Equation of a straight line Revision Notes for AQA GCSE Further Maths

Equation of a straight line

Understanding different forms of line equations

When working with straight lines in coordinate geometry, there are several ways to write their equations. Each form is useful in different situations, and understanding when to use each one will help you solve problems more efficiently.

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Choosing the right form of equation depends on what information you're given in the problem. This flexibility is what makes coordinate geometry so powerful!

The most common forms you need to know are:

Gradient-intercept form: $y = mx + c$
Point-slope form: $y - y\_1 = m(x - x\_1)$
Two-point form: Used when you have two coordinates

Method 1: Gradient-intercept form ( $y = mx + c$ )

This is the most recognisable form of a line equation. Here, m represents the gradient (how steep the line is), and c represents where the line crosses the y-axis (the y-intercept).

This form is particularly useful when:

You know the gradient and where the line crosses the y-axis
You need to quickly identify the gradient and y-intercept from an equation
You're sketching graphs

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Remember: In $y = mx + c$ , the coefficient of $x$ is always the gradient, and the constant term is always the y-intercept.

Method 2: Point-slope form

When you know the gradient of a line and one point it passes through, you can use the point-slope form: $y - y\_1 = m(x - x\_1)$

In this formula:

m is the gradient
(x₁, y₁) is the known point the line passes through

This method is especially helpful when you don't immediately know the y-intercept but have other information about the line.

Method 3: Two-point form

When you have two points that a line passes through, you can find its equation using: $\frac{y - y\_1}{x\_2 - x\_1} = \frac{y\_2 - y\_1}{x\_2 - x\_1}$

This formula essentially finds the gradient first, then uses it to create the equation. It's based on the fact that the gradient between any two points on a straight line is constant.

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The two-point form is particularly useful in geometry problems where you're given coordinates of vertices or intersection points.

Worked example: Finding the line of symmetry

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Worked Example: Finding the Line of Symmetry

Problem: An isosceles triangle has vertices at A(2,3), B(8,5) and C(4,9). Find the equation of the line of symmetry.

Solution approach: The line of symmetry in an isosceles triangle runs from one vertex to the midpoint of the opposite side.

Step 1: Find the midpoint of BC Midpoint D = $\left(\frac{8+4}{2}, \frac{5+9}{2}\right) = (6,7)$

Step 2: Use the two-point form with A(2,3) and D(6,7) $\frac{y - 3}{7 - 3} = \frac{x - 2}{6 - 2}$

Step 3: Simplify $\frac{y - 3}{4} = \frac{x - 2}{4}$ $y - 3 = x - 2$

Answer: $y = x + 1$

Worked example: Perpendicular lines and area calculations

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Worked Example: Perpendicular Lines and Area Calculations

Problem: The line $5x - 4y = 40$ intersects the x-axis at P and the y-axis at Q. Find the area of triangle OPQ and the equation of the line through Q perpendicular to PQ.

Step 1: Find the intercepts

For x-intercept (point P): Set $y = 0$ → $5x = 40$ → $x = 8$ , so P(8,0)
For y-intercept (point Q): Set $x = 0$ → $-4y = 40$ → $y = -10$ , so Q(0,-10)

Step 2: Calculate area Area of triangle OPQ = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 10 = 40$ units²

Step 3: Find the perpendicular line First, find the gradient of PQ: Gradient = $\frac{0-(-10)}{8-0} = \frac{10}{8} = \frac{5}{4}$

For perpendicular lines, the gradients multiply to give -1: Perpendicular gradient = $-\frac{4}{5}$

Since the line passes through Q(0,-10): Answer: $y = -\frac{4}{5}x - 10$

Understanding parallel and perpendicular lines

Understanding the relationship between line gradients is crucial for solving many coordinate geometry problems.

Parallel lines:

Have the same gradient
Never intersect
Have equations like $y = 2x + 3$ and $y = 2x - 1$

Perpendicular lines:

Their gradients multiply to give -1
They meet at right angles
If one line has gradient $m$ , the perpendicular line has gradient $-\frac{1}{m}$

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Key Relationship: For perpendicular lines with gradients $m\_1$ and $m\_2$ : $m\_1 \times m\_2 = -1$

This means if you know one gradient, you can find the perpendicular gradient by taking the negative reciprocal.

Common exam techniques and tips

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Essential Problem-Solving Strategies

Always identify what form of equation is most suitable for the given information
When finding intercepts, substitute $x = 0$ for y-intercept and $y = 0$ for x-intercept
For perpendicular lines, remember that if one gradient is $m$ , the other is $-\frac{1}{m}$
Sketch graphs when possible - visual representation often makes problems clearer
Check your answer by substituting known points back into your equation

The key to success in coordinate geometry is choosing the right method for each problem. Don't be afraid to try a different approach if your first attempt becomes too complicated.

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Key Points to Remember:

The three main forms of line equations are $y = mx + c$ , $y - y\_1 = m(x - x\_1)$ , and the two-point form
Perpendicular lines have gradients that multiply to give -1, while parallel lines have identical gradients
To find intercepts, substitute zero for the opposite variable ( $x = 0$ for y-intercept, $y = 0$ for x-intercept)
The midpoint formula $\left(\frac{x\_1 + x\_2}{2}, \frac{y\_1 + y\_2}{2}\right)$ is essential for problems involving symmetry
Always choose the most appropriate method based on the information given in the problem

Equation of a straight line (AQA GCSE Further Maths): Revision Notes