Straight Lines Class 11 NCERT: A Complete Guide

Understanding the Slope (Gradient) of a Line

The most fundamental property of a non-vertical straight line is its 'slope' or 'gradient'. It's a number that measures the line's direction and steepness. The slope, usually denoted by 'm', is defined as the tangent of the angle θ (theta) that the line makes with the positive direction of the x-axis, measured anti-clockwise. So, m = tan θ.

• If θ is acute (0° < θ < 90°), the slope 'm' is positive, and the line rises from left to right.
• If θ is obtuse (90° < θ < 180°), the slope 'm' is negative, and the line falls from left to right.
• If θ = 0°, the line is horizontal, and m = tan 0° = 0.
• If θ = 90°, the line is vertical. Its slope is undefined because tan 90° is undefined.

We can also calculate the slope if we know any two points on the line, say P(x₁, y₁) and Q(x₂, y₂). The slope 'm' is the ratio of the change in the y-coordinate (rise) to the change in the x-coordinate (run). The formula is: m = (y₂ - y₁) / (x₂ - x₁). This formula is extremely useful and frequently applied in problems.

Various Forms of the Equation of a Line

Point-Slope Form

Used when you know the slope (m) and a single point (x₁, y₁) on the line. The equation is: y - y₁ = m(x - x₁).

Two-Point Form

Used when you know two points (x₁, y₁) and (x₂, y₂) on the line. The equation is derived from the point-slope form: y - y₁ = [(y₂ - y₁) / (x₂ - x₁)](x - x₁).

Slope-Intercept Form

Used when you know the slope (m) and the y-intercept (c). The y-intercept is the point where the line crosses the y-axis. The equation is: y = mx + c.

Intercept Form

Used when you know the x-intercept (a) and the y-intercept (b). The equation is: x/a + y/b = 1.

Normal Form

Represents the line in terms of the length of the perpendicular (p) from the origin to the line and the angle (ω) this perpendicular makes with the positive x-axis. The equation is: x cos ω + y sin ω = p.

Worked Examples on Straight Lines

• Example 1: Find the equation of the line passing through the point (-2, 3) and perpendicular to the line 3x - 4y + 7 = 0. Step 1: Find the slope of the given line. First, we rearrange the equation 3x - 4y + 7 = 0 into the slope-intercept form (y = mx + c). 4y = 3x + 7 y = (3/4)x + 7/4 The slope of this line (m₁) is 3/4. Step 2: Find the slope of the required line. Let the slope of the required line be m₂. Since the lines are perpendicular, the product of their slopes is -1. m₁ m₂ = -1 (3/4) m₂ = -1 m₂ = -4/3 Step 3: Use the point-slope form to find the equation. The required line passes through the point (-2, 3) and has a slope of -4/3. Using y - y₁ = m(x - x₁): y - 3 = (-4/3)(x - (-2)) y - 3 = (-4/3)(x + 2) 3(y - 3) = -4(x + 2) 3y - 9 = -4x - 8 4x + 3y - 1 = 0 Final Answer: The equation of the required line is 4x + 3y - 1 = 0.
• Example 2: Reduce the equation √3x + y - 8 = 0 into normal form. Find the values of p and ω. Step 1: Compare the given equation with the general form. The general equation is Ax + By + C = 0. Here, A = √3, B = 1, C = -8. To convert to normal form (x cos ω + y sin ω = p), we first move the constant term to the right side: √3x + y = 8. Since the right side (p) must be positive, this is correct. Step 2: Divide the equation by √(A² + B²). Calculate √(A² + B²) = √((√3)² + 1²) = √(3 + 1) = √4 = 2. Divide the entire equation by 2: (√3/2)x + (1/2)y = 8/2 (√3/2)x + (1/2)y = 4 Step 3: Compare with the normal form to find p and ω. The normal form is x cos ω + y sin ω = p. Comparing our equation with this form, we get: cos ω = √3/2 sin ω = 1/2 p = 4 Since both cos ω and sin ω are positive, the angle ω lies in the first quadrant. The angle for which cos ω = √3/2 and sin ω = 1/2 is ω = 30° or π/6 radians. Final Answer: The normal form of the equation is (√3/2)x + (1/2)y = 4. The values are p = 4 and ω = 30° (or π/6).

Exam Traps and Key Points

Success in Straight Lines comes from both understanding concepts and avoiding common pitfalls. Here are some key points to remember for your exams:

• Slope Conditions: Do not mix up the conditions for parallel and perpendicular lines. Parallel lines have equal slopes (m₁ = m₂) , while perpendicular lines have slopes whose product is -1 (m₁m₂ = -1). A common mistake is to assume the slope of a perpendicular line is the negative of the original, when it is the negative reciprocal.
• Undefined Slope: Remember that a vertical line has an undefined slope. If a question involves a line parallel to the y-axis, its equation will be of the form x = constant, and you cannot use formulas involving 'm'.
• Intercept Form: For the intercept form x/a + y/b = 1, 'a' is the x-intercept and 'b' is the y-intercept. A common error is to swap them. Also, the right side of the equation must be 1. Always divide by the constant to bring it to this form.
• Normal Form: When converting to normal form x cos ω + y sin ω = p, ensure that the perpendicular distance 'p' is always positive. If the constant term is negative after moving it to the RHS, you must multiply the entire equation by -1 before proceeding.

Practice Questions with Solutions

• Q: Find the equation of the line that passes through the point (2, 2) and has intercepts on the axes whose sum is 9. A: Step 1: Let the equation of the line in intercept form be x/a + y/b = 1. We are given that a + b = 9, so b = 9 - a. Step 2: Substitute b in the equation: x/a + y/(9-a) = 1. Step 3: Since the line passes through (2, 2), this point must satisfy the equation. Substitute x=2, y=2: 2/a + 2/(9-a) = 1. Step 4: Solve for 'a'. Take the LCM: [2(9-a) + 2a] / [a(9-a)] = 1. 18 - 2a + 2a = 9a - a². a² - 9a + 18 = 0. (a - 3)(a - 6) = 0. So, a = 3 or a = 6. Step 5: Find the corresponding values of 'b' and the equations. Case 1: If a = 3, then b = 9 - 3 = 6. The equation is x/3 + y/6 = 1, which simplifies to 2x + y = 6. Case 2: If a = 6, then b = 9 - 6 = 3. The equation is x/6 + y/3 = 1, which simplifies to x + 2y = 6. Final answer: The two possible equations are 2x + y - 6 = 0 and x + 2y - 6 = 0.
• Q: Find the angle between the lines y - √3x - 5 = 0 and √3y - x + 6 = 0. A: Step 1: Find the slopes of both lines by converting them to the slope-intercept form (y = mx + c). Line 1: y - √3x - 5 = 0 => y = √3x + 5. The slope is m₁ = √3. Line 2: √3y - x + 6 = 0 => √3y = x - 6 => y = (1/√3)x - 6/√3. The slope is m₂ = 1/√3. Step 2: Use the formula for the angle (θ) between two lines: tan θ = |(m₂ - m₁) / (1 + m₁m₂)|. Step 3: Substitute the values of m₁ and m₂. tan θ = |( (1/√3) - √3 ) / ( 1 + (√3)(1/√3) )| tan θ = |( (1-3)/√3 ) / ( 1 + 1 )| tan θ = |( -2/√3 ) / 2| = |-1/√3| = 1/√3. Step 4: Find the angle θ. Since tan θ = 1/√3, the acute angle θ is 30°. Final answer: The angle between the lines is 30°.
• Q: Find the distance of the point (3, -5) from the line 3x - 4y - 26 = 0. A: Step 1: Recall the formula for the perpendicular distance (d) of a point (x₁, y₁) from a line Ax + By + C = 0. The formula is d = |Ax₁ + By₁ + C| / √(A² + B²). Step 2: Identify the values from the given data. The point (x₁, y₁) is (3, -5). The line is 3x - 4y - 26 = 0, so A = 3, B = -4, and C = -26. Step 3: Substitute these values into the formula. d = |3(3) + (-4)(-5) - 26| / √(3² + (-4)²). d = |9 + 20 - 26| / √(9 + 16). d = |3| / √25. d = 3 / 5. Final answer: The distance of the point (3, -5) from the line is 3/5 units.
• Q: Find the equation of a line passing through the intersection of the lines x + y = 3 and 2x - y = 0, and parallel to the line 3x - y + 1 = 0. A: Step 1: Find the point of intersection of the first two lines. We have two equations: (1) x + y = 3 (2) 2x - y = 0 => y = 2x Substitute y=2x into (1): x + 2x = 3 => 3x = 3 => x = 1. Now find y: y = 2x = 2(1) = 2. The point of intersection is (1, 2). Step 2: Find the slope of the required line. The required line is parallel to 3x - y + 1 = 0. First, find the slope of this line. y = 3x + 1. The slope (m) is 3. Since the required line is parallel to this line, its slope is also 3. Step 3: Use the point-slope form to find the equation. The line passes through (1, 2) and has a slope of 3. y - y₁ = m(x - x₁) y - 2 = 3(x - 1) y - 2 = 3x - 3 3x - y - 1 = 0. Final answer: The equation of the required line is 3x - y - 1 = 0.

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