1The idea in one paragraph

A straight line is the graph of a relationship that changes at a constant rate: every step of 1 to the right moves the line up or down by the same amount. That fixed amount is the gradient, m. Know the gradient and one point, and the line is fixed; there is exactly one line through that point with that slope. Everything on this page follows from that. The three standard equations are three ways of writing the same line, chosen for what each one shows. Parallel lines share a gradient. Perpendicular lines have gradients that multiply to −1. And a slope in the real world, a road or a ramp, is the same number: vertical rise divided by horizontal distance.

2Gradient: rise over run

The gradient of a line measures how steep it is and which way it slopes. Take any two points on it, (x₁, y₁) and (x₂, y₂). The run is the horizontal change x₂ − x₁, the rise is the vertical change y₂ − y₁, and

m = (y₂ − y₁) ÷ (x₂ − x₁) · rise over run, from the formula booklet.

Figure 1 draws it for A(−2, 5) and B(4, −1).

Two things make this reliable. First, subtract in the same order on top and bottom: B minus A in both, or A minus B in both. Swapping the order in one place only flips the sign and gives +1, which is wrong. Second, it does not matter which two points on the line you pick; a straight line has the same gradient everywhere, which is exactly what makes it straight.

The sign of m tells you the direction, as Figure 2 shows. Reading from left to right, a line with m > 0 rises and a line with m < 0 falls. A horizontal line has m = 0: the rise is zero whatever the run. A vertical line has no gradient at all (the gradient is undefined), because the run is zero and you cannot divide by zero. A vertical line is written x = k and a horizontal line y = k.

The size of m tells you the steepness. A gradient of 3 means up 3 for every 1 across. A gradient of 0.2 means up 0.2 for every 1 across, a gentle slope. Gradients of 3 and −3 are equally steep; one rises and one falls.

3Gradient in the real world: roads, ramps and rates

The guide asks you to calculate gradients of inclines. A road, a ramp or a bridge has a gradient in exactly the sense above, with one care point: the run is the horizontal distance, not the distance along the slope.

A mountain road climbs 84 m while covering a horizontal distance of 1.2 km. Put both in metres first.

Figure 3 shows that triangle. A road sign reading 7% means exactly this: 7 m up for every 100 m across.

The gradient is also the tangent of the angle of inclination θ, the angle the slope makes with the horizontal, because tan θ = opposite ÷ adjacent = rise ÷ run. So tan θ = 0.07 and θ = arctan 0.07 ≈ 4.00°. That uses the right-angled trigonometry you already know, and Paper 2 likes to ask for it.

An access ramp works the same way. Suppose a design rule says the gradient of a wheelchair ramp may not exceed 1/12, and a ramp rises 0.45 m over a horizontal run of 5.4 m. Its gradient is 0.45 ÷ 5.4 = 0.0833…, which is exactly 1/12. It meets the rule, just.

A gradient is also a rate of change, and that is how it appears in models. A taxi firm charges a fixed €3.50 plus €2.40 per kilometre, so the cost of a journey of d km is C(d) = 2.4d + 3.5. The gradient 2.4 is the extra cost for each extra kilometre, in euros per km. The intercept 3.5 is the cost when d = 0, the fixed charge. A 12 km journey costs 2.4 × 12 + 3.5 = €32.30. When a question says interpret the gradient, the answer always has this shape: "for each extra one unit of x, y increases by m units", with the real units named.

4Three forms of one line

The formula booklet gives three forms. They are equivalent: each describes exactly the same set of points. Learn what each one is good for.

Work one line through all three. The line passes through (6, 1) with gradient −2/3.

Check with the known point: 2(6) + 3(1) − 15 = 12 + 3 − 15 = 0. It lies on the line, so the equation is right. That check takes five seconds and catches most sign slips. Figure 4 draws the line with all three forms beside it.

Going back from general form. Rearrange for y. From ax + by + d = 0 you get y = −(a/b)x − d/b, so the gradient is −a/b. Take 3x − 4y + 12 = 0: then 4y = 3x + 12, so y = (3/4)x + 3, with gradient 3/4 and y-intercept (0, 3). Do the rearrangement each time rather than memorising −a/b; the rearrangement is where the method mark lives.

A line through two points. Find the gradient first, then use either point in the point-gradient form. For P(1, 4) and Q(5, −4):

Using Q instead of P gives y + 4 = −2(x − 5), which simplifies to the same line. Check with the point you did not use.

5Intercepts

An intercept is where a line crosses an axis, and it is a point, so write it as coordinates.

• The y-intercept is where x = 0. Put x = 0 into the equation.
• The x-intercept is where y = 0. Put y = 0 into the equation.

For 2x + 3y − 15 = 0: with x = 0, 3y = 15 so y = 5, giving (0, 5). With y = 0, 2x = 15 so x = 7.5, giving (7.5, 0). Both are marked on Figure 4. For 3x − 4y + 12 = 0: the y-intercept is (0, 3) and the x-intercept, from 3x + 12 = 0, is (−4, 0).

In y = mx + c the y-intercept can be read straight off as (0, c). No other form hands you an intercept for free.

6Parallel and perpendicular lines

Two lines with gradients m₁ and m₂ are:

Parallel when m₁ = m₂. Perpendicular when m₁ × m₂ = −1, so m₂ = −1 ÷ m₁, the negative reciprocal.

Parallel lines rise at the same rate, so they never meet. That is the whole reason: same gradient, different intercept.

The perpendicular rule needs a picture, and Figure 5 gives it. Take a gradient triangle with run 4 and rise 3, so m₁ = 3/4. Turn the whole triangle a quarter turn about the origin. The run of 4 becomes a rise of 4, and the rise of 3 becomes a run of −3, pointing left. The new gradient is 4 ÷ (−3) = −4/3. The two numbers have swapped places (the reciprocal) and one sign has flipped (the negative). Multiply them: (3/4) × (−4/3) = −1. That works for any gradient, because turning through 90° always swaps run and rise and flips one sign.

The rule has one exception you must know. A horizontal line (m = 0) and a vertical line (no gradient) are perpendicular, but you cannot multiply an undefined gradient by anything. Treat that pair by eye: y = 3 and x = −1 meet at a right angle.

Now use both rules on the same line. Start from L: 3x − 4y + 12 = 0, whose gradient is 3/4 (section 4), and the point (2, −1). Figure 6 draws the results.

Parallel to L through (2, −1). Same gradient, 3/4. A quick route in general form: a parallel line has the same a and b, so it is 3x − 4y + d = 0 for some d. Substitute the point: 3(2) − 4(−1) + d = 0, so 6 + 4 + d = 0 and d = −10. The line is 3x − 4y − 10 = 0.

Perpendicular to L through (2, −1). Gradient −4/3.

Notice in general form that the perpendicular line swapped a and b and changed one sign: 3x − 4y became 4x + 3y. That is the negative reciprocal in disguise, and it makes a quick check.

The perpendicular bisector. The perpendicular bisector of a segment AB is the line through its midpoint at right angles to it. Every point on it is the same distance from A as from B, which is why it turns up in questions about the point equidistant from two towns or two masts. Figure 7 draws it for A(−1, 2) and B(5, 6).

7Where two lines meet

Two lines that are not parallel meet at exactly one point, and that point satisfies both equations at once. So finding it means solving the two equations simultaneously, which is prior learning.

Where do y = 2x − 1 and x + 3y = 11 meet? Substitute the first into the second.

On Paper 1 you do this by hand, and the numbers will be kind. On Paper 2 you may use your GDC, either its simultaneous-equation solver or by graphing both lines and using the intersect tool, which is the method 2.4 develops for curves too. Either way, give the answer as a point, (2, 3), not as two separate numbers.

If two lines have the same gradient and different intercepts they never meet: the equations have no solution. If they have the same gradient and the same intercept they are the same line, and every point is a solution.