NCERT Plus One (Class 11) Mathematics • Chapter 9

Straight Lines

Q: Find the slope and y-intercept of the line 6x + 3y − 5 = 0 .

Rewrite as 3y = −6x + 5 ⇒ y = −2x + 5/3 (or use m = −A/B = −6/3 = −2 , c = −C/B = 5/3 ). Answer: Slope m = −2 , y-intercept c = 5/3 .

Q: Find the equation of the line intersecting the x-axis at a distance of 3 units to the left of the origin with slope −2 .

Point on x-axis 3 units left of origin is (−3, 0) (so x-intercept d = −3 ). Using y = m(x − d) : y = −2(x − (−3)) = −2x − 6 ⇒ 2x + y + 6 = 0 .

Q: Find the equation of the line passing through (−3, 5) and perpendicular to the line through the points (2, 5) and (−3, 6) .

1) Slope of given line: m₁ = (6 − 5)/(−3 − 2) = −1/5 . 2) Perpendicular slope: m₂ = −1/(−1/5) = 5 . 3) Point-Slope through (−3, 5) : y − 5 = 5(x + 3) ⇒ 5x − y + 20 = 0 .

Q: If the angle between two lines is π/4 and slope of one of the lines is 1/2 , find the slope of the other line.

tan(π/4) = 1 = |(m − 1/2)/(1 + m/2)| . • Case 1 ( +1 ): m − 1/2 = 1 + m/2 ⇒ m/2 = 3/2 ⇒ m = 3 . • Case 2 ( −1 ): m − 1/2 = −1 − m/2 ⇒ (3/2)m = −1/2 ⇒ m = −1/3 . Both 3 and −1/3 are required!

Q: Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3) .

Equal intercepts means a = b , so Intercept Form is x/a + y/a = 1 ⇒ x + y = a . Pass through (2, 3) : 2 + 3 = a ⇒ a = 5 . Equation: x + y = 5 (or x + y − 5 = 0 ).

Q: Reduce the equation √3 x + y − 8 = 0 into normal form and find the values of p and ω .

1) Write with positive RHS: √3 x + y = 8 . 2) Divide by √((√3)² + 1²) = 2 : (√3/2)x + (1/2)y = 4 . 3) Compare with x cos ω + y sin ω = p : cos ω = √3/2, sin ω = 1/2 ⇒ ω = 30° (π/6) , p = 4 .

Q: Find the distance between the parallel lines 15x + 8y − 34 = 0 and 15x + 8y + 31 = 0 .

Both have identical A = 15, B = 8 and constants on the LHS: C₁ = −34, C₂ = 31 . d = |C₁ − C₂| / √(A² + B²) = |−34 − 31| / √(15² + 8²) = 65 / √289 = 65/17 units.

Q: Find the coordinates of the foot of perpendicular from the point (−1, 3) to the line 3x − 4y − 16 = 0 .

Slope of given line = 3/4 ⇒ slope of perpendicular = −4/3 . Equation of perpendicular through (−1, 3) : y − 3 = −4/3 (x + 1) ⇒ 4x + 3y − 5 = 0 . Solve 3x − 4y = 16 and 4x + 3y = 5 simultaneously: 9x − 12y = 48 and 16x + 12y = 20 ⇒ 25x = 68 ⇒ x = 68/25 ; substitute to get y = −49/25 . Foot of perpendicular: (68/25, −49/25) .

Slope & angle, parallel/perpendicular lines, point-slope/intercept/normal forms & point-to-line distance.

📄 Official NCERT Textbook PDF

Topic 9.2.1 – 9.2.2: Slope m = tan θ, Parallel & Perpendicular Slopes

Picture cycling up the hairpin bends of Munnar vs gliding on Fort Kochi's flat beach road! A flat road has zero steepness (m = 0), a hill climb has positive steepness, and a vertical cliff drop has undefined steepness! Measure the anti-clockwise angle θ from the positive x-axis (0° ≤ θ < 180°) — its tangent m = tan θ captures a straight line's exact steepness in one single number!

INCLINATION & SLOPEm = tan θ (0° ≤ θ < 180°, anti-clockwise from +x axis; undefined at 90°!)

TWO-POINT SLOPEm = y₂ − y₁x₂ − x₁ (Rise ÷ Run, when x₁ ≠ x₂)

PARALLEL & PERPParallel: m₁ = m₂ | Perpendicular: m₁ · m₂ = −1

🧠 Memory Trick: Parallel lines copy slopes (m₁ = m₂); Perpendicular lines FLIP & NEGATE (m₂ = −1/m₁) — e.g., 2/3 → −3/2!

💡 Simple Kid-Check: Walk left to right on a graph! Climbing uphill means acute θ and positive slope (m > 0). Sliding downhill means obtuse θ (90° < θ < 180°) and negative slope (m < 0). Walking flat means m = 0!

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Horizontal Line (y = k): Parallel to x-axis → θ = 0° → m = tan 0° = 0.
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Vertical Line (x = h): Parallel to y-axis → θ = 90° → slope is undefined (÷ 0), yet it IS perpendicular to y = k!

🧒 Quick Slope & Perpendicular Check: Line passes through A(3, −2) and B(−1, 4).
• m_AB = 4 − (−2)−1 − 3 = 6−4 = −32 (downhill!).
• Any parallel line has slope −3/2; any perpendicular line has slope +2/3 (flip & negate!).

Watch THREE classic 1-mark traps: (1) Flipping Rise/Run upside down as (x₂ − x₁)/(y₂ − y₁) — ALWAYS put y on top! (2) Subtracting in mismatched order like (y₂ − y₁)/(x₁ − x₂) flips the sign! (3) A vertical line has no defined slope (NEVER write 0 × ∞ = −1 — vertical x = h and horizontal y = k are perpendicular by geometry)!

Topic 9.2.3 – 9.2.4: Angle Between Two Lines & Collinearity Condition

When two highways cross at a busy Thrissur roundabout, they create BOTH a sharp acute turn θ and a wide obtuse partner 180° − θ! With slopes m₁ and m₂, one clean formula locks onto the sharp angle θ immediately — and if θ = 0° between segments AB and BC, all three points sit strictly on one single straight track (collinear)!

ANGLE FORMULAtan θ = |m₂ − m₁1 + m₁m₂| (assuming 1 + m₁m₂ ≠ 0)

ACUTE vs OBTUSE|...| gives acute angle θ; obtuse partner is 180° − θ!

COLLINEARITY TESTA, B, C are collinear ⇔ Slope(AB) = Slope(BC)!

🧠 Memory Trick: Check the denominator 1 + m₁m₂ FIRST! If it equals 0, lines are perpendicular (90°). If the numerator m₂ − m₁ = 0, lines are parallel (0°)!

💡 Simple Kid-Check: Why does |...| (modulus) appear in tan θ = |(m₂ − m₁)/(1 + m₁m₂)|? Because taking absolute value forces tan θ ≥ 0, guaranteeing you land on the acute angle θ! If a question asks for an unknown line making angle θ with a known line, removing |...| gives ± tan θ — producing TWO valid lines!

🔎 Quick Angle Example: Find the angle between lines with slopes m₁ = 1/2 and m₂ = 3:
tan θ = |3 − 1/21 + 3(1/2)| = |5/25/2| = 1 ⇒ θ = 45° (and obtuse partner 135°)!

📏 2-Mark Collinearity Check: Without distance formula, find x if A(x, −1), B(2, 1), C(4, 5) are collinear:
Set Slope(AB) = Slope(BC): 1 − (−1)2 − x = 5 − 14 − 2 ⇒ 22 − x = 2 ⇒ 2 − x = 1 ⇒ x = 1!

Watch TWO high-frequency board exam traps: (1) Solving "find the slope of a line making 45° with slope 1/2" and dropping modulus without writing ± 1 — you MUST solve BOTH +1 (giving m = 3) AND −1 (giving m = −1/3)! (2) Plugging a vertical line directly into the formula: if one line is vertical (x = k) and the other has slope m, draw a triangle — tan θ = |1/m| directly!

Topic 9.3.1 – 9.3.3: Horizontal/Vertical, Point-Slope & Two-Point Equations

How do you write an algebraic equation so ONLY points standing right on your straight track pass the test? On a horizontal shelf at height 4, every point obeys y = 4! On a vertical flagpole at x = −2, every point obeys x = −2! For any slanted ramp through station (x₀, y₀) with slope m, every rider (x, y) has slope (y − y₀)/(x − x₀) = m!

AXIS-PARALLEL LINESHorizontal: y = b | Vertical: x = a (x-axis is y = 0; y-axis is x = 0!)

POINT-SLOPE FORMThrough (x₀, y₀) with slope m: y − y₀ = m(x − x₀)

TWO-POINT FORMThrough (x₁, y₁), (x₂, y₂): y − y₁ = y₂ − y₁x₂ − x₁(x − x₁)

🧠 Memory Trick: Every equation of a tilted line starts from ONE root idea: Slope from fixed point to (x, y) = Fixed slope m! Point-Slope and Two-Point are literally the SAME rule!

💡 Simple Kid-Check: Wait — why is the x-axis equation y = 0 instead of x = 0? Stand anywhere on the x-axis: (1, 0), (5, 0), (−3, 0). Look at which coordinate stays frozen at zero every single time — y is always 0! Similarly, on the y-axis, x stays frozen at 0 (x = 0)!

Given Clues	Use Form	Equation Formula
Horizontal at height `b`	Horizontal	`y = b` (or `y = 0` for x-axis)
Vertical at distance `a`	Vertical	`x = a` (or `x = 0` for y-axis)
One point `(x₀, y₀)` & slope `m`	Point-Slope	`y − y₀ = m(x − x₀)`
Two points `(x₁, y₁), (x₂, y₂)`	Two-Point	`y − y₁ = [(y₂ − y₁)/(x₂ − x₁)](x − x₁)`

🧒 Quick Two-Point Example: Find the line through (−1, 1) and (2, −4):
• First grab slope: m = −4 − 12 − (−1) = −53.
• Plug into Point-Slope at (−1, 1): y − 1 = −53(x + 1) ⇒ 3y − 3 = −5x − 5 ⇒ 5x + 3y + 2 = 0!

Watch TWO super-common traps: (1) Mixing up the axes — writing x = 0 for the x-axis! Remember: x-axis is y = 0 and y-axis is x = 0! (2) Plugging vertical points like (3, 1) and (3, 5) into Two-Point Form divides by 3 − 3 = 0! Whenever x₁ = x₂ = 3, skip fractions and write x = 3 immediately!

Topic 9.3.4 – 9.3.6: Slope-Intercept (y = mx + c), Intercept (x/a + y/b = 1) & Normal Forms

Every straight line wears THREE super-useful exam outfits depending on which clue is handed to you: (1) its slope plus where it slices the y-axis (or x-axis!), (2) BOTH axis slices a and b simultaneously, or (3) its shortest perpendicular distance p from the origin at angle ω!

SLOPE-INTERCEPTy-intercept c: y = mx + c | x-intercept d: y = m(x − d)!

INTERCEPT FORMMeets axes at (a, 0) & (0, b): xa + yb = 1

NORMAL FORMp > 0 perpendicular from (0,0) at angle ω: x cos ω + y sin ω = p

🧠 Memory Trick: Watch c vs d! y = mx + c uses y-intercept (0, c), whereas y = m(x − d) uses x-intercept (d, 0)! Never swap them!

💡 Simple Kid-Check: In Intercept Form x/a + y/b = 1, the triangle trapped between the line and the two coordinate axes ALWAYS has area ½ |ab|! And in Normal Form x cos ω + y sin ω = p, picture shining a spotlight straight from origin (0, 0) perpendicular onto the line: p is the beam's positive length and ω is the beam's anti-clockwise angle from +x!

🏆 Classic 3-Mark Intercept Puzzle: Find the equation of the line through (2, 2) cutting off intercepts on axes whose sum is 9.
1) Since a + b = 9, put b = 9 − a in xa + y9 − a = 1.
2) Plug in (2, 2): 2a + 29 − a = 1 ⇒ 18 = a(9 − a) ⇒ a² − 9a + 18 = 0 ⇒ a = 6 or a = 3.
3) So TWO lines work: x/6 + y/3 = 1 (x + 2y = 6) and x/3 + y/6 = 1 (2x + y = 6)!

📏 Quick Normal Form Check: Perpendicular from origin has length p = 5 and makes ω = 30° with +x axis.
Plug straight in: x cos 30° + y sin 30° = 5 ⇒ √32x + 12y = 5 ⇒ √3 x + y = 10!

Watch THREE high-trap mistakes: (1) Writing y = mx + d when given x-intercept d — since it passes through (d, 0), it MUST be y = m(x − d)! (2) Allowing p < 0 in Normal Form — p is ALWAYS a strictly non-negative length from origin (flip all signs if RHS is negative)! (3) Remember x/a + y/b = 1 CANNOT represent lines passing through the origin (0, 0) (where a = b = 0)!

Topic 9.3.7: General Equation Ax + By + C = 0 Conversion Tricks

No matter how wildly a line equation starts out, clearing fractions and moving everything to one side ALWAYS packs it into ONE universal suitcase: Ax + By + C = 0 (with A, B not both zero)! Think of this section as the 3-button Swiss Army decoder that cracks ANY Ax + By + C = 0 back into Slope-Intercept, Intercept, or Normal form in 10 seconds!

TO SLOPE-INTERCEPTB ≠ 0: m = −AB, y-int c = −CB

TO INTERCEPT FORMC ≠ 0: x-int a = −CA, y-int b = −CB

TO NORMAL FORMMove C right, make RHS > 0, divide by √(A² + B²)!

🧠 Memory Trick for Normal Form: Put constant on RHS and force RHS POSITIVE first! Only then divide by +√(A² + B²) so cos ω and sin ω immediately land in the RIGHT quadrant!

💡 Simple Kid-Check: Wait — what if A = 0 or B = 0 or C = 0 in Ax + By + C = 0? Easy! If A = 0, no x appears → horizontal line y = −C/B. If B = 0, no y appears → vertical line x = −C/A. If C = 0, plugging (0, 0) gives 0 = 0 → line passes straight through the origin!

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Quick Parallel / Perpendicular Line Shortcut: Line parallel to Ax + By + C = 0 is just Ax + By + k = 0 (keep Ax + By identical, swap only C → k!). Line perpendicular swaps A & B and flips one sign: Bx − Ay + k = 0!

🏆 Complete 3-Form Conversion on x − √3y + 8 = 0:
• Slope-Intercept: √3y = x + 8 ⇒ y = 1√3x + 8√3 → m = 1/√3 (θ = 30°), c = 8/√3.
• Intercept Form: x − √3y = −8 ⇒ x−8 + y8/√3 = 1 → a = −8, b = 8/√3.
• Normal Form: Move 8 right and make RHS +8: −x + √3y = 8. Divide by √((−1)² + (√3)²) = 2:
(−12)x + (√32)y = 4 → cos ω = −1/2, sin ω = +√3/2 (Quadrant II!) → ω = 120°, p = 4!

Watch TWO sneaky sign traps: (1) Forgetting the minus inside B when using m = −A/B! For 3x − 4y + 10 = 0, A = 3, B = −4, so m = −3/(−4) = +3/4 (NOT −3/4!). (2) In Normal Form conversion of x − √3y + 8 = 0, writing (1/2)x − (√3/2)y = −4 and calling p = −4 or ω = 300° — WRONG! Always flip signs so p > 0 FIRST, which gives Quadrant II ω = 120°!

Topic 9.4: Perpendicular Distance from Point to Line & Distance Between Parallel Lines

Standing at railway crossing P(x₁, y₁), how do you find the shortest walk to a straight track Ax + By + C = 0? Drop straight at 90°! Just plug (x₁, y₁) right into the line's left-hand side, wrap it in modulus bars, and divide by Pythagoras's √(A² + B²)! And the exact lane width between two parallel roads works the exact same way!

POINT TO LINEP(x₁, y₁) to Ax + By + C = 0: d = |Ax₁ + By₁ + C|√(A² + B²)

ORIGIN SPECIAL CASEFrom (0, 0) to Ax + By + C = 0: p = |C|√(A² + B²)

TWO PARALLEL LINESAx + By + C₁ = 0 & Ax + By + C₂ = 0: d = |C₁ − C₂|√(A² + B²)

🧠 Memory Trick: For parallel lines, check that A and B MATCH IDENTICALLY (same side, same coefficients!) BEFORE subtracting |C₁ − C₂|!

💡 Simple Kid-Check: Notice how the two formulas are actually twins! Plugging P(x₁, y₁) into Ax + By + C₁ = 0 when (x₁, y₁) sits on the parallel line Ax₁ + By₁ + C₂ = 0 means Ax₁ + By₁ = −C₂, turning |Ax₁ + By₁ + C₁| straight into |C₁ − C₂|!

🧒 Quick Point-to-Line Example: Find distance of point (3, −5) from 3x − 4y − 26 = 0:
d = |3(3) − 4(−5) − 26|√(3² + (−4)²) = |9 + 20 − 26|√25 = 35 units!

🏆 Parallel Lines Trap-Buster: Distance between 3x + 4y = 5 and 6x + 8y + 15 = 0.
1) Write BOTH in = 0 form: 3x + 4y − 5 = 0 and 6x + 8y + 15 = 0.
2) Match coefficients! Divide the second by 2: 3x + 4y + 15/2 = 0 (so A = 3, B = 4 match!).
3) Now apply formula: d = |−5 − 15/2|√(3² + 4²) = 25/25 = 52 units!

Watch the #1 3-mark DHSE exam trap: applying |C₁ − C₂|/√(A² + B²) between 3x − 4y + 7 = 0 and 6x − 8y + 1 = 0 WITHOUT scaling one equation so A and B match identically! Also make sure BOTH constants sit on the SAME side (Ax + By + C = 0) before reading C₁ and C₂, and NEVER include C² inside the square root!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (2024 DHSE Model • 1 Marks): Find the slope and y-intercept of the line 6x + 3y − 5 = 0.

Rewrite as 3y = −6x + 5 ⇒ y = −2x + 5/3 (or use m = −A/B = −6/3 = −2, c = −C/B = 5/3).
Answer: Slope m = −2, y-intercept c = 5/3.

Q2 (2023 DHSE • 2 Marks): Find the equation of the line intersecting the x-axis at a distance of 3 units to the left of the origin with slope −2.

Point on x-axis 3 units left of origin is (−3, 0) (so x-intercept d = −3).
Using y = m(x − d): y = −2(x − (−3)) = −2x − 6 ⇒ 2x + y + 6 = 0.

Q3 (2022 DHSE • 2 Marks): Find the equation of the line passing through (−3, 5) and perpendicular to the line through the points (2, 5) and (−3, 6).

1) Slope of given line: m₁ = (6 − 5)/(−3 − 2) = −1/5.
2) Perpendicular slope: m₂ = −1/(−1/5) = 5.
3) Point-Slope through (−3, 5): y − 5 = 5(x + 3) ⇒ 5x − y + 20 = 0.

Q4 (2021 DHSE • 3 Marks): If the angle between two lines is π/4 and slope of one of the lines is 1/2, find the slope of the other line.

tan(π/4) = 1 = |(m − 1/2)/(1 + m/2)|.
• Case 1 (+1): m − 1/2 = 1 + m/2 ⇒ m/2 = 3/2 ⇒ m = 3.
• Case 2 (−1): m − 1/2 = −1 − m/2 ⇒ (3/2)m = −1/2 ⇒ m = −1/3.
Both 3 and −1/3 are required!

Q5 (2020 DHSE • 3 Marks): Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

Equal intercepts means a = b, so Intercept Form is x/a + y/a = 1 ⇒ x + y = a.
Pass through (2, 3): 2 + 3 = a ⇒ a = 5.
Equation: x + y = 5 (or x + y − 5 = 0).

Q6 (NCERT Ex 9.3 • 3 Marks): Reduce the equation √3 x + y − 8 = 0 into normal form and find the values of p and ω.

1) Write with positive RHS: √3 x + y = 8.
2) Divide by √((√3)² + 1²) = 2: (√3/2)x + (1/2)y = 4.
3) Compare with x cos ω + y sin ω = p: cos ω = √3/2, sin ω = 1/2 ⇒ ω = 30° (π/6), p = 4.

Q7 (2019 DHSE • 3 Marks): Find the distance between the parallel lines 15x + 8y − 34 = 0 and 15x + 8y + 31 = 0.

Both have identical A = 15, B = 8 and constants on the LHS: C₁ = −34, C₂ = 31.
d = |C₁ − C₂| / √(A² + B²) = |−34 − 31| / √(15² + 8²) = 65 / √289 = 65/17 units.

Q8 (NCERT Misc • 4 Marks): Find the coordinates of the foot of perpendicular from the point (−1, 3) to the line 3x − 4y − 16 = 0.

Slope of given line = 3/4 ⇒ slope of perpendicular = −4/3.
Equation of perpendicular through (−1, 3): y − 3 = −4/3 (x + 1) ⇒ 4x + 3y − 5 = 0.
Solve 3x − 4y = 16 and 4x + 3y = 5 simultaneously:
9x − 12y = 48 and 16x + 12y = 20 ⇒ 25x = 68 ⇒ x = 68/25; substitute to get y = −49/25.
Foot of perpendicular: (68/25, −49/25).