This note covers straight-line graphs and their equations, how to find and use gradients, plotting and reading common curves, and interpreting distance-time and speed-time graphs. These skills link algebra to pictures and to real situations such as journeys, costs and rates of change.
Key ideas
Gradient. The gradient of a straight line measures how steep it is. Between two points \((x_1, y_1)\) and \((x_2, y_2)\): \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] A positive gradient slopes up to the right, a negative gradient slopes down, and a horizontal line has gradient 0.
Equation of a straight line. Written as \(y = mx + c\), where \(m\) is the gradient and \(c\) is the \(y\)-intercept. To find the equation from a point \((x_1, y_1)\) and a gradient \(m\), use \(y - y_1 = m(x - x_1)\). Vertical lines have equations \(x = k\) and have no gradient value.
Parallel and perpendicular lines.
- Parallel lines have equal gradients.
- Perpendicular lines (Higher) have gradients that multiply to \(-1\). If one line has gradient \(m\), a perpendicular line has gradient \(-\dfrac{1}{m}\).
Midpoint and length of a line segment. For points \(A(x_1,y_1)\) and \(B(x_2,y_2)\): \[ \text{midpoint} = \left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right), \qquad AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \]
Plotting curves. Make a table of \(x\)-values, work out each \(y\), plot the points and join them with a smooth curve (never with a ruler).
- Quadratic \(y = ax^2 + bx + c\): a U-shaped parabola (an inverted U when \(a \lt 0\)). The roots are where it crosses the \(x\)-axis. For \(y = x^2 - 2x - 3\), the roots are \(x=-1\) and \(x=3\) and the lowest point is \((1,-4)\).
- Cubic: an S-shaped curve that can have up to two turning points.
- Reciprocal \(y = \dfrac{k}{x}\): two separate branches that never touch the axes.
- \(y = k^x\): an exponential curve that passes through \((0,1)\) and rises steeply.
Solving equations from graphs. The solutions of \(f(x) = g(x)\) are the \(x\)-coordinates of the points where the two graphs meet. To solve \(x^2 - 2x - 3 = 2\), draw the line \(y = 2\) on the same axes and read off where it meets the curve.
Gradient of a curve (Higher). The gradient at a point on a curve is the gradient of the tangent there. Draw a tangent with a ruler so that it just touches the curve, then pick two points far apart on the tangent to work out the gradient.
Real-life graphs.
- Distance-time graph: the gradient is the speed. A horizontal section means the object is stationary. A steeper line means faster travel.
- Speed-time graph: the gradient is the acceleration. A negative gradient means deceleration. The area under the graph is the distance travelled. Use triangles, rectangles and trapeziums to find the area.
- For a curved speed-time graph, estimate the area by splitting it into strips (trapeziums).
- Average speed is total distance divided by total time.
- Conversion graphs and cost graphs use the same ideas: the gradient is the rate (for example pounds per hour), and the intercept is a fixed amount.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(m = \dfrac{y_2-y_1}{x_2-x_1}\) | Gradient of the line through two points. |
| \(y = mx + c\) | Straight line with gradient \(m\) and \(y\)-intercept \(c\). |
| \(y - y_1 = m(x - x_1)\) | Line through \((x_1,y_1)\) with gradient \(m\). |
| \(m_1 = m_2\) | Condition for parallel lines. |
| \(m_1 m_2 = -1\) | Condition for perpendicular lines (Higher). |
| Midpoint | Average of the \(x\)-coordinates and average of the \(y\)-coordinates. |
| \(AB = \sqrt{(\Delta x)^2 + (\Delta y)^2}\) | Length of a line segment, from Pythagoras' theorem. |
| Gradient of distance-time graph | Speed. |
| Gradient of speed-time graph | Acceleration. |
| Area under speed-time graph | Distance travelled. |
Worked examples
Example 1 \(A\) is the point \((2, 3)\) and \(B\) is the point \((6, 11)\). (a) Find the equation of line \(AB\). (b) (Higher) Find the equation of the line through \(C(4, 5)\) perpendicular to \(AB\). (c) Find the midpoint and length of \(AB\).
(a) \(m = \dfrac{11-3}{6-2} = \dfrac{8}{4} = 2\).
Use \(y - 3 = 2(x - 2)\). Then \(y - 3 = 2x - 4\), so \(y = 2x - 1\).
Check with \(B\): \(2(6) - 1 = 11\). Correct.
(b) The perpendicular gradient is \(-\dfrac{1}{2}\). Use \(y = -\tfrac12 x + c\) with the point \((4,5)\): \(5 = -2 + c\), so \(c = 7\).
The line is \(y = -\tfrac12 x + 7\).
(c) Midpoint: \(\left(\dfrac{2+6}{2}, \dfrac{3+11}{2}\right) = (4, 7)\).
Length: \(AB = \sqrt{4^2 + 8^2} = \sqrt{80} = 8.94\) (3 s.f.).
Answer: (a) \(y=2x-1\) (b) \(y=-\tfrac12x+7\) (c) midpoint \((4,7)\), length \(8.94\)
Example 2 A car starts from rest and accelerates uniformly to 20 m/s in 8 s. It then travels at 20 m/s for 30 s, then slows uniformly to rest in 12 s. Find (a) the acceleration in the first stage, (b) the deceleration in the last stage, (c) the total distance, (d) the average speed.
(a) Gradient of the first section: \(\dfrac{20 - 0}{8} = 2.5\) m/s².
(b) Gradient of the last section: \(\dfrac{0 - 20}{12} = -1.67\) m/s². The deceleration is 1.67 m/s² (3 s.f.).
(c) Find the area under the graph in three parts.
Stage 1 (triangle): \(\tfrac12 \times 8 \times 20 = 80\) m.
Stage 2 (rectangle): \(30 \times 20 = 600\) m.
Stage 3 (triangle): \(\tfrac12 \times 12 \times 20 = 120\) m.
Total distance \(= 80 + 600 + 120 = 800\) m.
(d) Total time \(= 8 + 30 + 12 = 50\) s. Average speed \(= \dfrac{800}{50} = 16\) m/s.
Answer: (a) 2.5 m/s² (b) 1.67 m/s² (c) 800 m (d) 16 m/s
Example 3 A cyclist rides 12 km from home in 40 minutes, rests for 20 minutes, then rides straight back home in 60 minutes. Find the speed on each journey and the average speed for the whole trip, in km/h.
Outward speed: 40 minutes is \(\tfrac{2}{3}\) h, so speed \(= \dfrac{12}{2/3} = 18\) km/h.
The rest is a horizontal line on the distance-time graph, so the speed is 0 km/h.
Return speed: 60 minutes is 1 h, so speed \(= \dfrac{12}{1} = 12\) km/h.
Whole trip: distance \(= 12 + 12 = 24\) km. Time \(= 40 + 20 + 60 = 120\) min \(= 2\) h. The rest is included in the total time.
Average speed \(= \dfrac{24}{2} = 12\) km/h.
Answer: 18 km/h out, 12 km/h back, average speed 12 km/h
Common errors
- Common error: subtracting the coordinates in different orders, for example \(y_2 - y_1\) on top but \(x_1 - x_2\) on the bottom. This gives the wrong sign.
- Common error: giving the perpendicular gradient as the negative only, or the reciprocal only. It needs both: flip and change sign.
- Common error: reading the gradient from a graph with unequal scales by counting squares instead of using the axis values.
- Common error: leaving out the rest period when finding average speed over a whole journey.
- Common error: joining points on a curve with straight segments or drawing a flat bottom on a parabola. Draw a smooth curve through every point.
Exam tip: When you find the area under a speed-time graph, split it into simple shapes and write the area of each shape separately, so that a slip in one part does not hide the rest of your method.
Check yourself
- Find the equation of the line through \((1, 4)\) and \((5, -4)\).
Answer
\(m = \dfrac{-4-4}{5-1} = -2\). Use \(y - 4 = -2(x - 1)\), so \(y = -2x + 6\). Check with \((5,-4)\): \(-10 + 6 = -4\). - (Higher) Are the lines \(y = 3x + 2\) and \(3y + x = 5\) perpendicular?
Answer
Rearrange the second line: \(y = -\tfrac13 x + \tfrac53\), so its gradient is \(-\tfrac13\). The product of the gradients is \(3 \times (-\tfrac13) = -1\), so the lines are perpendicular. - Find the length of the line from \((-2, 1)\) to \((4, 9)\).
Answer
\(\Delta x = 6\), \(\Delta y = 8\). Length \(= \sqrt{36 + 64} = \sqrt{100} = 10\). - A train slows from 30 m/s to 12 m/s in 9 s at a constant rate. Find the acceleration and the distance covered.
Answer
Acceleration \(= \dfrac{12-30}{9} = -2\) m/s², so the deceleration is 2 m/s². Distance is the area of a trapezium: \(\tfrac12(30+12) \times 9 = 189\) m.