Straight Lines

2.1 Core concepts

• Recap formulae carried into this chapter: distance PQ = √[(x₂–x₁)² + (y₂–y₁)²]; internal section formula ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)); midpoint ((x₁+x₂)/2, (y₁+y₂)/2); area of triangle = ½ |x₁(y₂–y₃) + x₂(y₃–y₁) + x₃(y₁–y₂)|; three points are collinear iff this area is zero (NCERT §9.1, pp. 151–152). These are the building blocks for every straight-line formula that follows.
• Inclination and slope: the angle θ measured anticlockwise from the positive x-axis to a line is its inclination, with 0° ≤ θ ≤ 180°. The slope is m = tan θ (θ ≠ 90°). Lines parallel to the x-axis have inclination 0° (slope 0); vertical lines have inclination 90° and undefined slope (NCERT §9.2, pp. 152–153). The slope is a real number that captures "rise over run".
• Slope from two points: for points (x₁, y₁) and (x₂, y₂) on a non-vertical line, m = (y₂ – y₁)/(x₂ – x₁), valid for both acute and obtuse inclinations (NCERT §9.2.1, pp. 153–154). Order of points does not matter because both numerator and denominator change sign simultaneously.
• Parallelism and perpendicularity: two non-vertical lines are parallel iff m₁ = m₂; they are perpendicular iff m₁m₂ = –1, i.e. their slopes are negative reciprocals (NCERT §9.2.2, pp. 154–155). These two tests cover almost every "find the line through P parallel/perpendicular to L" CUET problem.
• Angle between two lines: the acute angle θ between non-vertical lines with slopes m₁ and m₂ satisfies tan θ = |(m₂ – m₁)/(1 + m₁m₂)|, provided 1 + m₁m₂ ≠ 0. The obtuse angle is φ = 180° – θ (NCERT §9.2.3, pp. 156–157). When 1 + m₁m₂ = 0, the lines are perpendicular and θ = 90°.
• Collinearity via slopes: three points A, B, C are collinear iff slope of AB = slope of BC (NCERT Summary, p. 174 and §9.2 Example/Exercise discussion). Equivalently, the area of triangle ABC is 0.
• Horizontal and vertical lines: a horizontal line at distance a from the x-axis has equation y = a or y = –a; a vertical line at distance b from the y-axis has equation x = b or x = –b (NCERT §9.3.1, pp. 159–160). These are the two cases that "elude" the general slope-intercept form.
• Point-slope form: the line through (x₀, y₀) with slope m is y – y₀ = m(x – x₀) (NCERT §9.3.2, pp. 160–161). Use when a point and slope are known directly.
• Two-point form: the line through (x₁, y₁) and (x₂, y₂) is y – y₁ = [(y₂ – y₁)/(x₂ – x₁)] (x – x₁) (NCERT §9.3.3, p. 161). Use when two points are known but slope is not separately given.
• Slope-intercept form: with slope m and y-intercept c, the line is y = mx + c; with x-intercept d, the line is y = m(x – d) (NCERT §9.3.4, pp. 161–162). This is the form in which graphs of linear functions are most naturally read.
• Intercept form: a line with x-intercept a and y-intercept b is x/a + y/b = 1 (NCERT §9.3.5, p. 163). a and b are signed — negative intercepts produce negative a or b.
• General form Ax + By + C = 0: any equation of this form (A, B not simultaneously zero) represents a line; its slope is –A/B (when B ≠ 0); used to derive standard forms by reduction (NCERT §9.3.5, p. 163 and §9.4 application). Every line in the plane can be expressed in this form.
• Distance of a point from a line: the perpendicular distance d from P(x₁, y₁) to the line Ax + By + C = 0 is d = |Ax₁ + By₁ + C|/√(A² + B²) (NCERT §9.4, pp. 164–166). The modulus is essential; distance is non-negative.
• Distance between two parallel lines: for y = mx + c₁ and y = mx + c₂, d = |c₁ – c₂|/√(1 + m²); for Ax + By + C₁ = 0 and Ax + By + C₂ = 0, d = |C₁ – C₂|/√(A² + B²) (NCERT §9.4.1, pp. 166–167). The A, B coefficients must match exactly before applying.
• Reductions between forms: Ax + By + C = 0 converts to slope-intercept y = –(A/B)x – C/B, to intercept form by dividing by –C: x/(–C/A) + y/(–C/B) = 1, and so on. CUET often tests these conversions directly.

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Fig 9.1 (p. 151): Plot of (6, –4) and (3, 0) — orients the recap of coordinate axes and distance.
• Fig 9.2 (p. 153): Inclination θ of a line with the positive x-axis; visualises 0° ≤ θ ≤ 180° and the convention that vertical lines get θ = 90°.
• Fig 9.3(i)–(ii) (pp. 153–154): Right-triangle construction (PMQ) used to derive m = (y₂ – y₁)/(x₂ – x₁) for both acute and obtuse θ.
• Fig 9.4 (p. 154): Two parallel lines having equal inclinations α = β, hence m₁ = m₂. The visual proof is one of the cleanest in coordinate geometry.
• Fig 9.5 (p. 155): Perpendicular lines with β = α + 90°, leading to m₁m₂ = –1 via tan(α + 90°) = −cot α.
• Fig 9.6 (p. 156): Angle between two intersecting lines L₁, L₂ — adjacent angles θ and φ with θ + φ = 180°. Exactly one of θ, φ is acute.
• Fig 9.7 (p. 158): Two possible slopes (3 and –1/3) in Example 2, illustrating why the acute-angle formula uses an absolute value — two distinct lines can both make the same angle with a given line.
• Fig 9.8 (p. 160): Horizontal lines y = ±a and vertical lines x = ±b.
• Fig 9.10 (p. 160): Point-slope construction for the line through P₀(x₀, y₀) with slope m.
• Fig 9.11 (p. 161): Two-point form via collinearity of P, P₁, P₂.
• Fig 9.12 (p. 162): Slope-intercept derivation through (0, c).
• Fig 9.13 (p. 163): Intercept form using (a, 0) and (0, b).
• Fig 9.14 (p. 165): Perpendicular PM from P(x₁, y₁) to Ax + By + C = 0, with auxiliary triangle PQR used to derive d = |Ax₁ + By₁ + C|/√(A² + B²).
• Fig 9.15 (p. 166): Two parallel lines y = mx + c₁ and y = mx + c₂; distance taken from A(–c₁/m, 0) to the second line.
• Process — find equation of a line: identify which data you have (point + slope ⇒ point-slope; two points ⇒ two-point; slope + intercept ⇒ slope-intercept; both intercepts ⇒ intercept). Use the matching form, then simplify to Ax + By + C = 0 if requested.
• Process — distance computation: write the line in Ax + By + C = 0 form, substitute the point's coordinates into the numerator, take absolute value, divide by √(A² + B²).

2.4 Common confusions / NTA trap points

• Sign inside the modulus: in d = |Ax₁ + By₁ + C|/√(A² + B²), forgetting the absolute value (and writing a negative distance) is the commonest trap (NCERT §9.4, p. 166).
• Acute vs obtuse angle: tan θ = (m₂ – m₁)/(1 + m₁m₂) without the modulus gives the signed tangent — only after taking |·| do you get the acute angle. NTA distractors often drop the modulus and pick the obtuse value (NCERT §9.2.3, p. 157).
• Vertical lines have no slope: writing m = ∞ instead of "undefined" can trick statement-based questions about whether m₁m₂ = –1 implies perpendicularity (this assumes both lines are non-vertical) (NCERT §9.2, p. 153; §9.2.2, p. 155).
• Slope from general form: for Ax + By + C = 0, slope is –A/B (not A/B and not –B/A). A frequent distractor (NCERT §9.3.5 and Example 9 setup, p. 167).
• Intercept-form sign of intercepts: in x/a + y/b = 1, "a" and "b" are signed intercepts. NCERT Example 8 uses a = –3, b = 2 to give 2x – 3y + 6 = 0 (NCERT §9.3.5, p. 163).
• Distance between two parallel lines requires equal A, B coefficients: if the equations are written with different leading coefficients (e.g. 3x – 4y + 7 = 0 vs 6x – 8y + 5 = 0), they must be rescaled to match before applying d = |C₁ – C₂|/√(A² + B²) (NCERT §9.4.1, p. 167).
• Parallel lines have slope ratio 1, perpendicular −1. Many students invert these. The "negative reciprocal" rule applies to perpendicular slopes, not parallel.
• Inclination > 180°. θ is always in [0°, 180°]; 200° is not a valid inclination. Reduce by ±180° if needed.
• Mistaking the y-intercept of Ax + By + C = 0. It is −C/B, not −C. The factor of B is easy to miss when B ≠ 1.
• Forgetting that horizontal lines have slope 0 (not undefined). Undefined slope is only for vertical lines.
• Two parallel lines coincide if intercept constants match. The "distance 0" case occurs only when both lines are identical.
• Wrong order of subtraction in slope formula. Either (y₂ − y₁)/(x₂ − x₁) or (y₁ − y₂)/(x₁ − x₂); both give the same slope. But mixing the orders (numerator from one, denominator from the other) flips the sign.
• Misapplying the angle formula to a horizontal-vertical pair. When one line is horizontal (m₁ = 0) and the other is vertical (m₂ undefined), the formula breaks down; the angle is 90° by inspection.
• Forgetting modulus in the perpendicular distance formula. The distance is always |Ax₁ + By₁ + C|/√(A² + B²); the sign inside indicates which side of the line the point lies on but never reverses the magnitude.
• Confusing "intercept" with "intercept on axis". "x-intercept = a" means the line crosses the x-axis at (a, 0), not that the x-coordinate of any point is a.

2.5 Key formulas & theorems

2.6 Solved examples (NCERT-grounded)

Example A (NCERT Example 1(a), p. 155). Slope of line through (3, −2) and (−1, 4)?

Step 1 — apply formula: m = (4 − (−2))/(−1 − 3) = 6/(−4). Step 2 — simplify: m = −3/2. Step 3 — state: slope = −3/2.

Example B (NCERT Example 3, p. 158). The line through (−2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (x, 24). Find x.

Step 1 — slope m₁: (8 − 6)/(4 − (−2)) = 2/6 = 1/3. Step 2 — slope m₂: (24 − 12)/(x − 8) = 12/(x − 8). Step 3 — perpendicularity: m₁ m₂ = −1 ⇒ (1/3) · 12/(x − 8) = −1 ⇒ 12/(x − 8) = −3 ⇒ x − 8 = −4 ⇒ x = 4. Answer: x = 4.

Example C (NCERT Example 5, p. 161). Equation of line through (−2, 3) with slope −4?

Step 1 — apply point-slope: y − 3 = −4(x − (−2)) = −4(x + 2). Step 2 — expand: y − 3 = −4x − 8 ⇒ 4x + y + 5 = 0. Step 3 — state: 4x + y + 5 = 0.

Example D (NCERT Example 9, p. 167). Distance of (3, −5) from 3x − 4y − 26 = 0?

Step 1 — identify A, B, C: A = 3, B = −4, C = −26. Step 2 — plug into distance formula: numerator = |3(3) + (−4)(−5) + (−26)| = |9 + 20 − 26| = 3. Step 3 — denominator and divide: √(9 + 16) = 5; d = 3/5. Answer: 3/5.

Example E (NCERT Example 10, p. 167). Distance between 3x − 4y + 7 = 0 and 3x − 4y + 5 = 0?

Step 1 — verify parallel: coefficients of x, y match; lines are parallel. Step 2 — apply parallel-distance formula: d = |7 − 5|/√(9 + 16) = 2/5. Step 3 — state: d = 2/5.