1. Gradient
What the number tells you
- The gradient measures how steep a line is and which way it slopes
- A gradient of 3 means the line rises 3 for every 1 step to the right, and a gradient of −4 means it falls 4 for every 1 step to the right
- An uphill line, running from bottom left to top right, has a positive gradient; a downhill line has a negative one
- Comparing steepness ignores the sign, so a gradient of −5 is steeper than one of 2
- A horizontal line has gradient 0, and a vertical line has no gradient at all because there is no horizontal step to divide by
Measuring it
- On a printed grid, pick two points that sit on the line and complete a right-angled triangle between them
- The gradient is the vertical change divided by the horizontal change
- Choose two points that sit exactly on grid intersections and are far apart, because a short triangle magnifies any reading error
- Read the scales before counting squares: the two axes often use different scales, and one square is frequently not one unit
- Put the minus sign back in yourself for a downhill line, since the counting gives only the size
- Away from a grid, the gradient between (x₁, y₁) and (x₂, y₂) is (y₂ − y₁) ÷ (x₂ − x₁)
- Between (1, 3) and (5, 11): (11 − 3) ÷ (5 − 1) = 8 ÷ 4 = 2
- Subtract the two coordinates in the same order top and bottom, or the sign comes out wrong
Worked example
Finding a gradient from two points
A straight line passes through (1, 2) and (5, 10). Find its gradient.
Solution:
- Gradient is the change up the side divided by the change along the bottom
- Change in y: 10 − 2 = 8
- Change in x: 5 − 1 = 4
- Gradient = 8 ÷ 4 = 2
- Take the two coordinates in the same order in both subtractions; reversing both gives (2 − 10) ÷ (1 − 5) = −8 ÷ −4 = 2, the same answer
- Reversing only one of them gives −2, which is a line sloping the other way
2. The straight-line equation
Reading a line from its equation
- Any straight line that is not vertical can be written y = mx + c: the number m sets its steepness and the number c fixes where it meets the y-axis
- The line therefore passes through (0, c)
- y = 4x − 9 has gradient 4 and meets the y-axis at −9
- y = 11 − 2x has gradient −2 and meets the y-axis at 11, so take the number attached to x, not the one written first
- A horizontal line has the form y = c, since every point on it has the same y value
- A vertical line has the form x = k, and it cannot be written as y = mx + c at all
Other forms of the equation
- These papers also use the form ax + by = c, which has to be rearranged before the gradient can be read
- Get a single y on one side, then divide every term by whatever multiplies it
- 3x + 4y = 24 becomes 4y = −3x + 24, then y = −0.75x + 6, so the gradient is −0.75 and it crosses the y-axis at 6
- Leaving 4y on the left and reading the gradient as −3 is the standard error here
- Answers must be given fully simplified, so tidy fractions and collect any like terms before writing the final line
Worked example
Reading a line from its equation
Write down the gradient and the y-axis crossing of y = 3x − 4, and of 2x + 5y = 20.
Solution:
- The first is already in the form y = mx + c
- The number in front of x is the gradient, so the gradient is 3
- The constant is where the line crosses the y-axis, at (0, −4)
- The second is not in that form, so rearrange it before reading anything
- Subtract 2x: 5y = −2x + 20
- Divide every term by 5: y = −0.4x + 4
- The gradient is −0.4 and the line crosses the y-axis at (0, 4)
- Reading a gradient of 2 straight off 2x + 5y = 20 is the standard slip: nothing can be read until y stands alone
Worked example
Where a line meets a horizontal line
The lines y = 25 − 4x and y = −3 cross at the point P. Find the coordinates of P.
Solution:
- Every point on y = −3 has a y-coordinate of −3, so P does too
- Substitute y = −3 into the other equation: −3 = 25 − 4x
- Subtract 25 from both sides: −28 = −4x
- Divide both sides by −4: x = 7
- P is (7, −3)
- Check in y = 25 − 4x: 25 − 4 × 7 = 25 − 28 = −3
- Give both coordinates when the point is asked for: the −3 comes straight from the horizontal line
3. Drawing a straight line
From an equation
- Build a small table of values by substituting three x values that are spread across the axes you are given
- Three points, not two: a slip in one of three points shows up because it misses the line through the other two, whereas two points always look convincing
- Plot the points and join them with a ruled line that runs the full width of the grid
- Alternatively start at (0, c) and step out the gradient, going one right and m up each time, down instead if m is negative
- For a gradient written as a fraction, step across by the bottom number and up by the top number
From the ax + by = c form
- A quicker route for this form is to find where the line meets each axis
- Substitute x = 0 to get the point on the y-axis, then y = 0 to get the point on the x-axis
- Join those two points and extend the line across the grid
- For 3x + 4y = 24, setting x = 0 gives (0, 6) and setting y = 0 gives (8, 0)
Worked example
Drawing a line two ways
Draw y = 2x + 3 for x from −1 to 2, and draw 3x + 2y = 12.
Solution:
- For the first, build a short table of values
- x = −1 gives y = 1; x = 0 gives 3; x = 1 gives 5; x = 2 gives 7
- Plot (−1, 1), (0, 3), (1, 5) and (2, 7), and join them with a ruled line
- For the second, the cover-up method is quicker than rearranging
- Setting x = 0 gives 2y = 12, so y = 6 and the line passes through (0, 6)
- Setting y = 0 gives 3x = 12, so x = 4 and the line passes through (4, 0)
- Plot those two intercepts and rule the line through them
- Two points fix a straight line, but a third is worth plotting as a check
4. Finding the equation of a line
From a graph
- Work out the gradient from a triangle, then read where the line crosses the y-axis
- Put both numbers into y = mx + c and write the equation out
- If the crossing point is off the edge of the grid, use a point the line does pass through instead
From a gradient and a point
- Write y = mx + c with the gradient filled in and c left as a letter
- Substitute the coordinates of the known point and solve the resulting equation for c
- Rewrite the equation with that value in place
Worked example
A gradient and one point
Find the equation of the line with gradient 5 that passes through (3, 7). Give your answer in the form y = mx + c.
Solution:
- Start from y = 5x + c
- Substitute x = 3 and y = 7: 7 = 5 × 3 + c
- So 7 = 15 + c
- Subtract 15: c = −8
- The equation is y = 5x − 8
5. Parallel lines
Same gradient, different intercept
- Two different lines are parallel exactly when their gradients are equal (or both are vertical), and they then never meet
- y = 7x + 2 and y = 7x − 5 are parallel; y = 7x + 2 and y = 4x + 2 are not
- Check by rearranging both equations into y = mx + c first, since a shared gradient can be hidden by the form
- To find the equation of a line parallel to a given one, copy the gradient across and then find the new intercept from the point you are given
Worked example
Parallel through a given point
A line is parallel to y = 7x + 2 and passes through (3, 8). Find its equation.
Solution:
- Parallel means the same gradient, so the new line is y = 7x + c
- Substitute x = 3 and y = 8: 8 = 21 + c
- Subtract 21: c = −13
- The line is y = 7x − 13
6. Perpendicular lines
Negative reciprocal gradients
- Two lines are perpendicular when they cross at a right angle
- Their gradients multiply to give −1, so each is the negative reciprocal of the other
- 4 pairs with −0.25, and ⅔ pairs with −1.5
- To turn one gradient into the other, flip the fraction over and change the sign
- Rearrange both equations to y = mx + c before comparing, otherwise the gradients cannot be read
- The special case is a horizontal and a vertical line: y = c and x = k are perpendicular, and this is the one pair whose gradients cannot be multiplied
Building the equation
- Rearrange the given line and read its gradient
- Take the negative reciprocal to get the gradient of the new line
- Substitute the point you are given to find c, exactly as for a parallel line
Worked example
Perpendicular through a given point
A line passes through (6, 1) and is perpendicular to y = ⅔x + 4. Find its equation. Give your answer in the form y = mx + c.
Solution:
- The given gradient is ⅔
- Flip it and change the sign: the new gradient is −1.5
- Check: ⅔ × (−1.5) = −1
- Start from y = −1.5x + c
- Substitute x = 6 and y = 1: 1 = −9 + c
- Add 9: c = 10
- The equation is y = −1.5x + 10
The perpendicular bisector
- Bisecting a segment at right angles means passing through its midpoint at 90°, and the line that does so is its perpendicular bisector
- It needs both skills together: the midpoint gives the point it passes through, and the negative reciprocal gives its gradient
- Find the midpoint, find the gradient of the segment, flip and negate that gradient, then substitute the midpoint to find c
Exam tip
Write each step of a perpendicular line
Perpendicular lines, including perpendicular bisectors, are set in 8 of the 21 papers. Write the first gradient, then −1 ÷ it, then the point substituted into y = mx + c, each on its own line. In all eight schemes the last two lines earn a method mark each, and the substitution mark follows through from your own perpendicular gradient, so an early slip does not cost it. A gradient worked out from two points earns its own mark too.
Worked example
Perpendicular bisector of a segment
Find an equation for the perpendicular bisector of the segment from P(2, −1) to Q(10, 5).
Solution:
- Midpoint: ((2 + 10) ÷ 2, (−1 + 5) ÷ 2) = (6, 2)
- Gradient of PQ: (5 − (−1)) ÷ (10 − 2) = 6 ÷ 8 = 0.75
- Negative reciprocal: −4 ÷ 3
- Start from y = (−4 ÷ 3)x + c
- Substitute (6, 2): 2 = −8 + c, so c = 10
- The perpendicular bisector is y = (−4 ÷ 3)x + 10