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Mathematics — Class 12 — L14: Straight Lines

NIOS Code 311 · Module 4 · Co-ordinate Geometry

Notes extracted from NIOS Mathematics Course (311), Lesson 14 — Straight Lines (ilovepdf_merged (6).pdf). Content covers sections 14.1–14.x.

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Overview — Straight Lines (L14)

This NIOS Mathematics (311) lesson builds the equation of a straight line in the plane from geometry you already know: slope, intercepts, and points on the line. Every later circle/conic problem still uses these forms.

y = mx + c · (y−y₁)=m(x−x₁) · x/a + y/b = 1 · x cos α + y sin α = p

Four classic forms · general form Ax+By+C=0

Slope as rise over run

14.1 Lines parallel to the axes

A line parallel to the x-axis has equation y = k (constant ordinate). A line parallel to the y-axis has equation x = h.

Example: Through (−2, −3) parallel to the x-axis → y = −3. Through the same point parallel to the y-axis → x = −2.

14.2 Standard forms of the line

Slope–intercept form

If a line has slope m and y-intercept c, its equation is:

y = mx + c

m = tan θ · c is signed intercept on OY

If the line is vertical, slope is undefined; use x = constant instead of this form.

Point–slope form

Line through A(x₁, y₁) with slope m:

y − y₁ = m(x − x₁)

Fails only for vertical lines

Two-point form

Through distinct points (x₁,y₁) and (x₂,y₂) with x₂ ≠ x₁:

y − y₁ = [(y₂ − y₁)/(x₂ − x₁)](x − x₁)

Slope from the two points

Intercept form

If the line cuts intercepts a on OX and b on OY (a,b ≠ 0):

x/a + y/b = 1

Intercepts can be negative

Coordinate axes and quadrants

Normal (perpendicular) form

If the perpendicular from the origin to the line has length p and makes angle α with the positive x-axis:

x cos α + y sin α = p

p ≥ 0 · α measured from +x-axis

14.3 General first-degree equation

Every non-degenerate line can be written as Ax + By + C = 0 with A,B not both zero. Conversely, every such equation represents a straight line.

Reduce to slope–intercept (B ≠ 0): y = (−A/B)x + (−C/B), so slope = −A/B and y-intercept = −C/B.

14.4 Distance from a point to a line

Distance of point (x₁,y₁) from Ax+By+C=0 is:

d = |Ax₁ + By₁ + C| / √(A² + B²)

Always take absolute value

Distance of the origin is |C| / √(A² + B²).

14.5–14.7 Parallel and perpendicular lines

Line parallel to Ax+By+C=0: Ax + By + K = 0 (same A,B; new constant K).
Line perpendicular to Ax+By+C=0: Bx − Ay + K = 0 (swap coefficients, change one sign).

Fix K using a given point on the required line.

Exam checklist: Identify which form is natural from the data; convert to Ax+By+C=0 only when needed for distance or family of lines.

MCQ Quiz — L14 Straight Lines

0 / 10 correct

Flashcards — L14

1 / 15

Golden Rules — L14 Straight Lines

Most exam-important points from this chapter:

Know the definitions of L14

Start every answer with the key definition or standard form from Straight Lines.

Write the formula first

Copy the formula, then substitute values; never jump to the number alone.

Watch conditions

Check domain, quadrant, non-zero denominators, and applicability of the theorem.

Master result 1

Memorise and apply: y = b (parallel to x-axis)

Master result 2

Memorise and apply: where b is equal in absolute value to the distance from the x-axis to the opposite side.

y − y₁ = m(x − x₁)

y = mx + c

x/a + y/b = 1

ax + by + c = 0

d = |ax₁+by₁+c|/√(a²+b²)

Angle between lines

∥ : a₁/a₂ = b₁/b₂

⊥ : a₁a₂ + b₁b₂ = 0

1. Formulas & Definitions

Full Ch 14 — Straight Lines study guide: definitions, derivations, why each formula works, history, deep understanding, pencil diagrams, step-by-step use, and 4-level solved examples.

y − y₁ = m(x − x₁)

Definition: Point–slope form: line through (x₁,y₁) with slope m.

Derivation

From definition m = (y−y₁)/(x−x₁), rearrange.

Variables

m = slope · (x₁,y₁) = known point

Why it works

Any other point on the line keeps the same slope from the fixed point.

Historical context

Standard form in school algebra since 19th century texts.

Deep understanding

Fails for vertical lines (use x = x₁ instead).

2. Diagrams & Visuals

Pencil sketch · labelled · step-by-step breakdown below

Note given point and slope
Write y−y₁=m(x−x₁)
Expand if general form needed
Check with a second point

3. Solved Examples

Basic

Q: Through (1,2), slope 3.

Solution: y−2=3(x−1)

Answer: y = 3x − 1

Intermediate

Q: Through (−2,−3) parallel to x-axis.

Solution: m=0 → y=−3

Answer: y = −3

Advanced

Q: Through (2,3) and (4,7).

Solution: m=2 → y−3=2(x−2)

Answer: y = 2x − 1

Exam

Q: Line slope −1/2 through origin.

Solution: y=−(1/2)x

Answer: x + 2y = 0

x/a + y/b = 1

Definition: Intercept form: x-intercept a, y-intercept b (neither zero).

Derivation

Line meets axes at (a,0) and (0,b); two-point form reduces to this.

Variables

a = x-intercept · b = y-intercept

Why it works

Useful when intercepts are given or easy to read from a graph.

Historical context

Common in coordinate geometry for triangle area ½|ab| with axes.

Deep understanding

If a line misses an axis (parallel), intercept form does not apply.

2. Diagrams & Visuals

Pencil sketch · labelled · step-by-step breakdown below

Find intercepts a, b
Write x/a + y/b = 1
Clear denominators → ax form
Verify with a point

3. Solved Examples

Basic

Q: Intercepts 3 and 4.

Solution: x/3+y/4=1

Answer: 4x+3y=12

Intermediate

Q: 3x+2y=12 in intercept form.

Solution: x/4+y/6=1

Answer: a=4, b=6

Advanced

Q: Line with equal intercepts through (2,3).

Solution: x/a+y/a=1 → x+y=a; 5=a

Answer: x+y=5

Exam

Q: Area of triangle with axes for x/2+y/5=1.

Solution: ½·2·5

Answer: 5

d = |ax₁ + by₁ + c| / √(a² + b²)

Definition: Perpendicular distance from point (x₁,y₁) to line ax+by+c=0.

Derivation

Normal form or projection of vector from a point on the line to the given point onto the unit normal.

Variables

Line: ax+by+c=0 · point (x₁,y₁) · d ≥ 0

Why it works

Absolute value makes distance positive; denominator normalises the coefficients.

Historical context

Standard result in analytic geometry (Lagrange / textbook normal form).

Deep understanding

If d=0 the point lies on the line — quick membership test.

2. Diagrams & Visuals

Pencil sketch · labelled · step-by-step breakdown below

Write line as ax+by+c=0
Plug point into |ax+by+c|
Divide by √(a²+b²)
Simplify

3. Solved Examples

Basic

Q: Distance (0,0) to x+y−2=0.

Solution: |−2|/√2=√2

Answer: √2

Intermediate

Q: (3,−2) to 3x−4y+5=0.

Solution: |9+8+5|/5=22/5

Answer: 22/5

Advanced

Q: Find k if (1,2) is distance 1 from x−y+k=0.

Solution: |1−2+k|/√2=1 → |k−1|=√2

Answer: k=1±√2

Exam

Q: Distance between parallel x+y=1 and x+y=5.

Solution: |1−5|/√2=4/√2=2√2

Answer: 2√2

tan φ = |(m₁ − m₂)/(1 + m₁ m₂)|

Definition: Acute angle φ between two lines with slopes m₁, m₂ (if 1+m₁m₂ ≠ 0).

Derivation

From tan(A−B) formula on the two inclination angles.

Variables

m₁, m₂ = slopes · φ = angle between lines

Why it works

If 1+m₁m₂=0 lines are perpendicular (tan undefined / 90°).

Historical context

Comes from trigonometry of difference of inclinations.

Deep understanding

For general form ax+by+c=0 use cos θ formula with coefficients.

2. Diagrams & Visuals

Pencil sketch · labelled · step-by-step breakdown below

Find m₁, m₂
Check 1+m₁m₂
Apply tan φ formula
State acute angle if required

3. Solved Examples

Basic

Q: m₁=1, m₂=0.

Solution: tanφ=|1|=1 → φ=45°

Answer: 45°

Intermediate

Q: m₁=2, m₂=1/2.

Solution: tanφ=|(2−0.5)/(1+1)|=1.5/2=0.75

Answer: φ = arctan(3/4)

Advanced

Q: Lines y=2x and y=−x/2.

Solution: 1+m₁m₂=0 → perpendicular

Answer: 90°

Exam

Q: Angle between x−y=0 and x+y=0.

Solution: m=1 and m=−1 → ⊥

Answer: 90°

5. Special Features & Extras

Complete study guide — exam tips, common mistakes, memory aids, and quick reference.

Exam Tips & Tricks

Convert to slope form before finding angles unless using coefficient formula.
Distance formula needs ax+by+c=0 with consistent c sign.
Parallel to axes: y=k or x=k — write immediately.

Common Student Mistakes

Writing y=mx+c for vertical lines
Dropping absolute value in distance
Using 1−m₁m₂ instead of 1+m₁m₂ in denominator

Memory Aids & Mnemonics

Point–slope: ‘start at point, walk with slope m’.

Distance: |plug point| over √(a²+b²).

Which Formula When?

Point + slope → point–slope form
Intercepts known → intercept form
Point to line length → distance formula
Angle between → tan φ formula

Quick reference box

• y−y₁=m(x−x₁) · y=mx+c · x/a+y/b=1

• d=|ax+by+c|/√(a²+b²)

• tanφ=|(m₁−m₂)/(1+m₁m₂)|

PYQ — Previous Year Questions

English questions from NIOS Mathematics (311) public / sample papers (Sample 2024, Apr 2024, Oct 2024, Apr 2025; board-style where scanned papers had no extractable text). Mapped exclusively to this chapter — no cross-chapter duplicates. Use Model Answer for the marking key and Explanation for working.

L14 — Straight Lines

6 question(s) · Sources: Apr 2024, Sample QP 2024

PYQ1. The intercepts made by the line 3x + 2y − 12 = 0 on the coordinate axes are:

(A) (4 and −6)
(B) (−4 and 6)
(C) (−4 and −6)
(D) (4 and 6)

1 mark(s) · MCQ · Sample QP 2024 · Q2(i)

Model Answer

1) (4 and 6)

Explanation

1) Write x/4 + y/6 = 1 by dividing 3x+2y=12 by

2) 1

2. Hence x-intercept = 4, y-intercept = 6.

PYQ2. Find the equation of the line passing through (3, 7) and (−2, 5).

1 mark(s) · SA · Sample QP 2024 · Q2(ii) OR

Model Answer

1) 2x − 5y + 29 = 0

Explanation

1) m = (5−7)/(−2−3) = (−2)/(−5) = 2/

5. Then y−7=(2/5)(x−3) ⇒ 5(y−7)=2(x−3) ⇒ 5y−35=2x−6 ⇒ 2x−5y+29=0.

PYQ3. The angle between the lines 2x + 3y = 4 and 3x − 2y = 7 is:

1 mark(s) · SA · Sample QP 2024 · Q4

Model Answer

1) 90° (π/2 radians)

Explanation

1) m₁ = −2/3, m₂ = 3/

2. Product m₁m₂ = (−2/3)(3/2) = −1 ⇒ lines are perpendicular ⇒ angle = 90°.

PYQ4. The slope of the line x + 3y = 0 is:

(A) 0
(B) 1
(C) −3
(D) −1/3

1 mark(s) · MCQ · Apr 2024 · Q4(a)

Model Answer

1) −1/3

Explanation

1) x+3y=0 ⇒ y=−(1/3)x ⇒ m=−1/3.

PYQ5. Write the equation of the line with slope 3/2 passing through the point (−1, 2).

1 mark(s) · SA · Apr 2024 · Q4 OR (concept)

Model Answer

1) 3x − 2y + 7 = 0

Explanation

1) y−2=(3/2)(x+1) ⇒ 2(y−2)=3(x+1) ⇒ 2y−4=3x+3 ⇒ 3x−2y+7=0.

PYQ6. The x-intercept and the y-intercept of the line 4x − 3y − 6 = 0 are respectively:

(A) 3/2 and 2
(B) 3/2 and −2
(C) −3/2 and 2
(D) 2 and 3/2

1 mark(s) · MCQ · Apr 2024 · Q1

Model Answer

1) 3/2 and −2

Explanation

1) 4x−3y=6 ⇒ x/(3/2) + y/(−2)=

1. So x-intercept = 3/2, y-intercept = −2.

Problem Solving — L14 Straight Lines

10 English exam-style problems for this chapter (NIOS Mathematics 311). Each question states the formula, gives full step-by-step working with the final answer, plus formal/easy explanations. English only.

Question 1 of 10Parallel to axes

Find the equation of the line through (−2, −3) that is (i) parallel to the x-axis (ii) parallel to the y-axis.

y = k · x = h

1) (i) Parallel to x-axis ⇒ y = constant. Through (−2,−3) ⇒ y = −3.

2) (ii) Parallel to y-axis ⇒ x = constant. Through (−2,−3) ⇒ x = −2.

Answer: (i) y = −3 (ii) x = −2

Formula used

y = k · x = h

Textbook formal language

A horizontal line has equation y=k; a vertical line has equation x=h. Substituting the given point fixes k and h.

Easy language (same calculation)

Same height as the point → y=−3. Same left-right position → x=−2.

Why this formula

Vertical lines cannot be written as y=mx+c.

Exam tip

Parallel to x-axis means constant y, not constant x.

Common mistakes

Writing x=−3 for horizontal
Using y=mx+c with undefined m

Question 2 of 10Slope-intercept

Find the equation of the line with slope 4 and y-intercept 0.

y = mx + c

1) Slope-intercept form: y = mx + c.

2) m = 4, c = 0 ⇒ y = 4x + 0 ⇒ y = 4x.

Answer: y = 4x

Formula used

y = mx + c

Textbook formal language

Substituting m=4 and c=0 into y=mx+c yields y=4x.

Easy language (same calculation)

Slope 4 and crosses origin → y=4x.

Why this formula

c=0 means the line passes through the origin.

Exam tip

Do not write 4x+y=0 unless asked for another form (optional: 4x−y=0).

Common mistakes

Using x-intercept by mistake
Writing m=0

Question 3 of 10Point-slope

Find the equation of the line through (2, −1) with slope 2/3.

y − y₁ = m(x − x₁)

1) y − (−1) = (2/3)(x − 2).

2) y + 1 = (2/3)x − 4/3.

3) Multiply by 3: 3y + 3 = 2x − 4.

4) 2x − 3y − 7 = 0.

Answer: 2x − 3y − 7 = 0

Formula used

y − y₁ = m(x − x₁)

Textbook formal language

Using point-slope form and clearing denominators gives the Cartesian equation 2x−3y−7=0.

Easy language (same calculation)

Plug point and slope into y−y₁=m(x−x₁), then tidy.

Why this formula

Any non-vertical line has a unique equation up to a non-zero multiple.

Exam tip

Clear fractions at the end for a neat integer form.

Common mistakes

Dropping the sign of y₁
Forgetting to multiply every term by 3

Question 4 of 10Two-point

Find the equation of the line through (3, −7) and (−2, −5).

y−y₁ = [(y₂−y₁)/(x₂−x₁)](x−x₁)

1) m = (−5−(−7))/(−2−3) = (2)/(−5) = −2/5.

2) y − (−7) = (−2/5)(x − 3).

3) 5(y+7) = −2(x−3) ⇒ 5y + 35 = −2x + 6.

4) 2x + 5y + 29 = 0.

Answer: 2x + 5y + 29 = 0

Formula used

y−y₁ = [(y₂−y₁)/(x₂−x₁)](x−x₁)

Textbook formal language

Two-point form produces slope −2/5 and simplifies to 2x+5y+29=0.

Easy language (same calculation)

Find slope between the points, then use point-slope with either point.

Why this formula

Both given points must satisfy the final equation (quick check).

Exam tip

Verify by substituting (3,−7) and (−2,−5).

Common mistakes

Wrong sign in Δy
Using external section by mistake

Question 5 of 10Distance point–line

Find the perpendicular distance of the point (2, 3) from the line 3x + 2y + 4 = 0.

d = |Ax₁+By₁+C|/√(A²+B²)

1) Here A=3, B=2, C=4, (x₁,y₁)=(2,3).

2) Numerator |3·2 + 2·3 + 4| = |6+6+4| = 16.

3) Denominator √(9+4)=√13.

4) d = 16/√13.

Answer: 16/√13

Formula used

d = |Ax₁+By₁+C|/√(A²+B²)

Textbook formal language

The distance formula yields |6+6+4|/√13 = 16/√13.

Easy language (same calculation)

Plug into |Ax+By+C| over root A²+B².

Why this formula

Rationalise only if the question asks (optional: (16√13)/13).

Exam tip

Keep the absolute value so distance is positive.

Common mistakes

Dropping absolute value
Using √(A+B) wrongly

Question 6 of 10Parallel line

Find the equation of the line through (1, 2) parallel to 2x + 3y + 6 = 0.

Ax+By+K=0 parallel to Ax+By+C=0

1) Any line parallel to 2x+3y+6=0 has form 2x+3y+K=0.

2) Passes through (1,2): 2(1)+3(2)+K=0 ⇒ 2+6+K=0 ⇒ K=−8.

3) Equation: 2x + 3y − 8 = 0 (or 2x+3y=8).

Answer: 2x + 3y − 8 = 0

Formula used

Ax+By+K=0 parallel to Ax+By+C=0

Textbook formal language

Parallel lines share A and B; the constant is fixed by the given point.

Easy language (same calculation)

Same 2x+3y, choose K so the point fits: K=−8.

Why this formula

Parallel ⇒ equal slopes −A/B.

Exam tip

Do not change coefficients of x and y for parallel lines.

Common mistakes

Using Bx−Ay for parallel
Arithmetic error in K

Question 7 of 10Point-slope

Write the equation of the line through (2, −1) with slope 3.

y − y₁ = m(x − x₁)

1) y − (−1) = 3(x − 2).

2) y + 1 = 3x − 6.

3) y = 3x − 7 (or 3x − y − 7 = 0).

Answer: y = 3x − 7

Formula used

y − y₁ = m(x − x₁)

Textbook formal language

Point–slope form with m=3 and (2,−1) simplifies to y=3x−7.

Easy language (same calculation)

Start at the point, rise 3 for each unit run.

Why this formula

General form ax+by+c=0 is also acceptable.

Exam tip

Expand carefully with the minus signs.

Common mistakes

y=3x−1
y−2=3(x+1)

Question 8 of 10Intercepts

A line has x-intercept 5 and y-intercept 2. Write its equation.

x/a + y/b = 1

1) Intercept form: x/5 + y/2 = 1.

2) Multiply by 10: 2x + 5y = 10.

Answer: x/5 + y/2 = 1 (or 2x + 5y = 10)

Formula used

x/a + y/b = 1

Textbook formal language

Intercept form uses a=5, b=2.

Easy language (same calculation)

Meets axes at (5,0) and (0,2).

Why this formula

Clear denominators for general form.

Exam tip

a and b must be non-zero.

Common mistakes

x/2+y/5=1
xy=10

Question 9 of 10Distance to line

Find the perpendicular distance from the point (1, 2) to the line 3x − 4y + 5 = 0.

d = |ax₁+by₁+c|/√(a²+b²)

1) Numerator |3(1) − 4(2) + 5| = |3 − 8 + 5| = |0| = 0.

2) Distance = 0.

3) The point lies on the line.

Answer: 0 (point lies on the line)

Formula used

d = |ax₁+by₁+c|/√(a²+b²)

Textbook formal language

Zero distance means the point satisfies the line equation.

Easy language (same calculation)

Plug in: 3−8+5=0.

Why this formula

No need for the denominator when numerator is 0.

Exam tip

If the numerator is zero, the point is on the line.

Common mistakes

Forcing a non-zero distance
Wrong absolute value

Question 10 of 10Parallel

Show that the lines 2x − y + 3 = 0 and 4x − 2y − 7 = 0 are parallel.

Parallel lines: m₁ = m₂

1) Write slopes: y = 2x + 3 ⇒ m₁ = 2.

2) 4x − 2y − 7 = 0 ⇒ 2y = 4x − 7 ⇒ y = 2x − 7/2 ⇒ m₂ = 2.

3) m₁ = m₂ = 2 ⇒ the lines are parallel (and distinct since constants differ).

Answer: Parallel (both slopes = 2)

Formula used

Parallel lines: m₁ = m₂

Textbook formal language

Equal slopes imply parallel lines.

Easy language (same calculation)

Second line is twice the first’s coefficients of x,y but different constant.

Why this formula

If a₁/a₂ = b₁/b₂ ≠ c₁/c₂ they are parallel distinct.

Exam tip

Compare slopes or coefficient ratios.

Common mistakes

Calling them coincident
Using product of slopes