MathsCBSE Class 9 · Introduction to Linear Polynomials

Finding y = ax + b from Two Facts, and Its Graph: Class 9

Q: How does linear decay differ from linear growth on a graph?

In growth the slope is positive and the quantity increases steadily; in decay the slope is negative and the quantity decreases by a fixed amount per unit.

Q: Can I use any two points to find the line?

Yes. Any two distinct points on a non-vertical line give the same slope and intercept.

Updated 17 October 2026 · Fact-checked

A linear relationship has the form $y = ax + b$, where $a$ is the constant change in $y$ per unit increase in $x$ (the slope) and $b$ is the value of $y$ at $x = 0$ (the y-intercept). Two known pairs fix the relationship: $a = \frac{y_2 - y_1}{x_2 - x_1}$, then $b = y_1 - ax_1$. Its graph is a straight line that meets the y-axis at $(0, b)$ and the x-axis at $\left(-\frac{b}{a}, 0\right)$.

Starting at the y-intercept (0, 4), every 3 steps right the line falls 2, so it meets the x-axis at (6, 0), and the axes triangle has area 12.

Illustration drawn for QuestionBuddy.

Understand Finding Linear Relationships y = ax + b and Their Graphs

A linear pattern is a sequence in which each step adds the same amount. If matchstick squares are laid in a row, one square needs $4$ sticks, two need $7$ , three need $10$ : each new square adds $3$ , so the number of sticks is $3 n + 1$ for $n$ squares. The constant added each time is the rate of change; the starting adjustment ( $+ 1$ ) is the constant term.

Linear growth and decay. If $y = a x + b$ with $a > 0$ the quantity grows by $a$ per unit of $x$ (a taxi fare, savings with a regular deposit); if $a < 0$ it decays by $∣ a ∣$ per unit (a burning candle, a draining tank). A candle of length 30 cm that burns 2.5 cm each hour has length $L = 30 - 2.5 t$ after $t$ hours; it is 12.5 cm after 7 hours and burns out when $t = 12$ . The meaningful values of $x$ stop where $y$ would become negative.

**Forming $y = a x + b$ from two facts.** If $y = 7$ when $x = 2$ and $y = - 5$ when $x = - 1$ , write both as equations: $2 a + b = 7$ and $- a + b = - 5$ . Subtract: $3 a = 12$ , so $a = 4$ and $b = 7 - 8 = - 1$ . The relationship is $y = 4 x - 1$ . Equivalently, slope $a = \frac{7 - ( - 5 )}{2 - ( - 1 )} = 4$ . Always verify with the second point.

Graph. The graph of $y = a x + b$ is a straight line. The y-intercept is $(0, b)$ ; the x-intercept (the zero) is $(- \frac{b}{a}, 0)$ when $a \neq = 0$ . For $y = - \frac{2}{3} x + 4$ : the intercepts are $(0, 4)$ and $(6, 0)$ , the slope is $- \frac{2}{3}$ (the line falls $2$ units for every $3$ to the right), and the triangle formed with the axes has area $\frac{1}{2} \times 6 \times 4 = 12$ . A point lies on the line exactly when its coordinates satisfy the equation: $(9, - 2)$ does, since $- \frac{2}{3} (9) + 4 = - 2$ .

Parallel lines. Lines with the same $a$ are parallel; lines through the origin have $b = 0$ . Different $a$ means the lines cross once. Comparing two linear plans, such as $300 + 5 g$ and $100 + 15 g$ for $g$ gigabytes of data, the plans cost the same where the lines cross: $300 + 5 g = 100 + 15 g$ gives $g = 20$ and a cost of $400$ . Below that usage the plan with the lower fixed charge is cheaper; above it, the plan with the lower rate is.

Exam approach. Write the two equations, solve for $a$ then $b$ , verify with a second point, and state the relationship with the variables named. For graph questions find both intercepts first, then plot a third point to check.

Finding Linear Relationships y = ax + b and Their Graphs: formulas to remember

Linear relationship: $y = a x + b$; $a$ = rate of change (slope), $b$ = value at $x = 0$ .
Slope from two points: $a = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$; Then $b = y_{1} - a x_{1}$ .
Intercepts: $(0, b) and (- \frac{b}{a}, 0)$; y-intercept and x-intercept.
Linear pattern: $t_{n} = d n + (t_{1} - d)$; $d$ = common difference.
Parallel lines: $y = a x + b_{1}, y = a x + b_{2}$; Same slope, different intercepts: never meet.
Intersection of two lines: $a_{1} x + b_{1} = a_{2} x + b_{2}$; Solve for $x$ , then find $y$ .

How to solve Finding Linear Relationships y = ax + b and Their Graphs questions

Form the relationship from two facts, then use it for prediction or for the graph.

1Write the two facts as pairs $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ .
2Find the slope $a = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ .
3Find $b = y_{1} - a x_{1}$ using either point.
4Write $y = a x + b$ and verify with the other point.
5For the graph, find the y-intercept $b$ and the x-intercept $- \frac{b}{a}$ , and join them.
6Interpret $a$ and $b$ in words, and reject values of $x$ outside the meaningful range.

Finding Linear Relationships y = ax + b and Their Graphs in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Finding a linear relationship from two bills

Suppose a plumber charges a fixed visiting fee plus a fixed rate per hour, and two customers pay 650 rupees for 3 hours and 950 rupees for 5 hours.

Find the hourly rate and the visiting fee, and write the charge for $h$ hours.

Try it first, then show the solutionSolution

Rate $= \frac{950 - 650}{5 - 3} = \frac{300}{2} = 150$ rupees per hour.
Fee $= 650 - 150 \times 3 = 650 - 450 = 200$ rupees.
Charge $= 150 h + 200$ ; check at $h = 5$ : $750 + 200 = 950$ .

Answer: The rate is 150 rupees per hour and the visiting fee 200 rupees, so the charge is $150 h + 200$ .

Memory hook: Slope is rate; intercept is the fixed fee.

Tricks to remember Finding Linear Relationships y = ax + b and Their Graphs

Short memory aids. Say each one aloud once, then use it on a question the same day.

Rise over run
Slope is change in $y$ divided by change in $x$ between two points.
Use it when: Finding $a$ .
b is the value at zero
Put $x = 0$ : the y-intercept is the starting value or fixed charge.
Use it when: Interpreting the constant term.
Check with the other point
After finding $y = a x + b$ from two facts, substitute the unused fact; it must fit.
Use it when: Catching arithmetic slips.

Finding Linear Relationships y = ax + b and Their Graphs: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Two facts, one line · 3 marks · Short answerBoard level

The variable $y$ is linearly related to $x$ . When $x = 2$ , $y = 7$ ; when $x = - 1$ , $y = - 5$ . Find the relationship and the value of $y$ at $x = 3$ .

Try it first, then show the solutionSolution

Slope $a = \frac{7 - ( - 5 )}{2 - ( - 1 )} = \frac{12}{3} = 4$ . (1)
$b = 7 - 4 (2) = - 1$ , so $y = 4 x - 1$ . (1)
Check with $x = - 1$ : $4 (- 1) - 1 = - 5$ ; at $x = 3$ : $y = 11$ . (1)

Answer: $y = 4 x - 1$ , and $y = 11$ at $x = 3$ .

Example 2: Matchstick pattern · 3 marks · Short answerBoard level

Matchsticks form a row of squares: 1 square uses 4 sticks, 2 squares use 7, 3 squares use 10. Find the rule for $n$ squares, the number of sticks for 25 squares, and the number of squares that can be made with exactly 100 sticks.

Try it first, then show the solutionSolution

Each extra square adds 3 sticks, so the rule is $S = 3 n + c$ ; with $n = 1$ , $4 = 3 + c$ , so $c = 1$ and $S = 3 n + 1$ . (1)
For 25 squares: $S = 3 (25) + 1 = 76$ sticks. (1)
$3 n + 1 = 100$ gives $n = 33$ squares. (1)

Answer: $S = 3 n + 1$ ; 76 sticks for 25 squares; 33 squares from 100 sticks.

Example 3: Reading a line · 5 marks · Long answerHarder board level

For the line $y = - \frac{2}{3} x + 4$ : (i) find the slope and both intercepts; (ii) find the area of the triangle formed with the axes; (iii) test whether $(9, - 2)$ and $(6, 1)$ lie on it.

Try it first, then show the solutionSolution

(i) Slope $= - \frac{2}{3}$ ; y-intercept $(0, 4)$ . (1)
x-intercept: $- \frac{2}{3} x + 4 = 0$ gives $x = 6$ , so $(6, 0)$ . (1)
(ii) The triangle has legs 6 and 4, so area $= \frac{1}{2} \times 6 \times 4 = 12$ square units. (1)
(iii) $(9, - 2)$ : $- \frac{2}{3} (9) + 4 = - 2$ , so it lies on the line. (1)
$(6, 1)$ : $- \frac{2}{3} (6) + 4 = 0 \neq = 1$ , so it does not lie on the line. (1)

Answer: (i) slope $- \frac{2}{3}$ , intercepts $(0, 4)$ and $(6, 0)$ ; (ii) 12 square units; (iii) $(9, - 2)$ is on the line, $(6, 1)$ is not.

Example 4: Linear decay · 3 marks · Short answerHarder board level

A candle 30 cm long burns down by 2.5 cm every hour. Write its length after $t$ hours, find the length after 7 hours, the time when it is half its original length, and the time when it burns out.

Try it first, then show the solutionSolution

$L = 30 - 2.5 t$ . (½)
At $t = 7$ : $L = 30 - 17.5 = 12.5$ cm. (½)
Half length $= 15$ cm: $30 - 2.5 t = 15$ , $t = 6$ hours. (1)
Burns out when $L = 0$ : $t = \frac{30}{2.5} = 12$ hours. (1)

Answer: $L = 30 - 2.5 t$ ; 12.5 cm after 7 h; half-length at 6 h; burns out at 12 h.

Example 5: Comparing two plans · 4 marks · Case-basedHOTS / competency

Case: Plan A costs $300 + 5 g$ rupees and Plan B costs $100 + 15 g$ rupees for $g$ gigabytes (GB) of data in a month. (i) At what usage do the plans cost the same, and what is the common cost? (ii) Which plan is cheaper for 12 GB? (iii) Which plan is cheaper for 30 GB? (iv) What does the point where the two graphs cross show?

Try it first, then show the solutionSolution

(i) $300 + 5 g = 100 + 15 g$ gives $200 = 10 g$ , so $g = 20$ GB, with cost $300 + 100 = 400$ rupees. (1)
(ii) At 12 GB: A $= 360$ and B $= 280$ , so Plan B is cheaper. (1)
(iii) At 30 GB: A $= 450$ and B $= 550$ , so Plan A is cheaper. (1)
(iv) The graphs cross at $(20, 400)$ : at 20 GB both plans cost 400 rupees; below it the plan with the lower fixed charge is cheaper, above it the plan with the lower rate is. (1)

Answer: (i) 20 GB at 400 rupees; (ii) Plan B; (iii) Plan A; (iv) the break-even usage.

Where students lose marks in Finding Linear Relationships y = ax + b and Their Graphs

The slips examiners see most often, and what to write instead.

Mistake: Subtracting in different orders
Using $y_{2} - y_{1}$ over $x_{1} - x_{2}$ .
Wrong working
$a = \frac{7 - ( - 5 )}{- 1 - 2} = - 4$ .
Right working
$a = \frac{7 - ( - 5 )}{2 - ( - 1 )} = 4$ .
Do this: Subtract in the same order on top and bottom.
Usually costs: The whole relationship.
Mistake: Reading the slope as the intercept
Mixing $a$ and $b$ .
Wrong working
For $y = 3 x - 7$ the y-intercept is 3.
Right working
The y-intercept is $- 7$ ; 3 is the slope.
Do this: Substitute $x = 0$ to find the intercept.
Usually costs: 1 mark.
Mistake: Not verifying with the second point
Trusting the first calculation.
Wrong working
$y = 3 x + 1$ from $(2, 7)$ and $(- 1, - 5)$ without a check.
Right working
$3 (- 1) + 1 = - 2 \neq = - 5$ ; the correct line is $y = 4 x - 1$ .
Do this: Always substitute the unused point.
Usually costs: 2 marks.

Exam tips

Name the variables in your final relationship (for example $S = 3 n + 1$ ).
Check with the unused point; it is a free mark-saver.
For graphs mark both intercepts with coordinates and label the axes.
In context questions, state whether the slope means growth or decay and what the intercept means.

Practice questions on Finding Linear Relationships y = ax + b and Their Graphs

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The y-intercept of the line $y = 3 x - 7$ is
(a) 3
(b) $- 7$
(c) $\frac{7}{3}$
(d) $- \frac{7}{3}$
Hint
Put $x = 0$ .
Answer
(b) $- 7$ . At $x = 0$ , $y = - 7$ , so the line meets the y-axis at $(0, - 7)$ .
Q2 · 3 marks · Short answerBoard level
A line has slope 2 and passes through $(0, 5)$ . Write its equation and find the point where it meets the x-axis.
Hint
Use $y = a x + b$ with $a = 2$ and $b = 5$ .
Answer
$y = 2 x + 5$ . Setting $y = 0$ : $x = - \frac{5}{2}$ , so the line meets the x-axis at $(- \frac{5}{2}, 0)$ .
Q3 · 3 marks · Short answerHarder board level
Temperature in degrees Fahrenheit $F$ is linearly related to degrees Celsius $C$ : $F = 32$ when $C = 0$ and $F = 212$ when $C = 100$ . Find $F$ in terms of $C$ , and find the temperature at which both scales show the same number.
Hint
Slope $= \frac{212 - 32}{100 - 0}$ ; then set $F = C$ .
Answer
Slope $= \frac{180}{100} = 1.8$ , intercept $32$ : $F = 1.8 C + 32$ . Setting $F = C$ : $C = 1.8 C + 32$ , so $- 0.8 C = 32$ and $C = - 40$ . Both scales show $- 40$ .
Q4 · 5 marks · Long answerHarder board level
Find the equation of the line through $(1, 3)$ parallel to $y = 2 x - 5$ . Find its intercepts with the axes and the distance between these two intercept points. Do the two lines ever meet?
Hint
Parallel lines have the same slope; distance between $(p, 0)$ and $(0, q)$ is $p^{2} + q^{2}$ .
Answer
Slope 2: $y = 2 x + b$ , and $3 = 2 + b$ gives $b = 1$ , so $y = 2 x + 1$ . y-intercept $(0, 1)$ ; x-intercept: $2 x + 1 = 0$ , $x = - \frac{1}{2}$ , so $(- \frac{1}{2}, 0)$ . Distance $= \frac{1}{4} + 1 = \frac{5}{2}$ . The lines have equal slope and different intercepts ( $1 \neq = - 5$ ), so they never meet.
Q5 · 4 marks · Case-basedHOTS / competency
Case: a tank is being filled at a steady rate. The water level is 24 cm after 2 minutes and 45 cm after 5 minutes. (i) Find the rate of filling. (ii) Write the level $h$ after $t$ minutes. (iii) Find the level at the start. (iv) When will the level reach 80 cm?
Hint
Use the two readings as the two facts.
Answer
(i) Rate $= \frac{45 - 24}{5 - 2} = 7$ cm per minute. (ii) $h = 7 t + b$ with $24 = 14 + b$ , so $b = 10$ and $h = 7 t + 10$ . (iii) At $t = 0$ the level is 10 cm. (iv) $7 t + 10 = 80$ gives $t = 10$ minutes.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What does $a$ mean in $y = a x + b$ ?Show answer
The rate of change (slope).
2.y-intercept of $y = a x + b$ ?Show answer
$(0, b)$ .
3.x-intercept of $y = a x + b$ ?Show answer
$(- \frac{b}{a}, 0)$ .
4.Slope of a line through two points?Show answer
$\frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ .
5.Do lines with the same slope meet?Show answer
No (unless they are the same line).

Finding Linear Relationships y = ax + b and Their Graphs: frequently asked questions

How does linear decay differ from linear growth on a graph?

In growth the slope is positive and the quantity increases steadily; in decay the slope is negative and the quantity decreases by a fixed amount per unit.

Can I use any two points to find the line?

Yes. Any two distinct points on a non-vertical line give the same slope and intercept.

Does every linear relationship pass through the origin?

No. It does only when b = 0.