Gradient and Intercepts of ax + by = c

Write $ax + by = c$ in the form $y = mx + c$ to find its gradient and $y$ -intercept; find an unknown number in the equation; match equations to lines; find the triangle a line cuts off with the axes; and read a real setting.

Before you start

In lesson 2.2 you drew the graph of $ax + by = c$ , a straight line, from its intercepts: put $y = 0$ for the $x$ -intercept and $x = 0$ for the $y$ -intercept.
In Secondary 1 a straight line was written $y = ax + b$ , with gradient $a$ ( $\frac{\text{vertical change}}{\text{horizontal change}}\text{).}$ In this chapter $a$ and $b$ are already the numbers in $ax + by = c$ , so that form is written $y = mx + c$ : the gradient is $m$ and the line crosses the $y$ -axis at $(0, c)$ . This $c$ is a different number from the $c$ in $ax + by = c$ .

The Gradient and the y-Intercept

From ax + by = c to y = mx + c

To find the gradient and the $y$ -intercept of a line $ax + by = c$ (with $b \ne 0$ ), make $y$ the subject and write the result in the form $y = mx + c$ .

Then the gradient is $m$ , the number multiplying $x$ , and the $y$ -intercept is the number on its own.

The $y$ -intercept is not the number on the right of $ax + by = c$ : in $2x + 3y = 12$ it is $4$ , not $12$ .

Example 1: Rewriting in the form y = mx + c

Find the gradient and the $y$ -intercept of (a) $2x + 3y = 12$ , (b) $5x - 2y = 6$ .

Solution

(a) Subtract $2x$ : $3y = -2x + 12$ . Divide every term by $3$ : $y = -\frac{2}{3}x + 4$ . So $m = -\frac{2}{3}$ , and the $y$ -intercept is $4$ .

Check on the graph drawn in lesson 2.2, Example 2: from $(0, 4)$ to $(6, 0)$ the line goes $6$ across and $4$ down, so its gradient is $\frac{-4}{6} = -\frac{2}{3}$ .

(b) Subtract $5x$ : $-2y = -5x + 6$ . Divide every term by $-2$ : $y = \frac{5}{2}x - 3$ . So $m = \frac{5}{2}$ , and the $y$ -intercept is $-3$ .

Check with two points: $x = 0$ gives $-2y = 6$ , $y = -3$ ; $x = 2$ gives $10 - 2y = 6$ , $y = 2$ . From $(0, -3)$ to $(2, 2)$ : $5$ up for $2$ across, gradient $\frac{5}{2}$ .

(a) Gradient $-\frac{2}{3}$ , $y$ -intercept $4$ . (b) Gradient $\frac{5}{2}$ , $y$ -intercept $-3$ .

An unknown number in the equation can be found from one more fact about the line: a point on it (substitute its coordinates) or its gradient (write the equation in the form $y = mx + c$ and compare).

Example 2: An unknown coefficient

(a) The line $ax + 2y = 8$ passes through the point $(2, 1)$ . Find $a$ , the gradient of the line, and its $x$ -intercept. (b) The line $4x + by = 6$ has gradient $2$ . Find $b$ .

Solution

(a) The point is on the line, so its coordinates make the equation true: $a(2) + 2(1) = 8$ , so $2a + 2 = 8$ , $2a = 6$ and $a = 3$ .

The line is $3x + 2y = 8$ . Make $y$ the subject: $2y = -3x + 8$ , so $y = -\frac{3}{2}x + 4$ . The gradient is $-\frac{3}{2}$ .

$x$ -intercept: put $y = 0$ . $3x = 8$ , so $x = \frac{8}{3} = 2\frac{2}{3}$ .

Check $(2, 1)$ : $3(2) + 2(1) = 6 + 2 = 8\text{.}$

(b) Make $y$ the subject: $by = -4x + 6$ , so $y = -\frac{4}{b}x + \frac{6}{b}$ . The gradient is $-\frac{4}{b}$ , so $-\frac{4}{b} = 2$ .

Multiply both sides by $b$ : $-4 = 2b$ , so $b = -2$ .

Check: the line is $4x - 2y = 6$ , so $-2y = -4x + 6$ and $y = 2x - 3$ , whose gradient is $2$ .

(a) $a = 3$ ; the gradient is $-\frac{3}{2}$ and the $x$ -intercept is $2\frac{2}{3}$ . (b) $b = -2$ .

Matching Equations to Graphs

To tell which line belongs to which equation, find each equation's intercepts and the sign of its gradient (up from left to right: positive; down: negative), and compare them with the drawing.

Example 3: Which line is which?

The lines $A$ , $B$ and $C$ are the graphs of $x + 2y = 4$ , $2x - y = 4$ and $2x + y = -2$ . Match each equation to its line.

Solution

$x + 2y = 4$ : intercepts $4$ and $2$ ; $y = -\frac{1}{2}x + 2$ , gradient negative.

$2x - y = 4$ : intercepts $2$ and $-4$ ; $y = 2x - 4$ , gradient positive.

$2x + y = -2$ : intercepts $-1$ and $-2$ ; $y = -2x - 2$ , gradient negative.

Only $A$ goes up from left to right, so $A$ is $2x - y = 4$ ; it crosses the $x$ -axis at $2$ .

$B$ crosses the $y$ -axis at $2$ and the $x$ -axis at $4$ : $x + 2y = 4$ .

$C$ crosses the $x$ -axis at $-1$ and the $y$ -axis at $-2$ : $2x + y = -2$ .

$A$ is $2x - y = 4$ , $B$ is $x + 2y = 4$ and $C$ is $2x + y = -2$ .

The Triangle Cut Off by a Line

A line that crosses both axes away from the origin cuts off a right-angled triangle with the axes. Its two shorter sides lie along the axes, and their lengths are the sizes of the two intercepts.

Example 4: Area of the triangle

The line $3x + 4y = 24$ meets the $x$ -axis at $A$ and the $y$ -axis at $B$ . Find the coordinates of $A$ and $B$ , and the area of triangle $OAB$ , where $O$ is the origin.

Solution

$A$ : put $y = 0$ . $3x = 24$ , so $x = 8$ : $A(8, 0)$ .

$B$ : put $x = 0$ . $4y = 24$ , so $y = 6$ : $B(0, 6)$ .

The triangle has a right angle at $O$ . Its base $OA$ is $8$ units and its height $OB$ is $6$ units.

Area $= \frac{1}{2} \times 8 \times 6 = 24$ square units.

$A(8, 0)$ , $B(0, 6)$ , and the area of triangle $OAB$ is $24$ square units.

A Graph from a Real Setting

In a real setting the intercepts and the gradient have meanings. An intercept is the amount of one thing when there is none of the other. The gradient says how much one amount changes as the other goes up.

Example 5: Apples and pears

Mei Ling spends exactly $ $24$ on $x$ apples at $ $2$ each and $y$ pears at $ $3$ each. (a) Write an equation connecting $x$ and $y$ . (b) Draw its graph. (c) Use the graph to list the numbers of apples and pears she can buy, with at least one of each. (d) Write the equation in the form $y = mx + c$ . What do its gradient and its $y$ -intercept tell you?

Solution

(a) The apples cost $ $2x$ and the pears $ $3y$ : $2x + 3y = 24$ .

(b) Intercepts: $y = 0$ gives $x = 12$ ; $x = 0$ gives $y = 8$ . Neither $x$ nor $y$ can be negative, so draw the line only from $(0, 8)$ to $(12, 0)$ .

Only whole numbers of fruit make sense, so the useful points are the points of the line with whole-number coordinates. The line is drawn dashed to show that the points between them do not count.

(c) The line passes exactly through grid points at $x = 0, 3, 6, 9, 12$ . With at least one of each fruit, $(0, 8)$ and $(12, 0)$ are left out.

Check $(9, 2)$ : $2(9) + 3(2) = 18 + 6 = 24\text{.}$

(d) $3y = -2x + 24$ , so $y = -\frac{2}{3}x + 8$ . The $y$ -intercept $8$ is the number of pears she can buy with no apples: 8 pears cost $ $24$ .

The gradient $-\frac{2}{3}$ says that for every $3$ more apples she buys $2$ fewer pears: 3 apples cost $ $6$ , the same as 2 pears. On the graph, each whole-number point is $3$ across and $2$ down from the one before.

(a) $2x + 3y = 24$ . (c) 3 apples and 6 pears, 6 apples and 4 pears, or 9 apples and 2 pears. (d) $y = -\frac{2}{3}x + 8$ : with no apples she buys 8 pears, and every 3 more apples means 2 fewer pears.

Common Mistakes

Reading the $y$ -intercept off $ax + by = c$ . In $2x + 3y = 12$ the $y$ -intercept is $4$ , not $12$ . Make $y$ the subject first.
Losing a sign when rearranging. From $5x - 2y = 6$ : $-2y = -5x + 6$ , and dividing every term by $-2$ gives $y = \frac{5}{2}x - 3$ .
Taking the sign of the gradient from the signs in $ax + by = c$ . In $2x + 3y = 12$ both numbers are positive, but the gradient is $-\frac{2}{3}$ : the line goes down from left to right.
A negative length. An intercept may be negative, but a side of a triangle never is: for $B(0, -6)$ , $OB = 6$ units.

Practice Questions

Work each question on paper before you open its solution. The questions get harder as you go.

Question 1

Find the gradient and the $y$ -intercept of (a) $4x + 2y = 10$ , (b) $3x - 5y = 15$ , (c) $x + 3y = 2$ .

ShowHide solution

(a) $2y = -4x + 10$ , so $y = -2x + 5$ : gradient $-2$ , $y$ -intercept $5$ .

(b) $-5y = -3x + 15$ . Divide every term by $-5$ : $y = \frac{3}{5}x - 3$ : gradient $\frac{3}{5}$ , $y$ -intercept $-3$ .

(a) $-2$ and $5$ (b) $\frac{3}{5}$ and $-3$ (c) $-\frac{1}{3}$ and $\frac{2}{3}$

Question 2

Find the gradient of the line $4x - 3y = 12$ (a) by writing the equation in the form $y = mx + c$ , (b) by drawing the line through its intercepts and counting the step from one intercept to the other.

ShowHide solution

(a) $-3y = -4x + 12$ . Divide every term by $-3$ : $y = \frac{4}{3}x - 4$ . The gradient is $\frac{4}{3}$ .

(b) $y = 0$ gives $4x = 12$ , $x = 3$ : the point $(3, 0)$ . $x = 0$ gives $-3y = 12$ , $y = -4$ : the point $(0, -4)$ .

From $(0, -4)$ to $(3, 0)$ the line goes $3$ across and $4$ up, so its gradient is $\frac{4}{3}$ . The two ways agree.

The gradient is $\frac{4}{3}$ .

Question 3

The line $5x + by = 11$ passes through $(3, -2)$ . Find $b$ and the gradient of the line.

ShowHide solution

$5(3) + b(-2) = 11$ , so $15 - 2b = 11$ , $-2b = -4$ and $b = 2$ .

The line is $5x + 2y = 11$ : $2y = -5x + 11$ , $y = -\frac{5}{2}x + \frac{11}{2}\text{.}$ The gradient is $-\frac{5}{2}$ .

Check: $5(3) + 2(-2) = 15 - 4 = 11\text{.}$

$b = 2$ ; the gradient is $-\frac{5}{2}$ .

Question 4

The line $2x + 3y = k$ passes through $(-3, 4)$ . Find $k$ and the $x$ -intercept of the line.

ShowHide solution

$k = 2(-3) + 3(4) = -6 + 12 = 6\text{.}$ The line is $2x + 3y = 6$ .

$x$ -intercept: $y = 0$ gives $2x = 6$ , so $x = 3$ .

$k = 6$ ; the $x$ -intercept is $3$ .

Question 5

The line $ax + 3y = 12$ has gradient $-2$ . Find $a$ and the $x$ -intercept of the line.

ShowHide solution

$3y = -ax + 12$ , so $y = -\frac{a}{3}x + 4$ . The gradient is $-\frac{a}{3}$ , so $-\frac{a}{3} = -2$ and $a = 6$ .

The line is $6x + 3y = 12$ . $x$ -intercept: $y = 0$ gives $6x = 12$ , so $x = 2$ .

Check: $3y = -6x + 12$ gives $y = -2x + 4$ , whose gradient is $-2$ .

$a = 6$ ; the $x$ -intercept is $2$ .

Question 6

The lines $P$ , $Q$ and $R$ are the graphs of $x + y = 3$ , $x - y = 3$ and $x + y = -3$ . Match each equation to its line.

ShowHide solution

$x - y = 3$ is $y = x - 3$ : gradient $1$ , positive. Only $P$ rises, so $P$ is $x - y = 3$ (it crosses the $x$ -axis at $3$ and the $y$ -axis at $-3$ ).

$x + y = 3$ : intercepts $3$ and $3$ . $Q$ crosses both axes at $3$ .

$x + y = -3$ : intercepts $-3$ and $-3$ . $R$ crosses both axes at $-3$ .

$P$ is $x - y = 3$ , $Q$ is $x + y = 3$ and $R$ is $x + y = -3$ .

Question 7

The line $2x - y = 6$ meets the $x$ -axis at $A$ and the $y$ -axis at $B$ . Find the area of triangle $OAB$ .

ShowHide solution

$A$ : $y = 0$ gives $2x = 6$ , $x = 3$ : $A(3, 0)$ . $B$ : $x = 0$ gives $-y = 6$ , $y = -6$ : $B(0, -6)$ .

The lengths are $OA = 3$ units and $OB = 6$ units: a length is never negative, although $B$ is below the $x$ -axis.

Area $= \frac{1}{2} \times 3 \times 6 = 9$ square units.

The area of triangle $OAB$ is $9$ square units.

Question 8

The line $2x + y = k$ , where $k > 0$ , cuts off a triangle of area $16$ square units with the axes. Find $k$ .

ShowHide solution

$y = 0$ gives $2x = k$ , so the $x$ -intercept is $\frac{k}{2}$ . $x = 0$ gives $y = k$ , so the $y$ -intercept is $k$ .

Since $k > 0$ , both intercepts are positive, and they are the lengths of the two shorter sides of the triangle. Area $= \frac{1}{2} \times \frac{k}{2} \times k = \frac{k^{2}}{4}\text{.}$

$\frac{k^{2}}{4} = 16$ , so $k^{2} = 64$ . Since $k > 0$ , $k = 8$ .

Check: $2x + y = 8$ meets the axes at $(4, 0)$ and $(0, 8)$ , and $\frac{1}{2} \times 4 \times 8 = 16$ .

$k = 8$

Question 9

A class spends exactly $ $48$ on $x$ plates of chicken rice at $ $4$ each and $y$ bowls of noodles at $ $3$ each. (a) Write an equation in $x$ and $y$ and draw its graph. (b) If they buy 6 plates of chicken rice, how many bowls of noodles do they buy? (c) List every possible order with at least one of each.

ShowHide solution

(a) $4x + 3y = 48$ . Intercepts: $y = 0$ gives $x = 12$ ; $x = 0$ gives $y = 16$ . Draw from $(0, 16)$ to $(12, 0)$ , dashed, since only whole numbers count.

(b) Up from $x = 6$ to the line, then across: $y = 8$ . Check: $4(6) + 3(8) = 24 + 24 = 48\text{.}$

(c) $y = \frac{48 - 4x}{3}$ is a whole number when $x$ is a multiple of $3$ . With $x$ and $y$ at least $1$ : $x = 3, 6, 9$ .

(a) $4x + 3y = 48$ (b) 8 bowls of noodles (c) 3 plates and 12 bowls, 6 plates and 8 bowls, or 9 plates and 4 bowls.

Question 10

A hall is set up for exactly $240$ guests, with $x$ long tables that seat $8$ each and $y$ round tables that seat $10$ each. (a) Show that $4x + 5y = 120$ . (b) Write the equation in the form $y = mx + c$ . (c) What does the $y$ -intercept tell you? (d) What does the gradient tell you? (e) If $15$ long tables are used, how many round tables are needed?

ShowHide solution

(a) The long tables seat $8x$ guests and the round tables $10y$ , so $8x + 10y = 240$ . Divide every term by $2$ : $4x + 5y = 120$ .

(b) $5y = -4x + 120$ . Divide every term by $5$ : $y = -\frac{4}{5}x + 24$ .

(d) The gradient is $-\frac{4}{5}$ : for every $5$ more long tables, $4$ fewer round tables are needed. Five long tables seat $40$ guests, the same as four round tables.

(e) $x = 15$ : $y = -\frac{4}{5}(15) + 24 = -12 + 24 = 12\text{.}$ Check: $8(15) + 10(12) = 120 + 120 = 240\text{.}$

(b) $y = -\frac{4}{5}x + 24$ (c) With no long tables, 24 round tables are needed. (d) Every 5 more long tables means 4 fewer round tables. (e) 12 round tables

Key Points

To find the gradient and the $y$ -intercept of $ax + by = c$ (with $b \ne 0$ ), make $y$ the subject and write it as $y = mx + c$ : the gradient is $m$ and the $y$ -intercept is the number on its own.
The $y$ -intercept is not the number on the right of $ax + by = c$ : in $2x + 3y = 12$ it is $4$ , not $12$ .
An unknown number in the equation: if a point is on the line, substitute its coordinates; if the gradient is given, write $y = mx + c$ and compare.
To match equations to lines, compare the intercepts and the sign of the gradient (up from left to right: positive; down: negative).
A line crossing both axes away from the origin cuts off a right-angled triangle; its area is $\frac{1}{2} \times$ (size of $x$ -intercept) $\times$ (size of $y$ -intercept).
In a real setting, an intercept is the amount of one thing when there is none of the other, and the gradient says how much one amount changes as the other goes up.