Skip to content# A-Level : Straight Line Equation | Practice Questions with Examples

## The Four Basic Forms of a Straight-Line Equation

## Slope-Intercept Form

**Formula **

## Understanding the Slope (Gradient)

## Understanding the Y-Intercept

## Point-Slope Form

**Formula **

## Two-Point Form

**Formula **

## Standard Form

**Formula **

## Summary

## Conclusion

## Practice Questions and Answers on Straight Line Equation

## Solutions

In this tutorial, we will explore one of the most fundamental concepts in algebra: the four basic forms that are important for A-Level : straight-line equation.

**Video Tutorial on A-Level : Straight Line Equation**

Watch this Video Tutorial as we explain all types of Straight Line Equations for A-Level Maths

The Four Basic forms of a Straight-Line Equation are essential for solving linear equations and understanding the geometry of straight lines in A-level mathematics.

- The Four Forms We will discuss are:

**Slope-Intercept Form****Point-Slope Form****Two-Point Form****Standard Form**

*They are very important in practicing questions for coordinate geometry as well.*

Here is one more link to practice a few extra questions: Maths Genie Straight Line Equation Questions

**The Slope-Intercept form is the most commonly used equation of a straight line.**

It expresses the line in terms of **its slope and y-intercept,** making it easy to graph and analyze.

The slope-intercept form is written as:

**y = m x + c**

Where:

**m**is the**slope**or**gradient**of the line.- C is the
**y-intercept**, the point where the line crosses the y-axis.

The **slope** indicates **how steep the line is. **

It is calculated as the ratio of the vertical change (rise) to the horizontal change (run) between two points on the line.

- Slope (m) = Rise / Run

- Alternatively, m = Change in y / Change in x

A * positive slope means the line rises* as it moves from left to right, while a

The **y-intercept (c)** is the value of y where the line crosses the y-axis **(when x = 0)**.

It indicates the starting point of the line on the y-axis.

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**Solved Example 1**

**Question: **Given the equation of the line **y = -3 x + 5**, find **the gradient and the Y-intercept**

**Solution: **

**Slope (m): -3****Y-intercept (c): 5**

**Interpretation**:

- The line crosses the y-axis at the point (0, 5).
- The negative slope of -3 indicates that for every unit increase in x, y decreases by 3 units. The line slopes downward from left to right.

**Steps to Draw the Graph:**

- Starting at the y-intercept plot the point (0, 5)
- Move down 3 units and right 1 unit to plot the next point.
- Connect the points to draw the line.

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** **

**Solved Example 2: Finding the Equation from a Graph**

**Question: What is the Equation of the Straight Line shown below**

**Solution: **

**Step 1: Find the Y-Intercept (c)**

- Look at where the line crosses the y-axis.
- For example, if the line crosses the y-axis at y = -1, then c = -1.

**Step 2: Calculate the Slope (m)**

- Choose two clear points on the line where it crosses grid lines for accuracy.
- For instance, point A at (x₁, y₁) and point B at (x₂, y₂).
- Calculate the change in y: Δy = y₂ – y₁.
- Calculate the change in x: Δx = x₂ – x₁.
- Compute the slope: m = Δy / Δx.

**Calculation:**

- From point A to point B, y decreases by 2 units (Δy = -2), and x increases by 1 unit (Δx = 1).
- Therefore, m = -2 / 1 = -2.
- Write the Equation: Substitute the slope and y-intercept into the slope-intercept form:
**y = m x + c.**

Using **m = -2** and **c = -1**, the equation is **y = -2 x – 1.**

**The point-slope form is useful when you know the slope of a line and one point through which it passes.**

The point-slope form is written as:

**y – y₁ = m ( x – x₁ )**

Where:

**m**is the**slope**or**gradient**of the line.- ( x₁, y₁ ) is the coordinated of a known point on the line

** **

** **

**Solved Example 3: Finding the Equation of a line with point-slope form**

**Question: A line has a gradient of -4/3 and it passes through the point (15,18).Find the equation of the line.**

**Solution: **

**Given:**

- Slope
**(m): -4/3** - Point:
**(15, 18)**

**Step 1**: Substitute the given values into the formula:

**y – 18 = (-4/3)(x – 15)**

**Step 2:** Simplify the Equation:

- To eliminate the fraction, multiply both sides by 3:

**3(y – 18) = -4(x – 15)**

- Expand both sides:

**3y – 54 = -4x + 60**

**Step 3:** Rewriting into Slope-Intercept Form (Optional):

- Rearrange the equation to solve for y:

**3y = -4x + 114 **

**y = (-4/3)x + 38**

The * two-point* form is ideal when you have two points on a line and need to find its equation.

The two-point form is written as:

**( y – y₁ )/( x – x₁ ) = ( y**_{2}– y₁ )/( x_{2}– x₁ )

Where:

**( x₁, y₁ )**and**( x₂, y₂ )**are two known points on the line

** **

** **

**Solved Example 4: Finding the Equation of a line with two-point form**

**Question: The point A (-3,5) and the point B (1,-15) lie on the line L. Find the equation of the line L.**

**Solution: **

**Given:**

- Point A: (-3, 5)
- Point B: (1, -15)

**Step 1:** Calculate the Slope (m):

**m = ( y₂ – y₁ ) / ( x₂ – x₁ ) **

**m = ( -15 – 5 ) / ( 1 – (-3) ) **

**m = ( -20 ) / ( 4 ) = -5 **

**Step 2:** Use One Point in the Point-Slope Form:

- Using point A (-3, 5):

**y – 5 = -5 ( x – (-3) )**

**y – 5 = -5 ( x + 3 ) **

**y – 5 = -5 x – 15 **

**y = -5 x – 10 **

**Interpretation: The equation y = -5 x – 10 represents the line passing through the points (-3, 5) and (1, -15).**

The** standard form** presents the linear equation in a general way, often used for various algebraic manipulations.

The standard form is written as:

**ax + by + c = 0**

Where:

**a**,**b**, and**c**are constants.**x**and**y**are variables.

** **

** **

**Solved Example 5: Converting to Slope-Intercept Form**

**Question: A line has equation 6x + 2y + 9 = 0**

** (a) Find the gradient of the line. **

**(b) Find where the line crosses the y-axis**

**Solution: **

**Given:**

- Equation: 6 x + 2 y + 9 = 0

**Step 1: **Solve for y:

**2 y = -6 x – 9**

**y = (-6 x – 9) / 2**

- Simplify:

**y = -3 x – 4.5**

**Step 2: **Identify the Slope and Y-Intercept:

**Slope (m)**: -3**Y-intercept (c)**: -4.5

**Interpretation**:

**The line has a slope of -3, indicating it falls steeply from left to right.****It crosses the y-axis at (0, -4.5).**

Understanding these four forms allows you to tackle various problems involving straight lines.

**1. Slope-Intercept Form (y = m x + c):**

- Quick identification of slope and y-intercept.
- Ideal for graphing and analyzing linear relationships.

**2. Point-Slope Form (y – y₁ = m ( x – x₁ )):**

- Useful when you know one point and the slope.
- Simplifies finding the equation of a line with limited information.

**3. Two-Point Form (( y – y₁ ) / ( x – x₁ ) = ( y₂ – y₁ ) / ( x₂ – x₁ )):**

- Best when you have two points.
- Calculates the slope and forms the equation simultaneously.

**4. Standard Form (a x + b y + c = 0):**

- General representation.
- Useful for solving systems of equations and certain algebraic manipulations.

Understanding and applying the four basic forms of straight-line equations is crucial in algebra and coordinate geometry. Each form serves a specific purpose, and knowing when to use each one simplifies solving linear equations and graphing lines.

- Use the
**slope-intercept form**when you need to quickly graph a line or identify its slope and y-intercept. - Use the
**point-slope form**when you know a point on the line and its slope. - Use the
**two-point form**when you have two points on the line. - Use the
**standard form**for general purposes or when dealing with systems of equations.

**Practice regularly** with different types of problems to strengthen your understanding and proficiency in working with straight-line equations.

**Question 1:** Find the equation of the line that passes through the point (6, -2) and has a slope of 3.

**Question 2:** Given the points (0, 7) and (4, 15), find the equation of the line passing through them.

**Question 3: **Convert the standard form equation 5x + y – 10 = 0 into slope-intercept form and identify the slope and y-intercept.

**Question 4:**Find the equation of the line perpendicular to y = 1/2x – 3 and passing through the point (4, 1).

**Question 5: **Determine the equation of the line that passes through the points (-2, -5) and (3, 10).

**Question 6: **A line has a slope of 4 and a y-intercept of −7. Write its equation and convert it to standard form.

**Question 7:**Determine the equation of the line passing through the points (2, -3) and (-4, 9)**.**

**Question 8: **Given the line 7y − 14x = 21, find its slope and y-intercept.

**Question 9: **Find the equation of a line with an undefined slope passing through the point (4, -2).

**Question 10: **Find the equation of a line with zero slope passing through the point (-3, 7).

**Question 1: **

**Step 1: **Use the Point-Slope Form:

**y − y _{1} = m (x − x_{1})**

**y − (−2) = 3 (x − 6)**

- Simplify:

**y + 2 = 3x − 18**

**Step 2: **Simplify the Equation: Rearrange:

**y = 3x − 20**

**Answer:** The equation of the line is **y = 3x − 20.**

**Question 2: **

**Step 1: **Calculate the Slope (m):

**m = ( y₂ – y₁ ) / ( x₂ – x₁ ) **

**m = ( 15 – 7 ) / ( 4 – 0 ) **

**m = ( 8 ) / ( 4 ) = 2**

**Step 2: **Use the Point-Slope Form with (0, 7):

**y − 7 = 2 (x − 0) **

- Simplify:

**y − 7 = 2x**

**Step 3: **Simplify the Equation:

**y = 2x + 7**

**Answer**: The equation of the line is **y = 2x + 7**

**Question 3: **

**Step 1: **Solve for y:

**y = −5x + 10**

**Step 2: **Identify the Slope and Y-Intercept:

- Slope (m): − 5
- Y-intercept (c): 10

**Answer**:

- The slope-intercept form is
**y = −5x + 10** - The slope is
**−5,**and the y-intercept is**10.**

**Question 4: **

**Step 1: **Find the Slope of the Given Line:

- The slope of the given line is m
_{1}= 1/2

**Step 2: **Find the Slope of the Perpendicular Line:

**m**(negative reciprocal of 1/2)_{2} = −2

**Step 3: **Use the Point-Slope Form with (4, 1):

**y − 1 = −2 (x − 4) **

- Simplify:

**y − 1 = −2x + 8**

**Step 4: **Simplify the Equation:

**y = −2x + 9**

**Answer**:

- The equation of the perpendicular line is
**y = −2x + 9**

**Question 5: **

**Step 1: **Calculate the Slope (m):

**m = ( y₂ – y₁ ) / ( x₂ – x₁ ) **

**m = ( 10 – (-5) ) / ( 3 – (-2) ) **

**m = 15 / 5 = 3**

**Step 2: **Use the Point-Slope Form with (-2, -5):

**y − (−5) = 3 (x − (−2)) **

- Simplify:

**y + 5 = 3 (x + 2)**

**Step 3: **Simplify the Equation:

**y + 5 = 3x + 6 **

**y = 3x + 1**

**Answer**:

- The equation of the line is
**y = 3x + 1**

**Question 6: **

**Step 1: **Write the Slope-Intercept Form:

**4x − y = 7**

**Step 2: **Convert to Standard Form:

- Subtract 4x from both sides:

**−4x + y = −7**

- Multiply both sides by −1:

**4x − y = 7**

**Answer:**

- Slope-intercept form:
**y = 4x − 7** - Standard form:
**4x − y = 7**

**Question 7:**

**Step 1: **Calculate the Slope (m):

**m = ( 9 – (-3) ) / ( -4 – 2 ) **

**= (12) / (-6) = -2**

**Step 2: **Use the Point-Slope Form with one of the points:

- Using (2, -3):

**y – (-3) = -2 ( x – 2 )**

- Simplify:

**y + 3 = -2 x + 4**

**Step 3: **Simplify the Equation:

- Rearrange to slope-intercept form:

**y = -2 x + 1**

**Answer**:

- The equation of the line is
**y = -2 x + 1.**

**Question 8: **

**Step 1: **Convert to Slope-Intercept Form:

**7y = 14x + 21 **

**y = 2x + 3**

**Step 2: **Identify the Slope and Y-Intercept:

- Slope (m): 2
- Y-intercept (c): 3

**Answer**:

- The slope is
**2,**and the y-intercept is**3.**

**Question 9: **

**Step 1: **An undefined slope indicates a vertical line.

**Step 2: **Equation of a Vertical Line:

**x = 4**

**Answer**:

- The equation of the line is
**x = 4**

**Question 10: **

**Step 1: **A zero slope indicates a horizontal line.

**Step 2: **Equation of a Horizontal Line:

**y = 7**

**Answer**:

- The equation of the line is
**y = 7**

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