Coordinates in four quadrants
A point is written as (x, y). First go across (x), then go up or down (y). Remember: "along the corridor, then up the stairs".
The two axes cut the plane into four quadrants:
- Quadrant 1: x positive, y positive, e.g. (3, 2)
- Quadrant 2: x negative, y positive, e.g. (−4, 1)
- Quadrant 3: both negative, e.g. (−2, −3)
- Quadrant 4: x positive, y negative, e.g. (5, −1)
The midpoint of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2).
Straight lines: y = mx + c
To plot a line, make a table of values: pick some x values, work out y, plot the points and join them with a ruler. If one point is off the line, check your sums.
Every straight line (except a vertical one) can be written as y = mx + c.
- m = gradient = rise ÷ run = (y₂ − y₁) ÷ (x₂ − x₁). Positive m goes up to the right, negative m goes down, m = 0 is flat.
- c = y-intercept: the line cuts the y-axis at (0, c).
Special lines: y = 3 is horizontal; x = −2 is vertical; y = x goes through the origin at 45°.
To find the equation from two points: work out m, then put one point into y = mx + c to find c. If the equation looks different, like 2y − 4x = 6, rearrange it to make y the subject: y = 2x + 3.
Where a line cuts the axes
Put x = 0 to find the y-intercept. Put y = 0 to find the x-intercept (the root).
Parallel and perpendicular lines
Parallel lines never meet. They have the same gradient: y = 2x + 1 and y = 2x − 3 are parallel.
Perpendicular lines meet at a right angle. Their gradients multiply to −1: m₁ × m₂ = −1. So the perpendicular gradient is the negative reciprocal: 2 → −½, −3 → ⅓, ¾ → −4/3.
Example: the line through (4, 1) perpendicular to y = 2x + 5 has m = −½. Then 1 = −½ × 4 + c, so c = 3, and the line is y = −½x + 3.
Solving equations using graphs
Every point on a line makes its equation true. So the point where two lines cross makes both equations true at once. That point is the solution of the simultaneous equations.
Example: draw y = 2x + 1 and y = −x + 7. They cross at (2, 5), so x = 2, y = 5.
To solve 2x + 1 = 4 with a graph, draw y = 2x + 1 and y = 4 and read the x value where they meet (x = 1.5). Graph answers can be a little rough, so check by putting the values back into the equations.
Other graphs you should recognise
Not every graph is a straight line. Learn the shapes so you can spot them:
- Quadratic y = x² + bx + c: a U-shaped curve (parabola). It has a turning point. Where it cuts the x-axis are the roots. If x² has a minus sign, it is an upside-down U.
- Cubic y = x³: an S-shaped curve.
- Reciprocal y = 1/x: two separate curves that get close to the axes but never touch them.
- Exponential y = 2ˣ: grows faster and faster (like money with interest).
- Trig y = sin x and y = cos x: waves that repeat every 360°.
Moving graphs: y = f(x) + a moves the graph up by a. y = f(x + a) moves it left by a. y = −f(x) reflects it in the x-axis, and y = f(−x) reflects it in the y-axis.
Real-life graphs: gradient and area
In a real-life graph the axes have units, so the gradient is a rate:
- Distance–time graph: gradient = speed. A flat part means the object is not moving.
- Velocity–time graph: gradient = acceleration; the area under the graph = distance travelled. Split the area into rectangles and triangles (or trapeziums).
- Conversion graph (e.g. kg to pounds, rupees to dollars): gradient = the exchange rate.
- Cost graph: gradient = cost per unit, intercept = fixed charge.
For a curved graph, the gradient at a point is the gradient of the tangent there; the area under it can be estimated with strips.
Try it: a graph from your own walk
Walk at a steady pace and count how many steps you take every 10 seconds for one minute. Make a table (time, total steps) and plot it. Is it a straight line? Find the gradient: that is your steps per second. Now walk faster for the next minute: what happens to the line? Then check with the slider in the 3D: a bigger m means a steeper line.
Key formulas and definitions
- y = mx + c
- gradient m = (y₂ − y₁) ÷ (x₂ − x₁) = rise ÷ run
- parallel lines: m₁ = m₂
- perpendicular lines: m₁ × m₂ = −1
- midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2)
- velocity–time graph: area under graph = distance
Worked examples
1. Find the gradient and y-intercept of y = 3x − 5.
Compare with y = mx + c: m = 3 and c = −5. The line goes up 3 for every 1 across and cuts the y-axis at (0, −5).
2. Find the gradient of the line through (1, 2) and (4, 11).
m = (11 − 2) ÷ (4 − 1) = 9 ÷ 3 = 3.
3. Find the equation of the line through (2, 7) with gradient 4.
y = 4x + c. Put in (2, 7): 7 = 8 + c, so c = −1. The line is y = 4x − 1.
4. Rearrange 3y + 6x = 12 into y = mx + c and state the gradient.
3y = −6x + 12, so y = −2x + 4. Gradient −2, y-intercept 4.
5. Find the line perpendicular to y = 3x + 2 that passes through (6, 1).
Perpendicular gradient = −1/3. 1 = −1/3 × 6 + c = −2 + c, so c = 3. The line is y = −⅓x + 3.
6. A car speeds up from 0 to 20 m/s in 10 s, then keeps 20 m/s for 15 s. Find the distance from the velocity–time graph.
Triangle: ½ × 10 × 20 = 100 m. Rectangle: 15 × 20 = 300 m. Total distance = 400 m. Acceleration in the first part = gradient = 20 ÷ 10 = 2 m/s².
Common mistakes
- Reading coordinates the wrong way round: (x, y) means across first, then up.
- Writing gradient as run ÷ rise. It is rise ÷ run (change in y over change in x).
- Forgetting the minus sign for a line that goes down: it has a negative gradient.
- For perpendicular lines, only flipping the fraction or only changing the sign. You must do both: 2 → −½.