Everything from the 6 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
A point is written \((x, y)\): across first, then up or down. To draw a linear graph, choose some \(x\) values, work out \(y\) for each, plot the points and join them with ONE ruled straight line. If a point is off the line, one of your calculations is wrong.
Fig. 1: the table of values for y = -2x + 5 gives five points, and all five lie on one straight line
Worked example: Finding the fourth corner
\(A(-2, 3)\), \(B(4, 3)\) and \(C(4, -1)\) are three corners of a rectangle \(ABCD\). Find the coordinates of \(D\).
\(A\) and \(B\) have the same \(y\) value, so \(AB\) is horizontal. \(B\) and \(C\) have the same \(x\) value, so \(BC\) is vertical.
\(D\) must be directly below \(A\), so it has the same \(x\) value as \(A\): \(x = -2\).
\(D\) must be level with \(C\), so it has the same \(y\) value as \(C\): \(y = -1\).
Answer: \(D(-2, -1)\)
Choose \(x\), work out \(y\), plot, rule.
Coordinates: \((x, y)\), across first, then up or down.
Table: substitute each \(x\) carefully; brackets round negative numbers.
Pattern check: the \(y\) values change by the same amount each step.
Plot and rule: one straight line through every point, across the full range.
Special lines: \(x = k\) is vertical, \(y = k\) is horizontal.
Watch out!
Writing the coordinates the wrong way round. \((x, y)\) is always across first, then up: 'along the corridor, then up the stairs'. \((-1, 3)\) and \((3, -1)\) are in different quadrants.
Sign slips with negative \(x\): \(-2 \times (-1)\) is \(+2\), not \(-2\). If one value breaks the steady pattern in your table, recheck that column before you plot.
Mixing up \(x = 3\) and \(y = 3\). The line \(x = 3\) is VERTICAL (every point has \(x = 3\)), even though it crosses the \(x\)-axis. Picture a few points on it before you draw.
Gradient \(= \dfrac{\text{rise}}{\text{run}} = \dfrac{\text{change in } y}{\text{change in } x}\). Uphill from left to right is positive, downhill is negative, a horizontal line has gradient 0, and a vertical line has no gradient (it is undefined).
Fig. 1: from A(0, 1) to B(4, 7) the line rises 6 for a run of 4, so its gradient is 6 / 4 = 1.5
Worked example: Uphill line from a grid
A straight line passes through the grid points \((0, 1)\) and \((4, 7)\). Find its gradient.
Run: from \(x = 0\) to \(x = 4\) is 4 units across.
Rise: from \(y = 1\) to \(y = 7\) is 6 units up.
Gradient \(= \dfrac{6}{4} = \dfrac{3}{2}\).
Answer: \(\dfrac{3}{2}\) or \(1.5\) (Fig. 1)
Change in \(y\) over change in \(x\).
Grid: draw a right-angled triangle under the line between two exact grid points.
Coordinates (Extended): \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\), same order top and bottom.
Sign: uphill positive, downhill negative.
Special lines: horizontal 0, vertical undefined.
Meaning: the gradient is the change in \(y\) for each 1 unit of \(x\), with units \(y\)-units per \(x\)-unit.
Watch out!
Counting squares when the axes have different scales. If one square across is 2 units and one square up is 10 units, read the actual values off the axes, not the number of squares.
Taking the \(y\) values in one order and the \(x\) values in the other order. If you start with \(y_2\) on top, you must start with \(x_2\) underneath, or the sign flips. And it is \(y\) over \(x\), never \(x\) over \(y\).
A horizontal line has gradient 0 (no rise at all), but a VERTICAL line does not have gradient 0: its run is 0, and you cannot divide by 0, so its gradient is undefined.
Missing end point: repeat the step from the known end to the midpoint.
Accuracy: exact surd or 3 significant figures.
Watch out!
Adding before squaring. \(\sqrt{6^2 + 8^2}\) is 10, but \(6 + 8 = 14\) and \(\sqrt{6 + 8} = 3.74\) are both wrong. Square each step first, add, then take the square root of the total.
Halving the difference instead of the sum. For \((-3, 8)\) and \((9, -2)\), \(\dfrac{9 - (-3)}{2} = 6\) is half the run, not the midpoint's \(x\)-coordinate. The midpoint is an average: ADD, then halve.
Forgetting the negative square root. \((k - 2)^2 = 25\) gives \(k - 2 = \pm 5\): a point 13 units away can be on either side, so there are usually two answers.
In \(y = mx + c\), \(m\) is the gradient and \(c\) is the \(y\)-intercept (the line crosses the \(y\)-axis at \((0, c)\)). To find an equation: get the gradient, then substitute one point to find \(c\). A vertical line has equation \(x = k\).
Fig. 1: y = 2x - 3 crosses the y-axis at (0, -3) and rises 2 for every 1 across, so m = 2 and c = -3
Worked example: Reading the gradient and intercept
Write down the gradient and \(y\)-intercept of (a) \(y = 6x + 3\) (b) \(y = 5 - 2x\).
(a) The coefficient of \(x\) is 6 and the constant is 3.
(b) Rewrite as \(y = -2x + 5\): the coefficient of \(x\) is \(-2\) (the minus sign belongs to it).
Read: in \(y = mx + c\), \(m\) is the gradient and \(c\) the \(y\)-intercept.
Rearrange (Extended): make \(y\) the subject, dividing every term.
Gradient: from the graph, or \(\dfrac{y_2 - y_1}{x_2 - x_1}\) from two points.
Intercept: substitute one point into \(y = mx + c\) and solve for \(c\).
Check: the other point must fit; give the equation fully simplified.
Watch out!
Taking the first number as the gradient. In \(y = 5 - 2x\) the gradient is \(-2\), not 5: the gradient is always the number in front of \(x\), with its sign.
Dividing only some of the terms. From \(4y = -5x + 8\), the equation \(y = -5x + 2\) divides only the 8 by 4. Every term on the right must be divided.
Putting the coordinates in the wrong places: for the point \((2, 1)\), \(x = 2\) and \(y = 1\). Writing \(2 = 3 \times 1 + c\) gives the wrong \(c\). Always check your final equation with the point.
Parallel lines have equal gradients. To find the line parallel to \(y = mx + c\) through a point: keep the same \(m\), substitute the point to find the new \(c\).
Fig. 1: y = 2x + 1 and y = 2x - 5 both rise 2 for every 1 across, so they are parallel; the second one passes through (3, 1)
Worked example: Spotting parallel lines
Which of these lines are parallel? A: \(y = 3x + 2\) B: \(y = 2x + 3\) C: \(y = 3x - 5\) D: \(y = 5 - 3x\)
Gradients: A is 3, B is 2, C is 3.
D is \(y = -3x + 5\), so its gradient is \(-3\), not 3.
Only A and C have the same gradient.
Answer: A and C
Same \(m\), new \(c\).
Gradient: write the given line as \(y = mx + c\) and read \(m\).
Parallel: the new line has the same gradient \(m\).
Intercept: substitute the given point into \(y = mx + c\) to find the new \(c\).
Shortcut (Extended): parallel to \(ax + by = c\) is \(ax + by = k\); put the point in to find \(k\).
Proof: equal gradients show that two sides are parallel.
Watch out!
Using the point's \(y\)-coordinate as the intercept. The line through \((1, -3)\) is NOT \(y = 4x - 3\): the point is not on the \(y\)-axis, so you must substitute to find \(c\).
Reading the coefficient of \(x\) in \(2x + 3y = 6\) as the gradient. The gradient is \(-\dfrac{2}{3}\), not 2: rearrange to \(y = \ldots\) before you read it.
Dividing by a negative and losing the sign. From \(-2y = -6x + 5\), every term is divided by \(-2\): \(y = 3x - 2.5\). Check the sign of your gradient by picturing which way the line slopes.
If two lines are perpendicular, \(m_1 \times m_2 = -1\), so the perpendicular gradient is \(-\dfrac{1}{m}\): flip the fraction and change the sign. The perpendicular bisector of \(AB\) passes through the midpoint of \(AB\) with the perpendicular gradient.
Fig. 1: AB has gradient -2; its perpendicular bisector passes through the midpoint M(2, 2) with gradient 1/2, giving y = x/2 + 1
Worked example: Perpendicular gradients
Find the gradient of a line perpendicular to (a) \(y = 3x - 2\) (b) \(2y = 3x + 1\).
(a) The gradient is 3. Flip: \(\dfrac{1}{3}\). Change sign: \(-\dfrac{1}{3}\).
(b) Make \(y\) the subject: \(y = \dfrac{3}{2}x + \dfrac{1}{2}\), so the gradient is \(\dfrac{3}{2}\).
Gradient: write the given line as \(y = mx + c\) and read \(m\).
Perpendicular: new gradient \(-\dfrac{1}{m}\); check the product is \(-1\).
Through a point: substitute it into \(y = mx + c\) to find \(c\).
Bisector of \(AB\): midpoint of \(AB\) plus the gradient perpendicular to \(AB\).
Tidy: clear fractions if asked for the form \(ax + by = c\).
Watch out!
Doing only half the recipe. The perpendicular to gradient 3 is not \(-3\) (sign only) and not \(\dfrac{1}{3}\) (flip only). It is \(-\dfrac{1}{3}\). Multiply your two gradients: you must get \(-1\).
Using the original gradient by mistake when you substitute. Write the NEW gradient down on its own line before finding \(c\), so you cannot slip back to the old one.
Putting \(A\) or \(B\) into the equation instead of the midpoint. The bisector passes through \(M\), not through the end points, so substitute the midpoint to find \(c\).