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Coordinate Geometry - Parallel and Perpendicular Lines

Grade 12IGCSE

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

🔑Concepts

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Gradient (Slope): The measure of the steepness of a line, calculated as the change in y divided by the change in x.

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Parallel Lines: Two lines are parallel if and only if they have the same gradient (m1=m2m_1 = m_2). They never intersect.

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Perpendicular Lines: Two lines are perpendicular if the product of their gradients is −1-1 (m1imesm2=−1m_1 imes m_2 = -1). This is also known as the negative reciprocal.

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Equation Forms: Understanding how to extract the gradient from the slope-intercept form (y=mx+cy = mx + c) and the general form (ax+by+c=0ax + by + c = 0).

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Point-Gradient Formula: A method to find the equation of a line when given a specific point (x1,y1)(x_1, y_1) and the gradient mm.

📐Formulae

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

y=mx+cy = mx + c

y−y1=m(x−x1)y - y_1 = m(x - x_1)

Parallel: m1=m2m_1 = m_2

Perpendicular: m2=−1m1m_2 = -\frac{1}{m_1}

💡Examples

Problem 1:

Find the equation of the line that passes through the point (4,−2)(4, -2) and is parallel to the line y=3x+7y = 3x + 7.

Solution:

y=3x−14y = 3x - 14

Explanation:

Since the lines are parallel, they share the same gradient. The gradient of the given line is m=3m = 3. Using the point-gradient formula y−y1=m(x−x1)y - y_1 = m(x - x_1) with point (4,−2)(4, -2), we get: y−(−2)=3(x−4)⇒y+2=3x−12⇒y=3x−14y - (-2) = 3(x - 4) \Rightarrow y + 2 = 3x - 12 \Rightarrow y = 3x - 14.

Problem 2:

Find the equation of the line perpendicular to 2x−5y=102x - 5y = 10 that passes through the point (0,3)(0, 3).

Solution:

y=−2.5x+3y = -2.5x + 3

Explanation:

First, find the gradient of the given line by rearranging to y=mx+cy = mx + c: 5y=2x−10⇒y=25x−25y = 2x - 10 \Rightarrow y = \frac{2}{5}x - 2. The gradient m1=25m_1 = \frac{2}{5}. The perpendicular gradient m2m_2 is the negative reciprocal: m2=−52=−2.5m_2 = -\frac{5}{2} = -2.5. Since the line passes through (0,3)(0, 3), the y-intercept c=3c = 3. Thus, y=−2.5x+3y = -2.5x + 3.

Problem 3:

The line L1L_1 passes through (1,2)(1, 2) and (3,5)(3, 5). The line L2L_2 is perpendicular to L1L_1 and passes through (5,1)(5, 1). Determine where L2L_2 crosses the x-axis.

Solution:

x = 5.75

Explanation:

  1. Find gradient of L1L_1: m1=5−23−1=32m_1 = \frac{5-2}{3-1} = \frac{3}{2}. 2. Find gradient of L2L_2: m2=−23m_2 = -\frac{2}{3}. 3. Equation of L2L_2: y−1=−23(x−5)⇒y=−23x+103+1⇒y=−23x+133y - 1 = -\frac{2}{3}(x - 5) \Rightarrow y = -\frac{2}{3}x + \frac{10}{3} + 1 \Rightarrow y = -\frac{2}{3}x + \frac{13}{3}. 4. X-axis crossing (where y=0y=0): 0=−23x+133⇒23x=133⇒2x=13⇒x=6.50 = -\frac{2}{3}x + \frac{13}{3} \Rightarrow \frac{2}{3}x = \frac{13}{3} \Rightarrow 2x = 13 \Rightarrow x = 6.5.