3 without having a finite gradient. Delivery quotes C=8 at d=0 km and C=38 at d=12 km give rate (38−8)/12=2.5 currency units per kilometre and C=8+2.5d. At d=8, the predicted cost is 28.
General-form lines and contextual models
For a nonvertical line through two points, m=(y₂−y₁)/(x₂−x₁) and y−y₁=m(x−x₁)
Match each coordinate calculation to its geometric or contextual condition.
Find the gradient of the line through (−1,3) and (5,−1). Give it to 3 significant figures.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
m=(−1−3)/(5−(−1))=−4/6=−2/3.
Where it goes wrong
- For ax+by+c=0, the gradient is −a/b only if b≠0.
- Do not confuse the fixed charge with the price per kilometre.
- A line fitted to two quoted distances does not prove the same tariff applies at all distances.
Not true: Every straight line has a finite gradient and can be written as y=mx+c. A vertical line x=k has an undefined gradient and cannot be written as y=mx+c.
Find the gradient of its nonvertical perpendicular.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
m perpendicular=−1/(−2/3)=3/2.
A vertical line x=k has an undefined gradient and cannot be written as y=mx+c.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
True. A vertical line x=k has an undefined gradient and cannot be written as y=mx+c.
Applying it
- Interpret the cost intercept, gradient and nonnegative distance domain.
- Within 0≤d≤12, interpolation uses the observed quote range; beyond it, extrapolation needs a tariff assumption.
- Check both original points after rearranging a geometric line.
For C=8+2.5d, find the predicted cost at d=8.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
C(8)=8+2.5×8=28.
Match each line form to its use.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
From ax+by+c=0 the gradient is -a/b only when b≠0; horizontal and vertical cases are handled directly.
In your course
- AQA A-level Mathematics (7357), specification section C.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- General form: The form ax+by+c=0 for a straight line, with a and b not both zero.
- For a nonvertical line through two points, m=(y₂−y₁)/(x₂−x₁) and y−y₁=m(x−x₁).
- Watch for: for ax+by+c=0, the gradient is −a/b only if b≠0.