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.

**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 $m_1 = m_2$, where $m_1$ and $m_2$ 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 $m_1 \times m_2 = -1$.

**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 $R$, the reflection of $(-1,3)$ in the line $3y + 2x = 33.$

**Solution**:

**1. Equation of Perpendicular Line**:

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

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

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

**2. Perpendicular Line Equation**:

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

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

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

**3. Intersection Point**:

Equate: $11 - \frac{2}{3}x = \frac{3x + 9}{2}$.

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

**4. Reflection Point ( R )**:

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

**Example 2: Finding the Equation of a Parallel Line**

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

**Solution**:

**1. Identify the Gradient of the Original Line**:

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

**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: $y - y_1 = m(x - x_1)$

- Substituting $(4,1)$ for $(x_1, y_1)$ and $2$ for $m$: $y - 1 = 2(x - 4)$
- Expanding and rearranging to the slope-intercept form $y = mx + c :$ $y = 2x - 7$

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

**Example 3: Determining a Perpendicular Line**

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

**Solution**:

**1. Determine the Gradient of the Original Line**:

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

**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 $-\frac{1}{-1/3} = 3$.

**3. Form the Equation of the Perpendicular Line**:

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

- Substitute $(6,-2)$ for $(x_1, y_1)$ and $3$ for $m$: $y + 2 = 3(x - 6)$
- Simplifying into the slope-intercept form: $y = 3x - 20$

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

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.