Linear Law Worked Examples (Easy)

Q: How do I know which quantities to plot on the two axes?

Compare the relation with Y=mX+cY = mX + cY=mX+c. The quantity that changes in a straight-line pattern becomes the horizontal XXX, and the quantity read against it becomes the vertical YYY. For y=ax2+by = ax^{2}+by=ax2+b, that means Y=yY = yY=y and X=x2X = x^{2}X=x2.

Q: What do the gradient and the intercept tell me?

The gradient equals the coefficient attached to XXX, and the vertical-axis intercept equals the constant term. Reading these two numbers off the straight line is usually all you need to find the unknown constants.

Q: How do I find the equation of a line of best fit from two points?

Find the gradient with m=Y2−Y1X2−X1m = \frac{Y_{2} - Y_{1}}{X_{2} - X_{1}}m=X2−X1Y2−Y1, then substitute one of the points into Y=mX+cY = mX + cY=mX+c to solve for ccc. Always check your equation against the other point.

What these examples cover

These easy Linear Law examples build the moves the whole chapter rests on: reducing a relation to the straight-line form $Y = mX + c$ , reading the gradient and the vertical-axis intercept, finding the equation of a line of best fit from two points, and using that line to estimate a value. Each one uses small, clean numbers so you can follow every line without a calculator getting in the way.

The best way to use the set is to cover the solution, attempt the question in full on paper, and only then check line by line against our working. If your answer differs, find the exact step where the two solutions part company, that single line is usually where the real learning is.

Treat the comparison with $Y = mX + c$ as the move to slow down on, and these turn into quick, reliable marks.

Worked examples

Work through all four. Attempt each fully before you read the matching solution, and watch how the same routine, compare with $Y = mX + c$ , decide what to plot, then read the gradient and intercept, runs through every one.

Q1[3 marks]

A line of best fit passes through the points $(2, 7)$ and $(6, 19)$ . Find (a) the gradient of the line, and (b) the equation of the line in the form $Y = mX + c$ .

Show worked solution

(a) The gradient measures how much $Y$ rises for each unit increase in $X$ . Use the two given points in the gradient formula:

m=\frac{19-7}{6-2}=\frac{12}{4}=3

(b) Substitute $m = 3$ into $Y = mX + c$ , then use one known point, say $(2, 7)$ , to find the intercept $c$ :

7=3(2)+c \;\Rightarrow\; 7=6+c \;\Rightarrow\; c=1

Answer

The gradient is $3$ and the line is $Y = 3X + 1$ . Check with the second point: at $X = 6$ , $Y = 3(6) + 1 = 19$ , which matches the given point exactly.

Q2[4 marks]

The variables $x$ and $y$ are related by $y = ax^{2} + b$ , where $a$ and $b$ are constants. When $y$ is plotted against $x^{2}$ , a straight line of gradient $4$ and vertical-axis intercept $-3$ is obtained.

Find (a) the values of $a$ and $b$ , and (b) the value of $y$ when $x = 2$ .

Show worked solution

(a) Match the relation to the straight-line form. Writing $y = ax^{2} + b$ beside $Y = mX + c$ shows that the vertical axis carries $Y = y$ and the horizontal axis carries $X = x^{2}$ :

y = a\,x^{2} + b \quad\Longleftrightarrow\quad Y = mX + c

So the gradient stands for $a$ and the intercept stands for $b$ . Read the two numbers straight off the graph:

a = m = 4, \qquad b = c = -3

(b) The relation is therefore $y = 4x^{2} - 3$ . Substitute $x = 2$ , taking the square before the multiplication:

y = 4(2)^{2} - 3 = 4(4) - 3 = 16 - 3 = 13

Answer

$a = 4$ , $b = -3$ , and $y = 13$ when $x = 2$ . The key move is recognising that plotting $y$ against $x^{2}$ , not against $x$ , is what turns a curved relation into a straight line.

Q3[4 marks]

The variables $x$ and $y$ are related by $y = \frac{p}{x} + q$ . A straight-line graph of $y$ against $\frac{1}{x}$ has gradient $5$ and vertical-axis intercept $2$ .

Find (a) the values of $p$ and $q$ , and (b) the value of $y$ when $x = 5$ .

Show worked solution

(a) Rewrite the relation so the changing quantity is clear: $y = p\left(\frac{1}{x}\right) + q$ . Compared with $Y = mX + c$ , the vertical axis is $Y = y$ and the horizontal axis is $X = \frac{1}{x}$ :

y = p\left(\frac{1}{x}\right) + q \quad\Longleftrightarrow\quad Y = mX + c

The gradient is $p$ and the intercept is $q$ , so read them off directly:

p = 5, \qquad q = 2

(b) The relation is $y = \frac{5}{x} + 2$ . Substitute $x = 5$ :

y = \frac{5}{5} + 2 = 1 + 2 = 3

Answer

$p = 5$ , $q = 2$ , and $y = 3$ when $x = 5$ . Plotting against $\frac{1}{x}$ rather than $x$ is what makes the graph straight here.

Q4[4 marks]

A line of best fit for the graph of $y$ against $x$ passes through the points $(1, 4)$ and $(5, 12)$ . Use the line to estimate (a) the value of $y$ when $x = 3$ , and (b) the value of $x$ when $y = 20$ .

Show worked solution

First find the equation of the line. The gradient comes from the two points:

m = \frac{12 - 4}{5 - 1} = \frac{8}{4} = 2

Substitute $m = 2$ and the point $(1, 4)$ into $y = mx + c$ to find the intercept:

4 = 2(1) + c \;\Rightarrow\; c = 2 \;\Rightarrow\; y = 2x + 2

(a) To estimate $y$ when $x = 3$ , substitute $x = 3$ into the equation:

y = 2(3) + 2 = 8

(b) To estimate $x$ when $y = 20$ , set $y = 20$ and solve for $x$ :

20 = 2x + 2 \;\Rightarrow\; 2x = 18 \;\Rightarrow\; x = 9

Answer

When $x = 3$ , $y = 8$ ; when $y = 20$ , $x = 9$ . Reading values off the equation of the line of best fit is often quicker and steadier than measuring on the graph paper itself.

Q5[3 marks]

The variables $x$ and $y$ are related by $y = ax^{n}$ , where $a$ and $n$ are constants. When $\lg y$ is plotted against $\lg x$ , a straight line with gradient $2$ and vertical-axis intercept $1$ is obtained.

Find the values of $a$ and $n$ .

Show worked solution

Take $\lg$ of both sides of $y = ax^{n}$ and compare the result with $Y = mX + c$ :

\lg y = n\lg x + \lg a \quad\Longleftrightarrow\quad Y = mX + c

So the horizontal axis carries $X = \lg x$ and the vertical axis carries $Y = \lg y$ ; the gradient stands for $n$ and the intercept stands for $\lg a$ . Read the two values straight off the graph:

n = 2, \qquad \lg a = 1 \;\Rightarrow\; a = 10^{1} = 10

Answer

$n = 2$ and $a = 10$ , so $y = 10x^{2}$ . Check: at $x = 10$ , $\lg x = 1$ , so $\lg y$ should be $2(1) + 1 = 3$ , and indeed $y = 10(10)^{2} = 1000$ , where $\lg 1000 = 3$ .

Q6[3 marks]

The variables $x$ and $y$ are related by $y = ab^{x}$ , where $a$ and $b$ are constants. When $\lg y$ is plotted against $x$ , a straight line with gradient $-1$ and vertical-axis intercept $2$ is obtained.

Find the values of $a$ and $b$ .

Show worked solution

Take $\lg$ of both sides of $y = ab^{x}$ and compare with $Y = mX + c$ :

\lg y = (\lg b)x + \lg a \quad\Longleftrightarrow\quad Y = mX + c

Here the horizontal axis is simply $X = x$ ; the gradient stands for $\lg b$ and the intercept stands for $\lg a$ . Convert each back using powers of $10$ :

\lg b = -1 \;\Rightarrow\; b = 10^{-1} = \frac{1}{10}, \qquad \lg a = 2 \;\Rightarrow\; a = 10^{2} = 100

Answer

$a = 100$ and $b = \frac{1}{10}$ , so $y = 100\left(\frac{1}{10}\right)^{x}$ . Check at $x = 0$ : $y = 100$ , and $\lg 100 = 2$ , matching the given intercept exactly.

Q7[3 marks]

The variables $x$ and $y$ are related by $xy = px + q$ , where $p$ and $q$ are constants. When the equation is rearranged and a graph of $y$ against $\frac{1}{x}$ is plotted, a straight line with gradient $6$ and vertical-axis intercept $2$ is obtained.

Find the values of $p$ and $q$ .

Show worked solution

Divide both sides of $xy = px + q$ by $x$ so that $y$ stands alone on the left:

xy = px + q \;\Longrightarrow\; y = p + q\left(\frac{1}{x}\right)

Compare this with $Y = mX + c$ : the vertical axis is $Y = y$ and the horizontal axis is $X = \frac{1}{x}$ , so the gradient stands for $q$ and the intercept stands for $p$ :

q = 6, \qquad p = 2

Answer

$p = 2$ and $q = 6$ , so $xy = 2x + 6$ . Check: dividing back by $x$ gives $y = 2 + \frac{6}{x}$ , a gradient of $6$ and an intercept of $2$ exactly as given.

Q8[3 marks]

The variables $x$ and $y$ are related by $y = ax^{2} + bx$ , where $a$ and $b$ are constants. When $\frac{y}{x}$ is plotted against $x$ , a straight line with gradient $3$ and vertical-axis intercept $-2$ is obtained.

Find the values of $a$ and $b$ .

Show worked solution

Divide both sides of $y = ax^{2} + bx$ by $x$ to isolate a single power of $x$ on the right:

y = ax^{2} + bx \;\Longrightarrow\; \frac{y}{x} = ax + b

Compare with $Y = mX + c$ : the vertical axis is $Y = \frac{y}{x}$ and the horizontal axis is $X = x$ , so the gradient stands for $a$ and the intercept stands for $b$ :

a = 3, \qquad b = -2

Answer

$a = 3$ and $b = -2$ , so $y = 3x^{2} - 2x$ . Check at $x = 1$ : $\frac{y}{x} = 3(1) - 2 = 1$ , and directly, $y = 3(1)^{2} - 2(1) = 1$ , so $\frac{y}{x} = 1$ too.

Notice the single habit underneath all four: compare the given relation with $Y = mX + c$ , decide exactly what belongs on each axis, and only then read or use the gradient and intercept. Once that comparison is second nature, Linear Law becomes one of the most dependable sources of marks in Add Math.

Key method points

These four examples rehearse the everyday skills that open almost every Linear Law question in Add Math. Keep the following points in mind as you practise more.

Always compare the given relation with $Y = mX + c$ first, and write down clearly what $Y$ and $X$ each stand for.
The gradient of the straight line equals the coefficient attached to $X$ ; the vertical-axis intercept equals the constant term.
To find a gradient from two points, use $m = \frac{Y_{2} - Y_{1}}{X_{2} - X_{1}}$ , then substitute one point to find the intercept.
For $y = ax^{2} + b$ plot $y$ against $x^{2}$ ; for $y = \frac{p}{x} + q$ plot $y$ against $\frac{1}{x}$ .
Once you know the equation of the line, estimate a value by substituting into it rather than measuring on the grid.
Because marking is analytic, a correct comparison and substitution can still earn method marks even if the final arithmetic slips.

How a teacher helps

When a student loses a mark here, it is almost always one fixable habit, plotting against $x$ instead of $x^{2}$ , or mixing up which axis carries the gradient. In a one-to-one lesson our teacher watches the exact line where the slip happens and corrects it on the spot, before it settles into a routine.

Because our teachers are experienced, you work with someone who explains the why behind each step, not just the what. Lessons are taught in English, while SPM papers are set in both Malay and English, so we make sure the notation reads the same to you either way.

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Frequently asked questions

How do I know which quantities to plot on the two axes?

Compare the relation with $Y = mX + c$ . The quantity that changes in a straight-line pattern becomes the horizontal $X$ , and the quantity read against it becomes the vertical $Y$ .

For $y = ax^{2}+b$ , that means $Y = y$ and $X = x^{2}$ .

What do the gradient and the intercept tell me?

The gradient equals the coefficient attached to $X$ , and the vertical-axis intercept equals the constant term. Reading these two numbers off the straight line is usually all you need to find the unknown constants.

How do I find the equation of a line of best fit from two points?

Find the gradient with $m = \frac{Y_{2} - Y_{1}}{X_{2} - X_{1}}$ , then substitute one of the points into $Y = mX + c$ to solve for $c$ . Always check your equation against the other point.

Should I read values from the graph or from the equation?

Once you have the equation, substitute into it, that is faster and less error-prone than measuring on the grid. Use the graph mainly to draw the line and to sanity-check your answer.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Linear Law, Worked Examples (easy)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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