3.1 Straight lines, gradients and straight-line models — Practise

Work through these on your own. Before you calculate, decide what the question gives you: a gradient, a point, or two points. In a model, check what the gradient and the intercept mean in the situation before you quote a number.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Work out the gradient of the line through $(2, 5)$ and $(7, 11)$ .
Work out the gradient of the line through $(-3, 4)$ and $(5, -2)$ .
Find the equation of the line with gradient $4$ through $(3, -2)$ , in the form $y = mx + c$ .
Find the equation of the line through $(2, 5)$ and $(7, 11)$ . Give the gradient and the intercept.
Write $y = \dfrac{2}{3}x - 4$ in the form $ax + by + c = 0$ , where $a$ , $b$ and $c$ are whole numbers.
Work out where the line $2x + 3y = 12$ crosses each axis.
Work out the gradient of the line $6x - 5y + 13 = 0$ .
Decide whether the point $(5, 2)$ lies on the line $y = 3x - 13$ .
A taxi charges a fixed £3 plus £2 for each kilometre. Write a model for the cost of a journey of $d$ kilometres, and work out the cost of a journey of $8$ km.
A graph shows cost $C$ in pounds against the number $n$ of items. The line passes through $(2, 7)$ and $(6, 31)$ . Work out the gradient and say what it means.
The points $(1, 3)$ , $(2, 5)$ and $(3, 7)$ lie on a straight line. Find the equation of the line.
The points $(4, -1)$ and $(4, 7)$ lie on a vertical line. Explain why the gradient is undefined, and write the equation of the line.

Answers and concise solutions

1. The gradient

Answer: Subtract the first point from the second on the top and on the bottom: $m = \dfrac{11 - 5}{7 - 2} = \dfrac{6}{5}$ . The line rises $6$ units for every $5$ units across.

2. The gradient

Answer: $m = \dfrac{-2 - 4}{5 - (-3)} = \dfrac{-6}{8} = -\dfrac{3}{4}$ . The gradient is negative, so the line falls as it moves to the right.

3. The equation

Answer: Use $y - y_1 = m(x - x_1)$ with $m = 4$ and $(3, -2)$ : $y + 2 = 4(x - 3) = 4x - 12$ , so $y = 4x - 14$ . Check the point: $4 \times 3 - 14 = -2$ , which is the given $y$ -value.

4. The gradient and the intercept

Answer: The gradient is $\dfrac{11 - 5}{7 - 2} = \dfrac{6}{5}$ . Substitute the point $(2, 5)$ into $y - 5 = \frac{6}{5}(x - 2)$ : $y = \frac{6}{5}x - \frac{12}{5} + 5 = \frac{6}{5}x + \frac{13}{5}$ . The intercept is $\frac{13}{5}$ . Multiplying by $5$ gives the equivalent form $6x - 5y + 13 = 0$ .

5. The equation in general form

Answer: Multiply every term by $3$ : $3y = 2x - 12$ . Collect the terms on one side: $2x - 3y - 12 = 0$ . The line passes through $(0, -4)$ , from the original equation, and through $(6, 0)$ , where $y = 0$ ; both points satisfy the new form, so the two equations describe the same line.

6. The two intercepts

Answer: Put $y = 0$ : $2x = 12$ , so the line crosses the $x$ -axis at $(6, 0)$ . Put $x = 0$ : $3y = 12$ , so it crosses the $y$ -axis at $(0, 4)$ .

7. The gradient

Answer: Rearrange to make $y$ the subject: $5y = 6x + 13$ , so $y = \frac{6}{5}x + \frac{13}{5}$ . The gradient is the coefficient of $x$ , which is $\frac{6}{5}$ . Rearranging always beats memorising a formula: the coefficient of $x$ is the gradient only after $y$ stands alone.

8. The decision

Answer: Substitute $x = 5$ : $3 \times 5 - 13 = 2$ . That is the given $y$ -value, so $(5, 2)$ lies on the line. The zero in the check is the difference between the line's value and the point's $y$ -coordinate.

9. The model and the cost

Answer: The model is $C = 3 + 2d$ pounds. The intercept $3$ is the fixed charge, paid before the taxi moves. The gradient $2$ is the rate, in pounds per kilometre. For $d = 8$ : $C = 3 + 16 = 19$ , so the fare is £19.

10. The gradient and its meaning

Answer: The gradient is $\dfrac{31 - 7}{6 - 2} = \dfrac{24}{4} = 6$ . The vertical quantity is cost in pounds and the horizontal quantity is a number of items, so the gradient means £6 per item: each extra item adds £6 to the cost.

11. The equation

Answer: Take the first two points: the gradient is $\dfrac{5 - 3}{2 - 1} = 2$ . Substitute $(1, 3)$ into $y - 3 = 2(x - 1)$ : $y = 2x + 1$ . Check the third point: $2 \times 3 + 1 = 7$ , so all three points fit the same line.

12. The reason and the equation

Answer: The gradient is $\dfrac{7 - (-1)}{4 - 4} = \dfrac{8}{0}$ . The change in $x$ is $0$ , and division by zero is undefined, so the line has no gradient. Every point on the line has $x = 4$ , so its equation is $x = 4$ . The general form $ax + by + c = 0$ still works here, with $b = 0$ .