TutorChase logo
Decorative notebook illustration
CIE A-Level Maths Study Notes

1.3.3 Gradients of Parallel and Perpendicular Lines

In this section, we delve into the integral component of Coordinate Geometry: the relationship between the gradients of parallel and perpendicular lines.

Understanding Gradients

The concept of the gradient, often referred to as the slope in the context of a line in coordinate geometry, is a fundamental concept in mathematics, particularly in calculus and geometry. It provides a quantitative measure of the steepness and direction of a line.

Definition:

The gradient of a line is a measure of how much the line rises (or falls) vertically for each unit of horizontal movement. In simpler terms, it tells us how steep the line is.

Direction: The gradient can be positive, negative, zero, or undefined.

  • Positive Gradient: The line slopes upwards as it moves from left to right.
  • Negative Gradient: The line slopes downwards as it moves from left to right.
  • Zero Gradient: The line is horizontal, indicating no vertical change.
  • Undefined Gradient: The line is vertical, indicating no horizontal change.
gradient

Special Gradients

Parallel Lines

  • Definition: Lines that do not intersect and maintain a constant distance apart.
  • Gradient Relationship: If two lines are parallel, their gradients are equal. Expressed as m1=m2m_1 = m_2, where m1m_1 and m2m_2 represent the gradients of the two lines.

Perpendicular Lines

  • Definition: Lines that intersect at a right angle (90 degrees).
  • Gradient Relationship: The gradients of perpendicular lines are negative reciprocals of each other, denoted as m1×m2=1m_1 \times m_2 = -1.

Gradient at a Point on a Curve

  • The gradient at any point on a curve corresponds to the gradient of the tangent at that point.

Gradient of a Tangent at a Curve's Vertex

  • At the vertex of a curve (stationary point), the gradient of the tangent is zero.

Solving Geometrical Problems

Example 1: Finding the Reflection of a Point

Find the coordinates of point RR, the reflection of (1,3)(-1,3) in the line 3y+2x=33.3y + 2x = 33.

Solution:

1. Equation of Perpendicular Line:

Start with 3y+2x=333y + 2x = 33, rearranged to.

The gradient (m)(m) is 23-\frac{2}{3}.

For the perpendicular line, m1=32m_1 = \frac{3}{2} (since m×m1=1.m \times m_1 = -1.)

2. Perpendicular Line Equation:

General form: y=32x+cy = \frac{3}{2}x + c.

Substituting (1,3)(-1,3), find c=92c = \frac{9}{2}.

Final Equation: 2y=3x+92y = 3x + 9.

3. Intersection Point:

Equate: 1123x=3x+9211 - \frac{2}{3}x = \frac{3x + 9}{2}.

Solve to get x=3,y=9x = 3, y = 9

4. Reflection Point ( R ):

Vector change from (1,3)(-1,3) to (3,9)(3,9) gives RR at (7,15)(7,15).

Example 2: Finding the Equation of a Parallel Line

Determine the equation of the line that is parallel to y=2x+3y = 2x + 3and passes through the point (4,1)(4,1).

Solution:

1. Identify the Gradient of the Original Line:

The gradient of the given line, y=2x+3y = 2x + 3, is 2 (the coefficient of x x).

2. Gradient of the Parallel Line:

Parallel lines have the same gradient.

Thus, the gradient of our new line is also 2.

3. Formulate the Equation of the Parallel Line:

Using the point-slope form of a line equation: yy1=m(xx1)y - y_1 = m(x - x_1)

  • Substituting (4,1)(4,1) for (x1,y1)(x_1, y_1) and 22 for mm: y1=2(x4)y - 1 = 2(x - 4)
  • Expanding and rearranging to the slope-intercept form y=mx+c:y = mx + c : y=2x7y = 2x - 7

The equation of the line parallel to y=2x+3y = 2x + 3 and passing through 4,14,1 is y=2x7y = 2x - 7 .


Example 3: Determining a Perpendicular Line

Find the equation of the line that is perpendicular to y=13x+5y = -\frac{1}{3}x + 5 and goes through the point (6,2)(6,-2).

Solution:

1. Determine the Gradient of the Original Line:

The gradient of the line y=13x+5y = -\frac{1}{3}x + 5 is 13-\frac{1}{3}.

2. Gradient of the Perpendicular Line:

The gradient of a line perpendicular to another is the negative reciprocal of the original line's gradient.

Therefore, the gradient of our new line is 11/3=3-\frac{1}{-1/3} = 3.

3. Form the Equation of the Perpendicular Line:

Using the point-slope form: yy1=m(xx1) y - y_1 = m(x - x_1)

  • Substitute (6,2)(6,-2) for (x1,y1)(x_1, y_1) and 33 for mm: y+2=3(x6)y + 2 = 3(x - 6)
  • Simplifying into the slope-intercept form: y=3x20y = 3x - 20

The equation of the line perpendicular to y=13x+5y = -\frac{1}{3}x + 5 and passing through (6,2)(6,-2) is y=3x20 y = 3x - 20.


Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
LinkedIn
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

Hire a tutor

Please fill out the form and we'll find a tutor for you.

1/2 About yourself
Still have questions?
Let's get in touch.