Add Maths · 7 Straight-line graphs

7 Straight-line graphs

Before you start

Plotting points and reading coordinates, including negative ones.
Solving a pair of linear simultaneous equations by substitution or elimination.
Pythagoras' theorem for a right-angled triangle.
Simplifying surds: $\sqrt{45} = \sqrt{9 \times 5} = 3 \sqrt{5}$ .
The laws of logarithms from topic 6: $\lg (x y) = \lg x + \lg y$ , $\lg x^{k} = k \lg x$ , and that $\lg y = c$ means $y = 10^{c}$ while $\ln y = c$ means $y = e^{c}$ .

Syllabus objectives covered (4)

Use the equation of a straight line.
Know and use the condition for two lines to be parallel or perpendicular.
Solve problems involving midpoint and length of a line, including finding and using the equation of a perpendicular bisector.
Transform given relationships to and from straight-line form, including determining unknown constants by calculating the gradient or intercept of the transformed graph.

Part 1 of 8

Gradient from two points

The gradient $m$ of a line measures how steep it is: how much $y$ changes for every $1$ unit you move to the right. Between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ it is the change in $y$ divided by the change in $x$ .
$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$
None of the straight-line formulas in this lesson is on the List of formulas, so you must recall them. Each one comes from a simple picture, which is how you rebuild it if you forget.
For $A (- 1, 4)$ and $B (3, - 2)$ : $m = \frac{- 2 - 4}{3 - (- 1)} = \frac{- 6}{4} = - \frac{3}{2}$ . Take the two points in the same order on the top and the bottom. If you mix the order, the sign flips.
Read the sign as a direction. A positive gradient slopes up from left to right and a negative one slopes down. A horizontal line has gradient $0$ , because $y$ never changes.
A vertical line has no gradient: $x_{2} - x_{1} = 0$ and you cannot divide by zero. Its equation is $x = k$ , for example $x = 3$ . A horizontal line has equation $y = k$ .

Part 2 of 8

The equation of a line

One point and the gradient fix a line. Take the known point $(x_{1}, y_{1})$ and any other point $(x, y)$ on the line. The gradient between them is $m$ , so $\frac{y - y_{1}}{x - x_{1}} = m$ . Multiply both sides by $(x - x_{1})$ :
$y - y_{1} = m (x - x_{1})$
Reach for this form first, because it uses exactly what you are given: a point and a gradient. For the line through $A (- 1, 4)$ with gradient $- \frac{3}{2}$ , notice that $x - (- 1)$ becomes $x + 1$ .
$y - 4 = - \frac{3}{2} (x + 1)$
If the question asks for the form $a x + b y = c$ with integers, clear the fraction by multiplying every term by $2$ : $2 y - 8 = - 3 x - 3$ . Collect the $x$ and $y$ terms on the left.
$3 x + 2 y = 5$
Check with the second point $B (3, - 2)$ : $3 (3) + 2 (- 2) = 9 - 4 = 5$ . It works. A point lies on a line exactly when its coordinates satisfy the equation, so this check takes seconds and catches a sign slip.
In the form $y = m x + c$ , the coefficient of $x$ is the gradient and $c$ is the $y$ -intercept: the line crosses the $y$ -axis at $(0, c)$ . This reading only works once $y$ is on its own with coefficient $1$ .
So to find the gradient of $3 x + 2 y = 5$ , rearrange first: $2 y = - 3 x + 5$ , so $y = - \frac{3}{2} x + \frac{5}{2}$ . The gradient is $- \frac{3}{2}$ , not $3$ , and the $y$ -intercept is $\frac{5}{2}$ .

Part 3 of 8

Where two lines meet

The point where two lines cross lies on both of them, so its coordinates satisfy both equations at once. Find it by solving the two equations simultaneously.
For $3 x + 2 y = 5$ and $y = 2 x - 1$ , substitute the second into the first, because it already gives $y$ on its own. Keep the bracket, since the whole of $2 x - 1$ is multiplied by $2$ .
$3 x + 2 (2 x - 1) = 5 \Rightarrow 7 x - 2 = 5 \Rightarrow x = 1$
Then $y = 2 (1) - 1 = 1$ , so the lines meet at $(1, 1)$ . Check in the other equation: $3 (1) + 2 (1) = 5$ . Give both coordinates; finding $x$ is only half the point.
Two lines with the same gradient and different intercepts are parallel and never meet. If you try to solve their equations simultaneously, the $x$ and $y$ terms cancel and leave a false statement such as $0 = 4$ : there is no solution.

Part 4 of 8

Parallel and perpendicular lines

Parallel lines have the same steepness, so the condition is equal gradients.
$m_{1} = m_{2}$
Perpendicular lines meet at right angles. The condition is that their gradients multiply to $- 1$ .
$m_{1} m_{2} = - 1$
Why: a line with gradient $m$ goes $1$ across and $m$ up. Turn that step through $90^{\circ}$ anticlockwise and it goes $m$ to the left and $1$ up, which is a gradient of $\frac{1}{- m} = - \frac{1}{m}$ . And $m \times (- \frac{1}{m}) = - 1$ .
So the perpendicular gradient is the negative reciprocal: turn the gradient upside down and change its sign. $\frac{2}{3}$ becomes $- \frac{3}{2}$ , $- 4$ becomes $\frac{1}{4}$ , and $1$ becomes $- 1$ . Both changes are needed.
One pair the rule cannot describe: a horizontal line and a vertical line are perpendicular, but a vertical line has no gradient to multiply. Recognise this pair by sight, for example $y = 2$ and $x = 5$ .
To test two given lines, rearrange each one to find its gradient. $2 x + 3 y = 6$ gives $y = - \frac{2}{3} x + 2$ , gradient $- \frac{2}{3}$ . $3 x - 2 y = 1$ gives $y = \frac{3}{2} x - \frac{1}{2}$ , gradient $\frac{3}{2}$ . The product is $- \frac{2}{3} \times \frac{3}{2} = - 1$ , so they are perpendicular.
To find the line through $(4, - 1)$ parallel to $2 x + 3 y = 6$ , use the same gradient $- \frac{2}{3}$ with the new point. Multiplying by $3$ gives $3 y + 3 = - 2 x + 8$ , so $2 x + 3 y = 5$ . Check: $2 (4) + 3 (- 1) = 5$ .
$y + 1 = - \frac{2}{3} (x - 4)$
Notice that the left-hand side $2 x + 3 y$ did not change. Parallel lines written as $a x + b y = c$ differ only in the constant $c$ , which is a quick way to check your answer.

Part 5 of 8

Midpoint and length of a line segment

The midpoint is halfway along in both directions, so it is the average of the $x$ -coordinates and the average of the $y$ -coordinates.
$M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
The length comes from Pythagoras. The change in $x$ and the change in $y$ are the two shorter sides of a right-angled triangle, and the segment is its hypotenuse.
$d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$
For $P (- 2, 3)$ and $Q (4, - 5)$ , the midpoint is $(\frac{- 2 + 4}{2}, \frac{3 + (- 5)}{2}) = (1, - 1)$ . The changes are $6$ across and $- 8$ up, so $P Q = \sqrt{6^{2} + {(- 8)}^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10$ .
graph diagram — not drawn yet
Square-scaled axes for −3 ≤ x ≤ 5 and −6 ≤ y ≤ 4. Mark P(−2, 3) and Q(4, −5) with labelled dots and join them with a solid segment. Draw a dashed horizontal leg from P to (4, 3) labelled '6' and a dashed vertical leg from (4, 3) down to Q labelled '8', with a right-angle mark at (4, 3). Label the segment PQ '10'. Mark the midpoint M(1, −1) on PQ with a labelled dot.
The changes in $x$ and $y$ are the shorter sides of a right-angled triangle, so $P Q = \sqrt{6^{2} + 8^{2}} = 10$ , and $M$ sits halfway in both directions.
Squaring removes the sign, so it does not matter which point you start from. But square the whole difference: ${(- 8)}^{2} = 64$ . Without the bracket, $- 8^{2}$ means $- (8^{2}) = - 64$ .
When the length is not a whole number, give it as a simplified surd on the non-calculator paper. From $R (1, 2)$ to $S (4, 8)$ : $\sqrt{3^{2} + 6^{2}} = \sqrt{45} = \sqrt{9 \times 5} = 3 \sqrt{5}$ .
Run the midpoint idea backwards to find a missing endpoint. If $M (3, 1)$ is the midpoint of $A B$ and $A$ is $(- 1, 4)$ , then $B$ is as far past $M$ as $A$ is before it. In $x$ : $- 1$ to $3$ is up $4$ , so carry on to $7$ . In $y$ : $4$ to $1$ is down $3$ , so carry on to $- 2$ . So $B = (7, - 2)$ .
Check with the formula: $(\frac{- 1 + 7}{2}, \frac{4 + (- 2)}{2}) = (3, 1)$ . In symbols, $B = (2 x_{M} - x_{A}, 2 y_{M} - y_{A})$ .

Part 6 of 8

The perpendicular bisector

The perpendicular bisector of $A B$ is the line that cuts $A B$ in half at right angles. Each word fixes one of the two things a line needs: 'bisector' means it passes through the midpoint of $A B$ , and 'perpendicular' fixes its gradient.
So the method always runs in the same order: the midpoint of $A B$ , the gradient of $A B$ , the negative reciprocal of that gradient, then $y - y_{1} = m (x - x_{1})$ through the midpoint.
For $A (- 3, 2)$ and $B (5, 6)$ : the midpoint is $M = (1, 4)$ . The gradient of $A B$ is $\frac{6 - 2}{5 - (- 3)} = \frac{4}{8} = \frac{1}{2}$ , so the perpendicular gradient is $- 2$ .
$y - 4 = - 2 (x - 1) \Rightarrow y = - 2 x + 6$
The trap is to get one of the two ingredients wrong. Using the gradient of $A B$ itself gives a line parallel to $A B$ . Using $A$ or $B$ as the point gives a line that is perpendicular but misses the middle. The line must use the perpendicular gradient and pass through $M$ .
What makes this line useful: every point on it is the same distance from $A$ as from $B$ . Take $(3, 0)$ , which is on $y = - 2 x + 6$ because $- 2 (3) + 6 = 0$ . Its distance to $A$ is $\sqrt{6^{2} + 2^{2}} = \sqrt{40}$ , and its distance to $B$ is $\sqrt{2^{2} + 6^{2}} = \sqrt{40}$ .
graph diagram — not drawn yet
Square-scaled axes for −4 ≤ x ≤ 6 and −2 ≤ y ≤ 8. Mark A(−3, 2) and B(5, 6) with labelled dots and join them with a solid segment. Mark the midpoint M(1, 4) with a labelled dot and a right-angle mark where the line y = −2x + 6 crosses AB at M. Draw y = −2x + 6 across the grid, from about (−1, 8) to (4, −2), labelled with its equation. Mark the point (3, 0) on it and join it to A and to B with dashed segments, each labelled '√40'.
The perpendicular bisector passes through $M$ at right angles to $A B$ , and any point on it, such as $(3, 0)$ , is the same distance from $A$ as from $B$ .
So a question asking for a point that is equidistant from $A$ and $B$ and also lies on some other line is asking where the perpendicular bisector meets that other line. Here, the point on the $x$ -axis that is equidistant from $A$ and $B$ is $(3, 0)$ .
The centre of any circle through $A$ and $B$ lies on their perpendicular bisector too, because the centre is the same distance (the radius) from every point on the circle.

Part 7 of 8

Straight-line form: Y = mX + c

Some relationships between $x$ and $y$ draw a curve, but give a straight line if you plot the right expressions instead of $x$ and $y$ . The aim is to write the relationship as $Y = m X + c$ , where $X$ and $Y$ are expressions in the variables only, and the unknown constants appear only in $m$ and $c$ .
$Y = m X + c$
Take $y^{2} = A x^{3} + B$ , with constants $A$ and $B$ . Match it to $Y = m X + c$ : put $Y = y^{2}$ and $X = x^{3}$ . Plotting $y^{2}$ against $x^{3}$ gives a straight line with gradient $A$ and intercept $B$ on the vertical axis.
'Plot $Y$ against $X$ ' means $Y$ goes up the vertical axis and $X$ along the horizontal axis. Swapping them turns the gradient upside down.
Now go the other way, from a line to the relationship. Suppose $y^{3}$ plotted against $\ln x$ is a straight line through $(0.5, 4)$ and $(2, 1)$ . These are $(X, Y)$ values: $\ln x$ and $y^{3}$ , not $x$ and $y$ . The gradient is $\frac{1 - 4}{2 - 0.5} = \frac{- 3}{1.5} = - 2$ .
Find the intercept by putting one point into $Y = m X + c$ : $4 = - 2 (0.5) + c$ , so $c = 5$ . Do this by substitution rather than by reading a graph. If the horizontal axis does not start at $0$ , the place where the line meets the left edge of the graph is not the intercept.
Replace $Y$ and $X$ by what they stand for. The relationship is $y^{3} = - 2 \ln x + 5$ , so $A = - 2$ and $B = 5$ in $y^{3} = A \ln x + B$ . Check with the other point: $- 2 (2) + 5 = 1$ . If you are asked for $y$ in terms of $x$ , undo the cube:
$y = \sqrt[3]{5 - 2 \ln x}$
The same steps handle $e^{2 y} = A x^{2} + B$ : plot $e^{2 y}$ against $x^{2}$ , and the gradient is $A$ and the intercept is $B$ . To make $y$ the subject at the end, take $\ln$ of both sides: $2 y = \ln (A x^{2} + B)$ , so $y = \frac{1}{2} \ln (A x^{2} + B)$ .

Part 8 of 8

Using logarithms: y = Axⁿ and y = Abˣ

When the unknown constants are multiplied together or sit in a power, no choice of $X$ and $Y$ separates them directly. Taking logs turns multiplication into addition and brings powers down, which produces exactly the shape $Y = m X + c$ .
For $y = A x^{n}$ , take $\lg$ of both sides. The product law splits $A x^{n}$ into $\lg A + \lg x^{n}$ , and the power law brings $n$ down. So plot $\lg y$ against $\lg x$ : the gradient is $n$ and the intercept is $\lg A$ .
$\lg y = n \lg x + \lg A$
For $y = A b^{x}$ , the same two laws give a different shape. Here $x$ was the power, so it comes down as $x$ itself. Plot $\lg y$ against $x$ : the gradient is $\lg b$ and the intercept is $\lg A$ .
$\lg y = x \lg b + \lg A$
To tell them apart, ask where $x$ is. If $x$ is the base, as in $x^{n}$ , plot against $\lg x$ . If $x$ is the power, as in $b^{x}$ , plot against $x$ .
Here is why it works on data. The values $x = 1, 2, 3, 4$ with $y = 6, 18, 54, 162$ fit $y = 2 \times 3^{x}$ . To 3 decimal places, $\lg y$ is $0.778, 1.255, 1.732, 2.210$ . Each step of $1$ in $x$ raises $\lg y$ by about $0.477$ , so the points lie on a straight line.
graph diagram — not drawn yet
Axes with x from 0 to 4.5 horizontally and lg y from 0 to 2.4 vertically, the vertical axis labelled 'lg y'. Plot the points (1, 0.778), (2, 1.255), (3, 1.732) and (4, 2.210) as crosses and draw a ruled straight line through them, extended back to meet the vertical axis at (0, 0.301), which is labelled. Draw a gradient triangle between x = 1 and x = 3 with the horizontal side labelled '2' and the vertical side labelled '0.954'.
Plotting $\lg y$ against $x$ for $y = 2 \times 3^{x}$ gives a straight line with intercept $\lg 2 \approx 0.301$ and gradient $\lg 3 \approx 0.477$ .
Read the constants off the line, then undo the logs. The gradient is $\lg b = 0.477$ , so $b = 10^{0.477} = 3.00$ to 3 significant figures. The intercept is $\lg A = 0.301$ , so $A = 10^{0.301} = 2.00$ . Those match $y = 2 \times 3^{x}$ .
The trap is to stop one step early. The intercept is $\lg A$ , not $A$ , and for $y = A b^{x}$ the gradient is $\lg b$ , not $b$ . Only $n$ in $y = A x^{n}$ comes straight off the graph, because the power law has already brought it down in front of $\lg x$ as a plain multiplier.
$\ln$ works the same way, with $e$ in place of $10$ when you undo it: an intercept of $2$ on a graph of $\ln y$ means $\ln A = 2$ , so $A = e^{2}$ . Use whichever log the question names.
On the calculator paper, keep full accuracy until the end. If you round $A$ and $b$ and then use the rounded values to find $x$ or $y$ , the answer can drift in the third significant figure. Where you can, substitute into the straight-line equation itself, such as $\lg y = x \lg b + \lg A$ with the unrounded gradient and intercept.

Worked examples

Example 1. The points $A (- 2, 1)$ and $B (4, 4)$ lie on the line $l_{1}$ . The line $l_{2}$ passes through $C (5, - 3)$ and is perpendicular to $l_{1}$ . (a) Find the equation of $l_{2}$ in the form $a x + b y = c$ . (b) Find the coordinates of the point $F$ where $l_{1}$ and $l_{2}$ meet. (c) Hence find the exact shortest distance from $C$ to $l_{1}$ .

Show the working

Gradient of $l_{1}$ , from $A$ and $B$ , with the points in the same order top and bottom.
$m_{1} = \frac{4 - 1}{4 - (- 2)} = \frac{3}{6} = \frac{1}{2}$
$l_{2}$ is perpendicular, so its gradient is the negative reciprocal, $- 2$ . Check: $\frac{1}{2} \times (- 2) = - 1$ .
(a) Use $y - y_{1} = m (x - x_{1})$ with $C (5, - 3)$ : $y + 3 = - 2 (x - 5)$ , so $y + 3 = - 2 x + 10$ . Collect the terms.
$2 x + y = 7$
(b) To find where the lines meet you need the equation of $l_{1}$ as well. Through $A (- 2, 1)$ with gradient $\frac{1}{2}$ : $y - 1 = \frac{1}{2} (x + 2)$ . Multiply by $2$ : $2 y - 2 = x + 2$ .
$x - 2 y = - 4$
Solve simultaneously. From $2 x + y = 7$ , $y = 7 - 2 x$ . Substitute: $x - 2 (7 - 2 x) = - 4$ , so $5 x - 14 = - 4$ and $x = 2$ . Then $y = 7 - 4 = 3$ . Check on $l_{1}$ : $2 - 2 (3) = - 4$ .
(c) The shortest distance from a point to a line is along the perpendicular, which is exactly $C F$ . Use the length formula with $C (5, - 3)$ and $F (2, 3)$ .
$C F = \sqrt{{(5 - 2)}^{2} + {(- 3 - 3)}^{2}} = \sqrt{9 + 36} = \sqrt{45} = 3 \sqrt{5}$

Answer: (a) $2 x + y = 7$ (b) $F (2, 3)$ (c) $3 \sqrt{5}$

Example 2. The points $A (- 1, 6)$ and $B (5, 2)$ lie on a circle whose centre $P$ lies on the line $y = x$ . Find the equation of the perpendicular bisector of $A B$ , the coordinates of $P$ , and the exact radius of the circle.

Show the working

Midpoint of $A B$ : average the coordinates.
$M = (\frac{- 1 + 5}{2}, \frac{6 + 2}{2}) = (2, 4)$
Gradient of $A B$ : $\frac{2 - 6}{5 - (- 1)} = \frac{- 4}{6} = - \frac{2}{3}$ . The perpendicular gradient is its negative reciprocal, $\frac{3}{2}$ .
Line through the midpoint $M (2, 4)$ with gradient $\frac{3}{2}$ : $y - 4 = \frac{3}{2} (x - 2)$ . Multiply by $2$ : $2 y - 8 = 3 x - 6$ .
$2 y = 3 x + 2$
The centre is the same distance from $A$ as from $B$ , so it lies on the perpendicular bisector. It is also on $y = x$ , so replace $y$ by $x$ : $2 x = 3 x + 2$ , which gives $x = - 2$ . So $P = (- 2, - 2)$ .
The radius is the distance from $P$ to either point.
$P A = \sqrt{{(- 1 - (- 2))}^{2} + {(6 - (- 2))}^{2}} = \sqrt{1 + 64} = \sqrt{65}$
Check with $B$ : $P B = \sqrt{{(5 + 2)}^{2} + {(2 + 2)}^{2}} = \sqrt{49 + 16} = \sqrt{65}$ . The two distances agree, as they must.

Answer: Perpendicular bisector $2 y = 3 x + 2$ (or $3 x - 2 y = - 2$ ); $P (- 2, - 2)$ ; radius $\sqrt{65}$ .

Example 3. The variables $x$ and $y$ are related by $y = A x^{n}$ , where $A$ and $n$ are constants. When $\ln y$ is plotted against $\ln x$ , the result is a straight line passing through the points $(1, 5)$ and $(3, 11)$ . Find the value of $n$ and the exact value of $A$ .

Show the working

First turn the relationship into straight-line form, so you know what the gradient and intercept stand for. Take $\ln$ of both sides and use the product and power laws.
$\ln y = n \ln x + \ln A$
Match with $Y = m X + c$ : $Y = \ln y$ , $X = \ln x$ , the gradient is $n$ and the intercept is $\ln A$ . The points $(1, 5)$ and $(3, 11)$ are $(\ln x, \ln y)$ values.
Gradient from the two points.
$n = \frac{11 - 5}{3 - 1} = 3$
Intercept: substitute $(1, 5)$ into $\ln y = 3 \ln x + \ln A$ , so $5 = 3 + \ln A$ and $\ln A = 2$ .
Undo the log: $\ln A = 2$ means $A = e^{2}$ . Leave it exact; the question asks for it.
Check with the other point. $\ln x = 3$ means $x = e^{3}$ , and then $y = e^{2} {(e^{3})}^{3} = e^{11}$ , so $\ln y = 11$ .

Answer: $n = 3$ and $A = e^{2}$ , so $y = e^{2} x^{3}$ .

Example 4. The variables $x$ and $y$ , where $y > 0$ , are such that when $y^{2}$ is plotted against $x^{3}$ a straight line is obtained. The line passes through $(1, 7)$ and $(3, 13)$ . (a) Find $y^{2}$ in terms of $x$ . (b) Find the exact value of $y$ when $x = 2$ .

Show the working

The axes are $Y = y^{2}$ and $X = x^{3}$ , so the line is $y^{2} = m x^{3} + c$ . The two points are $(x^{3}, y^{2})$ values.
Gradient from the two points.
$m = \frac{13 - 7}{3 - 1} = 3$
Intercept: substitute $(1, 7)$ into $Y = 3 X + c$ , so $7 = 3 + c$ and $c = 4$ .
(a) Replace $Y$ and $X$ by $y^{2}$ and $x^{3}$ . Check with $(3, 13)$ : $3 (3) + 4 = 13$ .
$y^{2} = 3 x^{3} + 4$
(b) Put $x = 2$ : $y^{2} = 3 (8) + 4 = 28$ . Since $y > 0$ , take the positive root and simplify the surd.
$y = \sqrt{28} = \sqrt{4 \times 7} = 2 \sqrt{7}$

Answer: (a) $y^{2} = 3 x^{3} + 4$ (b) $y = 2 \sqrt{7}$

Where people go wrong

Mistake: Taking the perpendicular gradient to $\frac{2}{3}$ as $- \frac{2}{3}$ or as $\frac{3}{2}$ .

Instead: It is $- \frac{3}{2}$ : turn the fraction upside down and change the sign.

The condition is $m_{1} m_{2} = - 1$ . $\frac{2}{3} \times (- \frac{2}{3}) = - \frac{4}{9}$ and $\frac{2}{3} \times \frac{3}{2} = 1$ ; only $\frac{2}{3} \times (- \frac{3}{2}) = - 1$ .

Mistake: Reading the gradient of $3 x + 2 y = 5$ as $3$ .

Instead: Rearrange to $y = - \frac{3}{2} x + \frac{5}{2}$ . The gradient is $- \frac{3}{2}$ .

The coefficient of $x$ is the gradient only when $y$ is the subject with coefficient $1$ .

Mistake: Drawing the perpendicular bisector of $A B$ through $A$ , or giving it the gradient of $A B$ .

Instead: Use the midpoint of $A B$ as the point and the negative reciprocal of the gradient of $A B$ as the gradient.

'Bisector' means it cuts $A B$ in half, so it passes through the middle. 'Perpendicular' means the gradients multiply to $- 1$ . Each word fixes one of the two things a line needs.

Mistake: Giving the intercept of a graph of $\lg y$ against $\lg x$ as the value of $A$ .

Instead: The intercept is $\lg A$ , so $A = 10^{intercept}$ , or $A = e^{intercept}$ for a graph of $\ln y$ .

The vertical axis shows $\lg y$ , so every value read from it is a log. The same is true of the gradient of $\lg y$ against $x$ , which is $\lg b$ , not $b$ .

Mistake: For $y = A b^{x}$ , plotting $\lg y$ against $\lg x$ .

Instead: Plot $\lg y$ against $x$ .

$\lg (A b^{x}) = \lg A + x \lg b$ . The $x$ comes down from the power as $x$ itself, not as $\lg x$ . Plotting against $\lg x$ is right for $y = A x^{n}$ , where $x$ is the base.

Check yourself

Find the gradient of the line through $(- 3, 5)$ and $(1, - 7)$ , and its equation in the form $y = m x + c$ .
Answer
$m = \frac{- 7 - 5}{1 - (- 3)} = \frac{- 12}{4} = - 3$ . Then $y - 5 = - 3 (x + 3)$ , so $y = - 3 x - 4$ . Check: $- 3 (1) - 4 = - 7$ .
Find the gradient and the $y$ -intercept of the line $4 x - 5 y = 20$ .
Answer
Rearrange: $5 y = 4 x - 20$ , so $y = \frac{4}{5} x - 4$ . The gradient is $\frac{4}{5}$ and the $y$ -intercept is $- 4$ , at $(0, - 4)$ .
Find the equation of the line through $(2, - 1)$ that is perpendicular to $y = 4 x - 3$ . Give your answer in the form $a x + b y = c$ .
Answer
Perpendicular gradient $- \frac{1}{4}$ . $y + 1 = - \frac{1}{4} (x - 2)$ ; multiply by $4$ : $4 y + 4 = - x + 2$ , so $x + 4 y = - 2$ . Check: $2 + 4 (- 1) = - 2$ .
Determine whether the lines $3 x - 6 y = 7$ and $2 x + y = 1$ are parallel, perpendicular or neither.
Answer
$3 x - 6 y = 7$ gives $y = \frac{1}{2} x - \frac{7}{6}$ , gradient $\frac{1}{2}$ . $2 x + y = 1$ gives $y = - 2 x + 1$ , gradient $- 2$ . The product is $\frac{1}{2} \times (- 2) = - 1$ , so they are perpendicular.
For $P (- 3, 4)$ and $Q (5, - 2)$ , find (a) the length $P Q$ (b) the equation of the perpendicular bisector of $P Q$ in the form $a x + b y = c$ .
Answer
(a) $\sqrt{8^{2} + {(- 6)}^{2}} = \sqrt{100} = 10$ . (b) Midpoint $(1, 1)$ . Gradient of $P Q = \frac{- 6}{8} = - \frac{3}{4}$ , so the perpendicular gradient is $\frac{4}{3}$ . $y - 1 = \frac{4}{3} (x - 1)$ gives $3 y - 3 = 4 x - 4$ , so $4 x - 3 y = 1$ .
$M (2, - 1)$ is the midpoint of $A B$ , and $A$ is $(- 4, 3)$ . Find the coordinates of $B$ and the exact length of $A B$ .
Answer
$B = (2 \times 2 - (- 4), 2 \times (- 1) - 3) = (8, - 5)$ . $A B = \sqrt{12^{2} + {(- 8)}^{2}} = \sqrt{208} = \sqrt{16 \times 13} = 4 \sqrt{13}$ .
The variables $x$ and $y$ are related by $y = A b^{x}$ . (a) State what you would plot to get a straight line, and what its gradient and intercept stand for. (b) That straight line has gradient $0.6$ and intercept $1.2$ . Find $A$ and $b$ to 3 significant figures.
Answer
(a) Plot $\lg y$ against $x$ , since $\lg y = x \lg b + \lg A$ ; the gradient is $\lg b$ and the intercept is $\lg A$ . (b) $b = 10^{0.6} = 3.98$ and $A = 10^{1.2} = 15.8$ .
When $e^{2 y}$ is plotted against $x^{2}$ , a straight line is obtained through $(1, 5)$ and $(3, 9)$ . Find $e^{2 y}$ in terms of $x$ , and hence write $y$ in terms of $x$ .
Answer
Gradient $\frac{9 - 5}{3 - 1} = 2$ ; intercept from $5 = 2 (1) + c$ , so $c = 3$ . So $e^{2 y} = 2 x^{2} + 3$ , and taking $\ln$ : $y = \frac{1}{2} \ln (2 x^{2} + 3)$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. The points $A$ and $B$ have coordinates $(1, 1)$ and $(5, 3)$ .

(a) Find the equation of the perpendicular bisector of $A B$ , giving your answer in the form $y = m x + c$ . 4 marks No calculator

Mark scheme

B1 midpoint $(3, 2)$
B1 gradient of $A B = \frac{1}{2}$ soi
M1 line through their midpoint with gradient $- 1 \div$ their gradient of $A B$
A1 $y = - 2 x + 8$ cao

(b) The perpendicular bisector meets the $y$ -axis at the point $C$ . Show that the area of triangle $A B C$ is $15$ . 3 marks No calculator

Mark scheme

B1 $C = (0, 8)$
M1 $A B = \sqrt{20}$ and the distance from $C$ to the midpoint $= \sqrt{45}$ oe, or another complete method such as an enclosing rectangle or $\frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$
A1 $15$ reached with no errors from a complete method, e.g. $\frac{1}{2} \times 2 \sqrt{5} \times 3 \sqrt{5} = 15$ with the distance from $C$ to the midpoint as the perpendicular height, or $35 - 3.5 - 4 - 12.5 = 15$ , or $\frac{1}{2} | - 5 + 35 + 0 | = 15$

Question 2. The points $A$ and $B$ have coordinates $(1, 2)$ and $(7, 5)$ . The line through $B$ that is perpendicular to $A B$ meets the line $y = 1$ at the point $C$ .

(a) Find the coordinates of $C$ . 3 marks No calculator

Mark scheme

M1 perpendicular gradient $- 2$ from gradient of $A B = \frac{1}{2}$
M1 equation of $B C$ with their perpendicular gradient through $B$ , e.g. $y - 5 = - 2 (x - 7)$ , and $y = 1$ substituted
A1 $(9, 1)$

(b) $A B C D$ is a rectangle. Find the coordinates of $D$ . 2 marks No calculator

Mark scheme

M1 complete method, e.g. midpoint of $A C$ equated to midpoint of $B D$ , or the step from $B$ to $C$ applied to $A$
A1 $(3, - 2)$

Question 3. The variables $x$ and $y$ are related by $y = A b^{x}$ , where $A$ and $b$ are constants. When $\lg y$ is plotted against $x$ , a straight line is obtained that passes through the points $(1, 1.18)$ and $(5, 2.38)$ .

(a) Show that $\lg y = x \lg b + \lg A$ . 2 marks

Mark scheme

M1 takes $\lg$ of both sides and uses the product law: $\lg y = \lg A + \lg b^{x}$
A1 uses the power law, $\lg b^{x} = x \lg b$ , to reach the given result with no errors

(b) Find the value of $A$ and the value of $b$ . 4 marks

Mark scheme

M1 gradient $= \frac{2.38 - 1.18}{5 - 1}$ $[= 0.3]$ and equated to $\lg b$
M1 intercept found from a point, e.g. $1.18 = 0.3 \times 1 + c$ $[c = 0.88]$ , and equated to $\lg A$
A1 $b = 2.00$ awrt
A1 $A = 7.59$ awrt

(c) Calculate the value of $x$ when $y = 100$ . 2 marks

Mark scheme

M1 $\lg 100 = 2$ substituted into their $\lg y = 0.3 x + 0.88$ , or into $y = A b^{x}$ with their unrounded values
A1 $x = 3.73$ awrt

How this is examined

Turning $y = A x^{n}$ or $y = A b^{x}$ into a straight line with logs, then finding the constants from that line: a multi-part question: Show that the log form is linear, Find the constants from two given points on the line (or from a line you Plot from a table), then use the relationship to find $x$ or $y$ Logs taken of both sides with the product and power laws written out, e.g. $\lg y = x \lg b + \lg A$ ; the gradient worked out as $\frac{Δ Y}{Δ X}$ from the transformed values; the intercept found by substituting a point; then each log undone, $b = 10^{gradient}$ and $A = 10^{intercept}$ , to 3 significant figures on the calculator paper.
Finding the equation of a perpendicular bisector: a Find worth about four marks, followed by parts that use the line, such as where it meets an axis or another line The midpoint correct, the gradient of $A B$ correct, its negative reciprocal used with the midpoint in $y - y_{1} = m (x - x_{1})$ , and the equation given in the form the question asks for.
A line through a given point, parallel or perpendicular to a given line, and where it meets another line: a Find for an equation followed by a Find for a point of intersection, sometimes inside a triangle, rectangle or other shape The given line rearranged to read its gradient, the parallel gradient $m$ or perpendicular gradient $- \frac{1}{m}$ stated, $y - y_{1} = m (x - x_{1})$ with the given point, then the simultaneous equations solved to give both coordinates.
Going back from a straight line of transformed variables to the original relationship: a Find that gives a straight line such as $y^{2}$ against $x^{3}$ through two points, and asks for $y$ in terms of $x$ or for $y$ at a given $x$ The gradient and intercept found from the transformed values, not from $x$ and $y$ ; the relationship written with numbers in, e.g. $y^{2} = 3 x^{3} + 4$ ; then rearranged for $y$ if the question asks.
Exact lengths, and areas built from them: a Find or Show that for a length, or for the area of a triangle once its perpendicular height is known, on the non-calculator paper Pythagoras set up with both differences squared, the surd simplified (e.g. $\sqrt{45} = 3 \sqrt{5}$ ), and for a Show that, every step to the given value written down.

Words to know

Gradient: The change in $y$ divided by the change in $x$ between two points on a line: $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .
Intercept: The value where a line crosses the vertical axis; for $y = m x + c$ it is $c$ , at $(0, c)$ .
Negative reciprocal: The gradient of a perpendicular line: $- \frac{1}{m}$ , made by turning $m$ upside down and changing its sign.
Midpoint: The point halfway between two points: $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .
Perpendicular bisector: The line through the midpoint of $A B$ at right angles to $A B$ . Every point on it is the same distance from $A$ as from $B$ .
Equidistant: The same distance from two or more points.
Straight-line form: A relationship rewritten as $Y = m X + c$ , where $X$ and $Y$ are expressions in the variables, so that plotting $Y$ against $X$ gives a straight line.
lg: Shorthand for $\log_{10}$ .

All Add Maths lessons Next: 8 Coordinate geometry of the circle