Maths › Coordinate geometry › Straight lines
Straight lines
Two numbers pin down every straight line there is, one point and one gradient, and everything else follows. Gradients decide parallel and perpendicular at a glance, and lines earn their keep as models, where the gradient is a rate and the intercept is a starting value.
Builds on Simultaneous equations and inequalities.
IN THIS TOPIC
- Find the equation of a line from a point and a gradient, or from two points, in either standard form.
- Use gradient conditions for parallel and perpendicular lines.
- Build and interpret straight-line models, gradient as rate and intercept as starting value.
WHAT YOU PROBABLY THINK
Perpendicular gradients are negatives of each other.
Two points, one line
A line's gradient is its rate of climb, the change in y over the change in x between any two of its points, m = (y2 − y1)/(x2 − x1). Once one point (x1, y1) and the gradient are known, every other point (x, y) on the line satisfies
and this point-gradient form is on the must-learn list. The other standard form is ax + by + c = 0 with integer coefficients, which questions often demand for the final answer because it has no fractions in it.
WORKED EXAMPLE
From two points to both forms
Find the equation of the line through (2, 5) and (6, 3), giving the answer in the form ax + by + c = 0 with integer coefficients.
Gradient first. m = (3 − 5)/(6 − 2) = −2/4 = −1/2.
Point-gradient form with (2, 5): y − 5 = −1/2(x − 2).
Double both sides and tidy: 2y − 10 = −x + 2, so x + 2y − 12 = 0.
Checking the second point costs one substitution, 6 + 6 − 12 = 0, and catches most slips in the topic.
Parallel and perpendicular
Parallel lines never gain on each other, so they share a gradient. Perpendicular lines obey the sharper condition, also on the must-learn list,
so each gradient is the negative reciprocal of the other, flip the fraction and change the sign. That kills the lie above. The perpendicular to gradient 2 is −1/2, not −2; negating alone fails whenever the gradient is not ±1, and a quick product check, 2 × (−2) = −4, exposes it.
WORKED EXAMPLE
A perpendicular bisector
Find the equation of the perpendicular bisector of the segment joining (2, 5) and (6, 3).
Two ingredients. The midpoint is ((2 + 6)/2, (5 + 3)/2) = (4, 4), and the segment's gradient is −1/2.
The bisector is perpendicular, so its gradient is the negative reciprocal, 2.
Through (4, 4) with gradient 2: y − 4 = 2(x − 4), which is y = 2x − 4.
Every point of this line is equidistant from the two ends, which is what “bisector” promises and what makes it the working horse of circle questions next lesson.
YOUR TURN
A right angle, proved not eyeballed
The points A(1, 2), B(5, 4) and C(3, 8) form a triangle. Show that the angle at B is a right angle, before opening the working.
Show the working
Gradient of AB: (4 − 2)/(5 − 1) = 1/2. Gradient of BC: (8 − 4)/(3 − 5) = −2.
The product is 1/2 × (−2) = −1, so AB ⊥ BC and the angle at B is 90°.
A sketch can suggest a right angle; only the gradient product can confirm one. Diagrams in these questions are deliberately drawn not-to-scale.
Lines as models
A straight-line model is a claim that something changes at a steady rate. The gradient carries the rate, with units; the intercept carries the starting value. A van bought for £12 000 that loses £150 of value each month is the line V = 12000 − 150t, and reading it is the skill the course wants.
WORKED EXAMPLE
Reading a depreciation model
A van's value in pounds after t months is modelled by V = 12000 − 150t. Interpret the numbers, find when the value reaches £6000, and criticise the model for large t.
The intercept 12 000 is the purchase price. The gradient −150 says the van loses £150 per month.
Setting V = 6000: 150t = 6000, so t = 40 months.
At t = 80 the model gives V = 0 and beyond that a negative value, which no van achieves. A linear model is a local description, and stating its range of validity is part of the answer.
TRY IT UNSEEN
Two tariffs cross
Gym A charges £25 per month. Gym B charges an £80 joining fee plus £15 per month. After how many months does B become the cheaper total, and what does the crossing point mean on a graph?
Show the working
Total costs after m months: A is 25m, B is 80 + 15m.
Equal when 25m = 80 + 15m, so 10m = 80 and m = 8. Both totals are then £200.
B is cheaper for m > 8. On a graph the two cost lines cross at (8, 200); the steeper line wins early, the flatter one wins late.
This is the simultaneous-equations lesson priced in pounds. Intersection of graphs and solution of equations are the same event.
THE EXAM BIT
- Quote m = (y2 − y1)/(x2 − x1) and keep the subtraction order consistent top and bottom; swapping one of them flips the sign of the gradient.
- When the demanded form is ax + by + c = 0 with integer coefficients, clear fractions and gather everything on one side. Leaving y = mx + c loses the answer mark.
- Perpendicular means negative reciprocal. Write the product m1m2 and show it equals −1; the shown product is usually a mark on its own.
- Substitute a known point into your final equation as a five-second check before moving on.
- In modelling questions, interpret gradient and intercept with units, and expect a part asking where the model breaks down.
CHECK YOURSELF
Find the equation of the line through (−1, 4) perpendicular to the line 3x + y − 2 = 0, giving your answer in the form ax + by + c = 0.
Show a hint
Rearrange to read off the first gradient, then take the negative reciprocal.
Show the answer
3x + y − 2 = 0 is y = −3x + 2, gradient −3, so the perpendicular gradient is 1/3.
Through (−1, 4): y − 4 = 1/3(x + 1). Tripling and tidying gives x − 3y + 13 = 0.
Check with the point: −1 − 12 + 13 = 0, and the gradient product is −3 × 1/3 = −1, as it must be.
One point and one gradient determine a line; y − y₁ = m(x − x₁) turns them into its equation.
Parallel lines share m; perpendicular gradients multiply to −1.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device only.
- Find the equation of a line from a point and a gradient, or from two points, in either standard form.
- Use gradient conditions for parallel and perpendicular lines.
- Build and interpret straight-line models, gradient as rate and intercept as starting value.
No animated video for this topic yet; these notes stand alone.