Gradient of a Line Revision Notes for Grade 10 NSC Matric Mathematics

Gradient of a Line

What is gradient?

Gradient is a measure that tells us how steep a line is. It describes the slope or steepness of the line joining two points. Think of gradient as how much a line rises or falls as you move from left to right.

The gradient shows us the ratio of vertical change to horizontal change. When we look at different lines, some are steeper than others - this is what gradient measures.

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The concept of gradient is fundamental to understanding how lines behave in coordinate geometry. A positive gradient means the line slopes upward from left to right, while a negative gradient means it slopes downward.

Understanding the gradient formula

To find the exact gradient of a line, we use a right-angled triangle formed by any two points on the line. The gradient equals the length of the vertical side divided by the length of the horizontal side.

For any two points A(x₁; y₁) and B(x₂; y₂), the vertical change is the difference in y-values, and the horizontal change is the difference in x-values.

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Key formula: $m = \frac{y_2 - y_1}{x_2 - x_1} \text{ or } \frac{y_1 - y_2}{x_1 - x_2}$

Important: Always be consistent with your order. If you use (y₂ - y₁) on top, use (x₂ - x₁) on the bottom.

Calculating gradient between two points

Let's work through finding the gradient step by step using the rise over run method.

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Worked Example: Basic gradient calculation

Question: Find the gradient of the line between points E(2; 5) and F(-3; 9).

Solution: Step 1: Assign coordinates

Let E be (x₁; y₁) = (2; 5) and F be (x₂; y₂) = (-3; 9)

Step 2: Apply the gradient formula

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Step 3: Substitute the values

$m_{EF} = \frac{9 - 5}{-3 - 2} = \frac{4}{-5} = -\frac{4}{5}$

Answer: The gradient of EF is $-\frac{4}{5}$

Finding missing coordinates using gradient

When we know the gradient and one complete coordinate, we can find missing coordinate values by applying the gradient formula and solving algebraically.

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Worked Example: Finding a missing coordinate

Question: Given G(7; -9) and H(x; 0), with gradient = 3, find x.

Solution: Step 1: Assign known values

G(x₁; y₁) = (7; -9), H(x₂; y₂) = (x; 0), and m = 3

Step 2: Apply the gradient formula

$3 = \frac{0 - (-9)}{x - 7}$

Step 3: Solve for x

$3 = \frac{9}{x - 7}$

$3(x - 7) = 9$

$x - 7 = 3$

$x = 10$

Answer: The coordinates of H are (10; 0)

Straight lines and their equations

Definition: A straight line is a set of points with a constant gradient between any two of the points.

When points lie on a straight line, the gradient between any pair of points is always the same. This leads us to the general equation of a straight line.

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General formula for a straight line: $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$

This can be simplified to: $y - y_1 = m(x - x_1)$

Standard form: $y = mx + c$ where m is the gradient and c is the y-intercept.

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Worked Example: Finding the equation of a straight line

Question: Find the equation of the straight line through P(-1; -5) and Q(5; 4).

Solution:

Step 1: Assign coordinates

P(x₁; y₁) = (-1; -5), Q(x₂; y₂) = (5; 4)

Step 2: Use the general line formula

$\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$

Step 3: Substitute and simplify

$\frac{y - (-5)}{x - (-1)} = \frac{4 - (-5)}{5 - (-1)}$

$\frac{y + 5}{x + 1} = \frac{9}{6} = \frac{3}{2}$

Step 4: Rearrange to standard form

$2(y + 5) = 3(x + 1)$

$2y + 10 = 3x + 3$

$2y = 3x - 7$

$y = \frac{3}{2}x - \frac{7}{2}$

Answer: The equation is $y = \frac{3}{2}x - \frac{7}{2}$

Parallel lines

Key concept: Two lines that run parallel to each other always maintain the same distance apart and have equal gradients.

For parallel lines: $m_1 = m_2$

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Remember: If two lines have the same gradient, they will never intersect (unless they are the same line). This is the defining property of parallel lines in coordinate geometry.

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Worked Example: Proving lines are parallel

Question: Prove that line AB with A(0; 2) and B(2; 6) is parallel to line CD with equation 2x - y = 2.

Solution:

Step 1: Find gradient of line AB

$m_{AB} = \frac{6 - 2}{2 - 0} = \frac{4}{2} = 2$

Step 2: Find gradient of line CD

Rewrite 2x - y = 2 in standard form: y = 2x - 2

Therefore, $m_{CD} = 2$

Step 3: Compare gradients

Since $m_{AB} = m_{CD} = 2$ , the lines are parallel.

Perpendicular lines

Key concept: If two lines intersect perpendicularly, the product of their gradients equals -1.

For perpendicular lines: $m_1 \times m_2 = -1$

This means perpendicular lines have gradients that are negative inverses of each other.

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Common mistake to avoid: Don't confuse perpendicular lines with parallel lines. Perpendicular lines intersect at 90°, while parallel lines never intersect.

If one line has gradient $\frac{2}{3}$ , a perpendicular line will have gradient $-\frac{3}{2}$ .

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Worked Example: Finding coordinates using perpendicular condition

Question: Line AB is perpendicular to line CD. Find y given A(2; -3), B(-2; 6), C(4; 3) and D(7; y).

Solution:

Step 1: Use the perpendicular relationship

$m_{AB} \times m_{CD} = -1$

Step 2: Calculate gradient of AB

$m_{AB} = \frac{6 - (-3)}{-2 - 2} = \frac{9}{-4} = -\frac{9}{4}$

Step 3: Use the perpendicular condition

$-\frac{9}{4} \times \frac{y - 3}{7 - 4} = -1$

$-\frac{9}{4} \times \frac{y - 3}{3} = -1$

Step 4: Solve for y

$\frac{y - 3}{3} = \frac{4}{9}$

$y - 3 = \frac{4}{3}$

$y = 3 + \frac{4}{3} = \frac{13}{3} = 4\frac{1}{3}$

Answer: The coordinates of D are $(7; 4\frac{1}{3})$

Special cases: horizontal and vertical lines

Horizontal lines

A line parallel to the x-axis is horizontal and has gradient = 0. This happens because there's no vertical change.

$m = \frac{\text{change in y}}{\text{change in x}} = \frac{0}{\text{change in x}} = 0$

Vertical lines

A line parallel to the y-axis is vertical and has undefined gradient. This happens because there's no horizontal change, leading to division by zero.

$m = \frac{\text{change in y}}{\text{change in x}} = \frac{\text{change in y}}{0} = \text{undefined}$

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Memory aid: Think of horizontal lines as "flat" (no rise, so gradient = 0) and vertical lines as "standing up" (no run, so gradient is undefined).

Points on a line (collinearity)

Collinear points are points that lie on the same straight line. We can prove points are collinear by showing they have the same gradient between any pair of points.

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Worked Example: Proving collinearity using gradient method

Question: Prove that A(-3; 3), B(0; 5) and C(3; 7) are on a straight line.

Solution:

Step 1: Calculate gradients between different pairs

$m_{AB} = \frac{5 - 3}{0 - (-3)} = \frac{2}{3}$

$m_{BC} = \frac{7 - 5}{3 - 0} = \frac{2}{3}$

Step 2: Compare gradients

Since $m_{AB} = m_{BC} = \frac{2}{3}$ , the points A, B and C lie on a straight line.

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Key Points to Remember:

Gradient formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$ - always be consistent with order
Parallel lines: Have equal gradients ( $m_1 = m_2$ )
Perpendicular lines: Have gradients whose product is -1 ( $m_1 \times m_2 = -1$ )
Horizontal lines: Have gradient = 0 (no vertical change)
Vertical lines: Have undefined gradient (no horizontal change)
Collinear points: Have the same gradient between any pair of points
Memory aids: "Rise over run" for gradient, "Same gradient = parallel", "Product = -1 = perpendicular"

Gradient of a Line (Grade 10 NSC Matric Mathematics): Revision Notes