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Coordinate Geometry - Gradient of Straight Lines

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The gradient (or slope) mm of a straight line measures its steepness and direction. It is defined as the ratio of the 'rise' (vertical change) to the 'run' (horizontal change) between any two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) on the line: m=change in ychange in x=y2−y1x2−x1m = \frac{\text{change in } y}{\text{change in } x} = \frac{y_2 - y_1}{x_2 - x_1}

Graph showing rise and run between two points to calculate gradient.
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A positive gradient means the line slopes upwards from left to right. A negative gradient means the line slopes downwards from left to right. A horizontal line has a gradient of 00, while a vertical line has an undefined gradient.

Visual comparison of positive and negative gradients.
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In the equation y=mx+cy = mx + c, mm represents the gradient and cc represents the yy-intercept (where the line crosses the yy-axis).

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Parallel lines have identical gradients (m1=m2m_1 = m_2). Perpendicular lines have gradients that are negative reciprocals of each other (m1×m2=−1m_1 \times m_2 = -1).

📐Formulae

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}

y=mx+cy = mx + c

mparallel=mm_{parallel} = m

mperpendicular=−1mm_{perpendicular} = -\frac{1}{m}

💡Examples

Problem 1:

Find the gradient of the line passing through the points A(1,−3)A(1, -3) and B(4,6)B(4, 6).

Solution:

m=6−(−3)4−1=93=3m = \frac{6 - (-3)}{4 - 1} = \frac{9}{3} = 3

Explanation:

Label the points as (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2). Substitute these values into the gradient formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} and simplify.

Problem 2:

Determine the gradient of the line given by the equation 2y+5x=102y + 5x = 10.

Solution:

2y=−5x+10⇒y=−52x+52y = -5x + 10 \Rightarrow y = -\frac{5}{2}x + 5. Therefore, m=−2.5m = -2.5.

Explanation:

Rearrange the equation into the standard form y=mx+cy = mx + c. The gradient is the coefficient of xx once yy is isolated.

Problem 3:

Line L1L_1 has the equation y=4x−2y = 4x - 2. Find the gradient of a line L2L_2 that is perpendicular to L1L_1.

Solution:

m1=4m_1 = 4, so m2=−14m_2 = -\frac{1}{4}.

Explanation:

Identify the gradient of the first line (m1=4m_1 = 4). Since the lines are perpendicular, the product of their gradients must be −1-1. Thus, m2=−1/4m_2 = -1 / 4.

Problem 4:

Calculate the gradient of the line segment shown in the coordinate plane that connects the points P(2,1)P(2, 1) and Q(6,4)Q(6, 4).

Line segment PQ on a coordinate grid from (2,1) to (6,4).

Solution:

  1. Identify the coordinates: (x1,y1)=(2,1)(x_1, y_1) = (2, 1) and (x2,y2)=(6,4)(x_2, y_2) = (6, 4).
  2. Use the gradient formula: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}.
  3. Substitute the values: m=4−16−2m = \frac{4 - 1}{6 - 2}.
  4. Simplify: m=34m = \frac{3}{4} or 0.750.75.

Explanation:

The gradient is found by dividing the vertical distance (3 units) by the horizontal distance (4 units) between the two points.

Problem 5:

Identify the gradient of the line represented by the function f(x)=−2x+3f(x) = -2x + 3 and sketch its direction.

Graph of the line y = -2x + 3 showing a downward slope.

Solution:

  1. Compare the given equation y=−2x+3y = -2x + 3 with the standard form y=mx+cy = mx + c.
  2. The coefficient of xx is the gradient, so m=−2m = -2.
  3. The negative sign indicates that for every 1 unit move to the right, the line goes down by 2 units.

Explanation:

By putting the linear equation into the form y=mx+cy = mx + c, the gradient is immediately visible as the multiplier of xx.