Sec 4 A-Math · Linear law

Converting to a straight line questions with answers

Practice questions on converting to a straight line, from the basics up to exam level, each with a full worked solution. Try them on paper first, then open the solution to check your working.

Worked example

The variables $x$ and $y$ are related by the equation $y^{2} = ax + b$ , where $a$ and $b$ are constants. When $y^{2}$ is plotted against $x$ , a straight line passing through the points $(3, 8)$ and $(7, 20)$ is obtained. Find the value of $b$ .

Step 1 of 5

Write the equation as Y = mX + c, with Y = y² and X = x: what do you do?

Nothing (it is already a straight line).

Try a question: Converting to a straight line

Free · no sign-up

Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

Questions

Question 1Divide through or already linear
The variables $x$ and $y$ are related by the equation $y = ax^{2} + bx$ , where $a$ and $b$ are constants. When $\dfrac{y}{x}$ is plotted against $x$ , a straight line passing through the points $(1, 7)$ and $(3, 1)$ is obtained. Find the value of $a$ .
Show the worked solution
1. Step 1: Write the equation as Y = mX + c, with Y = y ÷ x and X = x: what do you do?Divide both sides by x.
2. Step 2: So the straight-line form is $\dfrac{y}{x} = ax + b$
3. Step 3: Compare with Y = mX + c $a$ is the gradient and $b$ is the $Y$ -intercept.
4. Step 4: Gradient of the line through the two points: put them into the gradient formula $a = \dfrac{Y_2 - Y_1}{X_2 - X_1}$ $a = \dfrac{1 - 7}{3 - 1}$ $= -3$
Question 2Divide through or already linear
The variables $x$ and $y$ are related by the equation $y = ax^{2} + bx$ , where $a$ and $b$ are constants. When $\dfrac{y}{x}$ is plotted against $x$ , a straight line passing through the points $(3, 17)$ and $(7, 33)$ is obtained. Find the value of $b$ .
Show the worked solution
1. Step 1: Write the equation as Y = mX + c, with Y = y ÷ x and X = x: what do you do?Divide both sides by x.
2. Step 2: So the straight-line form is $\dfrac{y}{x} = ax + b$
3. Step 3: Compare with Y = mX + c $a$ is the gradient and $b$ is the $Y$ -intercept.
4. Step 4: Gradient of the line through the two points: put them into the gradient formula $a = \dfrac{Y_2 - Y_1}{X_2 - X_1}$ $a = \dfrac{33 - 17}{7 - 3}$ $= 4$
5. Step 5: Y-intercept: put the first point and the gradient into Y = mX + c $17 = 4(3) + b$ $b = 5$
Question 3Multiply through or use 1/x and √x
The variables $x$ and $y$ are related by the equation $y = ax + \dfrac{b}{x}$ , where $a$ and $b$ are constants. When $xy$ is plotted against $x^{2}$ , a straight line passing through the points $(3, 8)$ and $(7, 20)$ is obtained. Find the value of $b$ .
Show the worked solution
1. Step 1: Write the equation as Y = mX + c, with Y = xy and X = x²: what do you do?Multiply both sides by x.
2. Step 2: So the straight-line form is $xy = ax^{2} + b$
3. Step 3: Compare with Y = mX + c $a$ is the gradient and $b$ is the $Y$ -intercept.
4. Step 4: Gradient of the line through the two points: put them into the gradient formula $a = \dfrac{Y_2 - Y_1}{X_2 - X_1}$ $a = \dfrac{20 - 8}{7 - 3}$ $= 3$
5. Step 5: Y-intercept: put the first point and the gradient into Y = mX + c $8 = 3(3) + b$ $b = -1$
Question 4Multiply through or use 1/x and √x
The variables $x$ and $y$ are related by the equation $y = ax + \dfrac{b}{x}$ , where $a$ and $b$ are constants. When $xy$ is plotted against $x^{2}$ , a straight line passing through the points $(1, 7)$ and $(4, 22)$ is obtained. Find the value of $a$ .
Show the worked solution
1. Step 1: Write the equation as Y = mX + c, with Y = xy and X = x²: what do you do?Multiply both sides by x.
2. Step 2: So the straight-line form is $xy = ax^{2} + b$
3. Step 3: Compare with Y = mX + c $a$ is the gradient and $b$ is the $Y$ -intercept.
4. Step 4: Gradient of the line through the two points: put them into the gradient formula $a = \dfrac{Y_2 - Y_1}{X_2 - X_1}$ $a = \dfrac{22 - 7}{4 - 1}$ $= 5$

More on linear law

Notes: Linear law Worksheet with answers (PDF)

Endless questions on converting to a straight line

A free account gives you new questions every time, step-by-step help, and tracks what you've mastered.

Start free

Worked example

The variables

x

and

y

are related by the equation

y^{2} = ax + b

, where

a

and

b

are constants. When

y^{2}

is plotted against

x

, a straight line passing through the points

(3, 8)

and

(7, 20)

is obtained. Find the value of

b

Step 1 of 5

Write the equation as Y = mX + c, with Y = y² and X = x: what do you do?

Nothing (it is already a straight line).

Questions

Question 1Divide through or already linear

The variables $x$ and $y$ are related by the equation $y = ax^{2} + bx$ , where $a$ and $b$ are constants. When $\dfrac{y}{x}$ is plotted against $x$ , a straight line passing through the points $(1, 7)$ and $(3, 1)$ is obtained. Find the value of $a$ .

Show the worked solution

Step 1: Write the equation as Y = mX + c, with Y = y ÷ x and X = x: what do you do?Divide both sides by x.
Step 2: So the straight-line form is $\dfrac{y}{x} = ax + b$
Step 3: Compare with Y = mX + c $a$ is the gradient and $b$ is the $Y$ -intercept.
Step 4: Gradient of the line through the two points: put them into the gradient formula $a = \dfrac{Y_2 - Y_1}{X_2 - X_1}$ $a = \dfrac{1 - 7}{3 - 1}$ $= -3$

Question 2Divide through or already linear

The variables $x$ and $y$ are related by the equation $y = ax^{2} + bx$ , where $a$ and $b$ are constants. When $\dfrac{y}{x}$ is plotted against $x$ , a straight line passing through the points $(3, 17)$ and $(7, 33)$ is obtained. Find the value of $b$ .

Show the worked solution

Step 1: Write the equation as Y = mX + c, with Y = y ÷ x and X = x: what do you do?Divide both sides by x.
Step 2: So the straight-line form is $\dfrac{y}{x} = ax + b$
Step 3: Compare with Y = mX + c $a$ is the gradient and $b$ is the $Y$ -intercept.
Step 4: Gradient of the line through the two points: put them into the gradient formula $a = \dfrac{Y_2 - Y_1}{X_2 - X_1}$ $a = \dfrac{33 - 17}{7 - 3}$ $= 4$
Step 5: Y-intercept: put the first point and the gradient into Y = mX + c $17 = 4(3) + b$ $b = 5$

Question 3Multiply through or use 1/x and √x

The variables $x$ and $y$ are related by the equation $y = ax + \dfrac{b}{x}$ , where $a$ and $b$ are constants. When $xy$ is plotted against $x^{2}$ , a straight line passing through the points $(3, 8)$ and $(7, 20)$ is obtained. Find the value of $b$ .

Show the worked solution

Step 1: Write the equation as Y = mX + c, with Y = xy and X = x²: what do you do?Multiply both sides by x.
Step 2: So the straight-line form is $xy = ax^{2} + b$
Step 3: Compare with Y = mX + c $a$ is the gradient and $b$ is the $Y$ -intercept.
Step 4: Gradient of the line through the two points: put them into the gradient formula $a = \dfrac{Y_2 - Y_1}{X_2 - X_1}$ $a = \dfrac{20 - 8}{7 - 3}$ $= 3$
Step 5: Y-intercept: put the first point and the gradient into Y = mX + c $8 = 3(3) + b$ $b = -1$

Question 4Multiply through or use 1/x and √x

The variables $x$ and $y$ are related by the equation $y = ax + \dfrac{b}{x}$ , where $a$ and $b$ are constants. When $xy$ is plotted against $x^{2}$ , a straight line passing through the points $(1, 7)$ and $(4, 22)$ is obtained. Find the value of $a$ .

Show the worked solution

Step 1: Write the equation as Y = mX + c, with Y = xy and X = x²: what do you do?Multiply both sides by x.
Step 2: So the straight-line form is $xy = ax^{2} + b$
Step 3: Compare with Y = mX + c $a$ is the gradient and $b$ is the $Y$ -intercept.
Step 4: Gradient of the line through the two points: put them into the gradient formula $a = \dfrac{Y_2 - Y_1}{X_2 - X_1}$ $a = \dfrac{22 - 7}{4 - 1}$ $= 5$