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Straight Lines - Slope of a Line

Grade 11CBSE

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

🔑Concepts

The inclination of a line is the angle θ\theta made by the line with the positive direction of the x-axis measured in the anti-clockwise direction. Visually, if the line tilts to the right, θ\theta is acute (0<θ<900 < \theta < 90^{\circ}), and if it tilts to the left, θ\theta is obtuse (90<θ<18090^{\circ} < \theta < 180^{\circ}).

The slope or gradient of a non-vertical line is denoted by mm and is defined as m=tanθm = \tan \theta, where θ\theta is the inclination. For a horizontal line, θ=0\theta = 0^{\circ} so m=0m = 0; for a vertical line, θ=90\theta = 90^{\circ} and the slope is undefined.

The slope of a line passing through two distinct points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is calculated as the ratio of the vertical change (rise) to the horizontal change (run). On a graph, this is represented by the steepness of the segment connecting the two points.

Two non-vertical lines are parallel if and only if their slopes are equal (m1=m2m_1 = m_2). Visually, parallel lines run in the same direction and never meet, regardless of how far they are extended.

Two non-vertical lines are perpendicular if and only if the product of their slopes is 1-1 (m1m2=1m_1 \cdot m_2 = -1). Geometrically, this means the lines intersect at a right angle (9090^{\circ}).

Three points A,B,A, B, and CC are collinear (lie on the same straight line) if the slope of segment ABAB is equal to the slope of segment BCBC. If you plot these points, they will form a single continuous path without any bends.

The acute angle θ\theta between two intersecting lines with slopes m1m_1 and m2m_2 is determined by the relative difference in their steepness. This relationship is captured using the tangent of the angle between them.

📐Formulae

m=tanθm = \tan \theta

m=y2y1x2x1,x1x2m = \frac{y_2 - y_1}{x_2 - x_1}, x_1 \neq x_2

m1=m2 (Condition for parallel lines)m_1 = m_2 \text{ (Condition for parallel lines)}

m1m2=1 (Condition for perpendicular lines)m_1 \cdot m_2 = -1 \text{ (Condition for perpendicular lines)}

tanθ=m2m11+m1m2 (Angle between two lines)\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \text{ (Angle between two lines)}

💡Examples

Problem 1:

Find the slope of a line passing through the points A(3,2)A(3, -2) and B(1,4)B(-1, 4).

Solution:

  1. Identify the coordinates: (x1,y1)=(3,2)(x_1, y_1) = (3, -2) and (x2,y2)=(1,4)(x_2, y_2) = (-1, 4).
  2. Use the slope formula: m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}.
  3. Substitute the values: m=4(2)13m = \frac{4 - (-2)}{-1 - 3}.
  4. Simplify: m=4+24=64m = \frac{4 + 2}{-4} = \frac{6}{-4}.
  5. Final result: m=32m = -\frac{3}{2}.

Explanation:

To find the slope between two points, we calculate the change in y-coordinates divided by the change in x-coordinates. A negative slope indicates the line falls from left to right.

Problem 2:

If the angle between two lines is π4\frac{\pi}{4} and the slope of one of the lines is 12\frac{1}{2}, find the slope of the other line.

Solution:

  1. Let m1=12m_1 = \frac{1}{2} and the unknown slope be m2m_2.
  2. The angle θ=π4\theta = \frac{\pi}{4}, so tanθ=tan(π4)=1\tan \theta = \tan(\frac{\pi}{4}) = 1.
  3. Use the formula: 1=m21/21+(1/2)m21 = \left| \frac{m_2 - 1/2}{1 + (1/2)m_2} \right|.
  4. Remove the absolute value: m21/21+m2/2=1\frac{m_2 - 1/2}{1 + m_2/2} = 1 or m21/21+m2/2=1\frac{m_2 - 1/2}{1 + m_2/2} = -1.
  5. Solve Case 1: m212=1+m22    m22=32    m2=3m_2 - \frac{1}{2} = 1 + \frac{m_2}{2} \implies \frac{m_2}{2} = \frac{3}{2} \implies m_2 = 3.
  6. Solve Case 2: m212=1m22    3m22=12    m2=13m_2 - \frac{1}{2} = -1 - \frac{m_2}{2} \implies \frac{3m_2}{2} = -\frac{1}{2} \implies m_2 = -\frac{1}{3}.
  7. The possible slopes are 33 or 13-\frac{1}{3}.

Explanation:

We use the tangent formula for the angle between two lines. Since the formula involve absolute values, there are typically two possible lines (slopes) that could form the given angle with the reference line.