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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 (m) is a measure of the steepness of a line, defined as 'rise over run'.

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A positive gradient indicates the line slopes upwards from left to right.

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A negative gradient indicates the line slopes downwards from left to right.

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Horizontal lines have a gradient of 0 because the change in y is zero.

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Vertical lines have an undefined gradient because the change in x is zero.

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Parallel lines have equal gradients (m1=m2m_1 = m_2).

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Perpendicular lines have gradients that are negative reciprocals of each other (m1imesm2=−1m_1 imes 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.