How do you find the distance from a point to a line?

For line ax + by + c = 0 and point (x₁, y₁), distance = |ax₁ + by₁ + c| / √(a² + b²).

What is the angle between two lines with slopes m₁ and m₂?

tan θ = |(m₁ − m₂)/(1 + m₁m₂)|. If m₁m₂ = −1, the lines are perpendicular (θ = 90°).

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

View Solutions →

import { FAQJsonLd } from "@/components/JsonLd"; import ExerciseLayout from "@/components/ExerciseLayout";

Two-Dimensional Analytical Geometry — Exercise Solutions

Q: What is the condition for two lines to be parallel?

Two lines are parallel if their slopes are equal: m₁ = m₂. In general form: a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

**Class 11	TN State Board	Maths	Chapter 6**

Complete solutions for all exercise questions on Two-Dimensional Analytical Geometry, covering locus, straight lines in various forms, angle between lines, distance from point to line, circles, and parabola.

1. How to Use This Page

Quick Recap (Section 2) — Revise every formula before attempting problems.
MCQs (Section 3) — Test conceptual clarity; each answer includes a reasoning note.
Fill in the Blanks (Section 4) — Fast recall of definitions and results.
True or False (Section 5) — Spot common misconceptions; false statements include corrections.
Match the Following (Section 6) — Connect equations to their geometric properties.
Brief Answers (Section 7) — 2-mark questions requiring short derivations or proofs.
Detailed Answers (Section 8) — 5-mark questions with full step-by-step worked examples.
Common Mistakes (Section 9) — Avoid pitfalls that cost marks in exams.
Exam Checklist (Section 10) — Final self-audit before you walk into the hall.

2. Key Concepts (Quick Recap)

2.1 Locus

A locus is the set of all points satisfying a given geometric condition. If the condition translates to an equation in x and y, that equation is the equation of the locus.

Concept	Result
Locus of a point equidistant from two fixed points	Perpendicular bisector of the segment joining them
Locus of a point at constant distance r from a fixed point (h, k)	Circle: (x − h)² + (y − k)² = r²
Locus of a point equidistant from a fixed point and a fixed line	Parabola

2.2 Slope of a Line

Formula	Expression
Slope from two points	m = (y₂ − y₁) / (x₂ − x₁)
Slope from angle with x-axis	m = tan θ (θ measured anti-clockwise from positive x-axis)
Slope from general form ax + by + c = 0	m = −a/b

Conditions:

Relation	Condition (slope form)	Condition (general form)
Parallel lines	m₁ = m₂	a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Perpendicular lines	m₁ × m₂ = −1	a₁a₂ + b₁b₂ = 0
Coincident lines	m₁ = m₂ & same y-intercept	a₁/a₂ = b₁/b₂ = c₁/c₂

2.3 Forms of a Straight Line

Form	Equation	Key Info Needed
Slope-Intercept	y = mx + c	slope m, y-intercept c
Point-Slope	y − y₁ = m(x − x₁)	one point (x₁, y₁), slope m
Two-Point	(y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁)	two points
Intercept	x/a + y/b = 1	x-intercept a, y-intercept b
Normal	x cos α + y sin α = p	p = perpendicular from origin, α = angle of normal

2.4 Angle, Distance, and Family

Formula	Expression
Angle between two lines	tan θ = \	(m₁ − m₂)/(1 + m₁m₂)\
Distance from (x₁, y₁) to ax + by + c = 0	d = \	ax₁ + by₁ + c\	/ √(a² + b²)
Distance between parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0	d = \	c₁ − c₂\	/ √(a² + b²)
Family of lines through intersection of L₁ = 0 and L₂ = 0	L₁ + λL₂ = 0, λ ∈ ℝ

2.5 Circle

Property	Equation
Standard form (centre h, k; radius r)	(x − h)² + (y − k)² = r²
General form	x² + y² + 2gx + 2fy + c = 0
Centre from general form	(−g, −f)
Radius from general form	r = √(g² + f² − c)

2.6 Parabola

Type	Equation	Focus	Directrix	Latus Rectum
y² = 4ax (opens right)	y² = 4ax	(a, 0)	x = −a	4a
y² = −4ax (opens left)	y² = −4ax	(−a, 0)	x = a	4a
x² = 4ay (opens up)	x² = 4ay	(0, a)	y = −a	4a
x² = −4ay (opens down)	x² = −4ay	(0, −a)	y = a	4a

3. I. Choose the Best Answer (MCQs)

Q1. The slope of the line 3x − 5y + 7 = 0 is:

(a) 3/5    (b) −3/5    (c) 5/3    (d) −5/3

Answer(a) 3/5

Rewriting: 5y = 3x + 7 → y = (3/5)x + 7/5. Slope m = 3/5. Alternatively, from ax + by + c = 0: m = −a/b = −(3)/(−5) = 3/5.

Q2. The distance of the point (3, 4) from the line 4x + 3y − 12 = 0 is:

(a) 0    (b) 24/5    (c) 12/5    (d) 4

Answer(c) 12/5

d = |4(3) + 3(4) − 12| / √(4² + 3²)
  = |12 + 12 − 12| / √(16 + 9)
  = |12| / 5
  = 12/5

Q3. If the lines 2x + 3y − 5 = 0 and 4x + ky + 10 = 0 are parallel, then k equals:

(a) 6    (b) −6    (c) 3    (d) −3

Answer(a) 6

For parallel lines: a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

2/4 = 3/k → 1/2 = 3/k → k = 6.

Check c₁/c₂ = −5/10 = −1/2 ≠ 1/2. Condition satisfied.

Q4. The centre and radius of the circle x² + y² − 6x + 4y − 12 = 0 are:

(a) (3, −2), 5    (b) (−3, 2), 5    (c) (3, −2), √5    (d) (−3, −2), 5

Answer(a) (3, −2), 5

Compare with x² + y² + 2gx + 2fy + c = 0: 2g = −6 → g = −3; 2f = 4 → f = 2; c = −12.

Centre = (−g, −f) = (3, −2).

Radius = √(g² + f² − c) = √(9 + 4 + 12) = √25 = 5.

Q5. The latus rectum of the parabola x² = −16y is:

(a) 16    (b) 4    (c) 8    (d) 32

Answer(a) 16

Comparing x² = −4ay with x² = −16y: 4a = 16 → a = 4.

Latus rectum =	4a	= 16.

Q6. The equation of the line passing through (2, 3) and making equal intercepts on the axes is:

(a) x + y = 5    (b) x − y = −1    (c) x + y = 1    (d) 3x + 2y = 12

Answer(a) x + y = 5

Equal intercepts → a = b. Intercept form: x/a + y/a = 1 → x + y = a.

Passing through (2, 3): 2 + 3 = a → a = 5. Equation: x + y = 5.

4. II. Fill in the Blanks

1. The slope of a line making an angle of 135° with the positive x-axis is −1 (since tan 135° = tan(180° − 45°) = −tan 45° = −1).

2. The distance between the parallel lines 3x + 4y − 10 = 0 and 3x + 4y + 5 = 0 is 3 (since	−10 − 5	/√(9 + 16) = 15/5 = 3).

3. The condition for the lines ax + by + c = 0 and bx − ay + d = 0 to be perpendicular is always satisfied (since a·b + b·(−a) = ab − ab = 0; they are perpendicular for all values of a, b).

4. The locus of a point that moves so that its distance from (3, −2) is always 4 units is the circle (x − 3)² + (y + 2)² = 16.

5. The normal form of the line √3 x + y + 2 = 0 is x cos 150° + y sin 150° = 1 (after dividing by √(3 + 1) = 2 and adjusting signs; p = 1, α = 150°).

6. The angle between the lines y = 2x + 1 and y = −3x + 4 is 45° (tan θ =	2 − (−3)	/	1 + (2)(−3)	= 5/5 = 1 → θ = 45°).

7. If the line ax + by + c = 0 passes through the origin, then c = 0.

8. The radius of the circle x² + y² + 8x − 6y + 24 = 0 is 1 (g = −4, f = 3, r = √(16 + 9 − 24) = √1 = 1).

9. The focus of the parabola y² = 12x is (3, 0) (4a = 12 → a = 3, focus = (a, 0) = (3, 0)).

10. The equation of the x-axis is y = 0.

11. The family of lines passing through the intersection of x + y = 3 and x − y = 1 is given by (x + y − 3) + λ(x − y − 1) = 0, λ ∈ ℝ.

12. The length of the perpendicular from (1, 1) to the line 3x + 4y − 5 = 0 is 2/5 (	3 + 4 − 5	/√(9 + 16) = 2/5).

5. III. True or False (with Corrections)

Q1. The slope of a horizontal line is 1.

False. The slope of a horizontal line is 0 (not 1). A horizontal line has equation y = c, so m = 0.

Q2. Two lines with slopes m₁ and m₂ are perpendicular if m₁ × m₂ = 1.

False. The correct condition is m₁ × m₂ = −1 (not 1). If the product of slopes is −1, the lines are perpendicular.

Q3. The distance of the origin from the line x cos α + y sin α = p is	p	.

True. Substituting (0, 0) into the distance formula gives	0 · cos α + 0 · sin α − p	/ √(cos²α + sin²α) =	p	/1 =	p	.

Q4. A circle can be drawn through any three non-collinear points.

True. Exactly one circle passes through three given non-collinear points. The centre is the intersection of the perpendicular bisectors of chords formed by the three points.

Q5. The directrix of the parabola y² = 4ax is the line x = a.

False. The directrix of y² = 4ax is x = −a (not x = a). The focus is (a, 0) and the directrix is x + a = 0, i.e., x = −a.

Q6. Two coincident lines have the same slope and the same y-intercept.

True. Coincident lines are essentially the same line. They have m₁ = m₂ and the same y-intercept (or in general form: a₁/a₂ = b₁/b₂ = c₁/c₂).

6. IV. Match the Following

S. No.	Column A	S. No.	Column B
1	y = mx + c	(e)	Slope-intercept form
2	x/a + y/b = 1	(f)	Intercept form
3	x cos α + y sin α = p	(a)	Normal form
4	(x − h)² + (y − k)² = r²	(d)	Circle with centre (h, k)
5	y² = 4ax	(b)	Parabola opening right
6	ax + by + c = 0	(c)	General form of a line

Answer1 → e, 2 → f, 3 → a, 4 → d, 5 → b, 6 → c

7. V. Answer Briefly (2-Mark Questions)

Q1. Find the slope of the line passing through the points (−2, 5) and (4, −1).

m = (y₂ − y₁) / (x₂ − x₁)
  = (−1 − 5) / (4 − (−2))
  = −6 / 6
  = −1

The slope of the line is −1.

Q2. Find the equation of the line passing through (3, 7) and making an angle of 60° with the positive x-axis.

m = tan 60° = √3
Using point-slope form:
y − 7 = √3(x − 3)
y − 7 = √3 · x − 3√3
y = √3 · x − 3√3 + 7

The equation is y = √3x + (7 − 3√3) or equivalently √3x − y + 7 − 3√3 = 0.

Q3. Find the distance between the parallel lines 5x + 12y − 7 = 0 and 5x + 12y + 23 = 0.

d = |c₁ − c₂| / √(a² + b²)
  = |−7 − 23| / √(25 + 144)
  = |−30| / √169
  = 30 / 13

The distance is 30/13 units.

Q4. Find the centre and radius of the circle x² + y² − 4x + 6y − 5 = 0.

Comparing with x² + y² + 2gx + 2fy + c = 0:
  2g = −4  →  g = −2
  2f = 6   →  f = 3
  c = −5

Centre = (−g, −f) = (2, −3)

Radius = √(g² + f² − c)
       = √(4 + 9 + 5)
       = √18
       = 3√2

Centre = (2, −3), Radius = 3√2 units.

Q5. Find the focus and directrix of the parabola y² = −8x.

Comparing y² = −4ax with y² = −8x:
  4a = 8  →  a = 2

Focus = (−a, 0) = (−2, 0)
Directrix: x = a, i.e., x = 2
Latus rectum = 4a = 8

Focus: (−2, 0); Directrix: x = 2.

8. VI. Answer in Detail (5-Mark Questions)

Q1. Find the equation of the line passing through the point of intersection of the lines 3x + 2y − 7 = 0 and x − y + 1 = 0 and also through the point (2, 3).

Step 1: Find the point of intersection.
  3x + 2y − 7 = 0   ...(1)
  x − y + 1 = 0      ...(2)

From (2): x = y − 1
Substitute in (1): 3(y − 1) + 2y − 7 = 0
  3y − 3 + 2y − 7 = 0
  5y − 10 = 0
  y = 2
  x = 2 − 1 = 1

Point of intersection: (1, 2)

Step 2: Equation of line through (1, 2) and (2, 3).
  Slope m = (3 − 2)/(2 − 1) = 1/1 = 1

Using point-slope form:
  y − 2 = 1(x − 1)
  y − 2 = x − 1
  x − y + 1 = 0

The required equation is x − y + 1 = 0.

Using family of lines method:

Family: (3x + 2y − 7) + λ(x − y + 1) = 0

For the line to pass through (2, 3):
  (3(2) + 2(3) − 7) + λ(2 − 3 + 1) = 0
  (6 + 6 − 7) + λ(0) = 0
  5 = 0

This cannot be satisfied for any λ, meaning the point
(2, 3) lies on the line x − y + 1 = 0 itself (check:
2 − 3 + 1 = 0). So the answer from both methods is
consistent: x − y + 1 = 0.

Q2. Find the equation of the circle passing through the points (1, 0), (0, −6), and (3, 3). Also find its centre and radius.

Step 1: Let the circle be x² + y² + 2gx + 2fy + c = 0.

Substituting (1, 0):  1 + 0 + 2g(1) + 2f(0) + c = 0
  → 2g + c = −1           ...(1)

Substituting (0, −6): 0 + 36 + 0 + 2f(−6) + c = 0
  → −12f + c = −36          ...(2)

Substituting (3, 3):  9 + 9 + 2g(3) + 2f(3) + c = 0
  → 6g + 6f + c = −18       ...(3)

Step 2: Solve the system.
From (1): c = −1 − 2g

Substitute in (2): −12f + (−1 − 2g) = −36
  −2g − 12f = −35           ...(4)

Substitute c in (3): 6g + 6f + (−1 − 2g) = −18
  4g + 6f = −17              ...(5)

From (5): g = (−17 − 6f)/4

Substitute in (4):
  −2(−17 − 6f)/4 − 12f = −35
  (34 + 12f)/4 − 12f = −35
  34 + 12f − 48f = −140
  −36f = −174
  f = 174/36 = 29/6

From (5): 4g + 6(29/6) = −17
  4g + 29 = −17
  4g = −46
  g = −23/2

From (1): c = −1 − 2(−23/2) = −1 + 23 = 22

Step 3: Write the circle equation.
x² + y² + 2(−23/2)x + 2(29/6)y + 22 = 0
x² + y² − 23x + (29/3)y + 22 = 0

Multiply by 3:
3x² + 3y² − 69x + 29y + 66 = 0

Step 4: Centre and radius.
Centre = (−g, −f) = (23/2, −29/6)
Radius = √(g² + f² − c) = √((529/4) + (841/36) − 22)
       = √(4761/36 + 841/36 − 792/36)
       = √(4810/36)
       = √4810/6

The circle equation is 3x² + 3y² − 69x + 29y + 66 = 0, centre (23/2, −29/6), radius √4810/6.

Q3. Derive the condition for the lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 to be (i) parallel, (ii) perpendicular, (iii) coincident. Hence find the angle between the lines 2x + 3y − 4 = 0 and x − y + 2 = 0.

Derivation:

Slope of L₁: m₁ = −a₁/b₁
Slope of L₂: m₂ = −a₂/b₂

(i) Parallel: m₁ = m₂
  −a₁/b₁ = −a₂/b₂
  a₁/b₁ = a₂/b₂
  ⟹ a₁/a₂ = b₁/b₂ (and c₁/c₂ need not be equal)
  Condition: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

(ii) Perpendicular: m₁ · m₂ = −1
  (−a₁/b₁)(−a₂/b₂) = −1
  a₁a₂/(b₁b₂) = −1
  a₁a₂ = −b₁b₂
  ⟹ a₁a₂ + b₁b₂ = 0

(iii) Coincident: same slope AND same intercept
  a₁/a₂ = b₁/b₂ = c₁/c₂

Numerical example:

L₁: 2x + 3y − 4 = 0  →  m₁ = −2/3
L₂: x − y + 2 = 0    →  m₂ = −(1)/(−1) = 1

Angle between lines:
  tan θ = |(m₁ − m₂)/(1 + m₁m₂)|
        = |(−2/3 − 1)/(1 + (−2/3)(1))|
        = |(−5/3)/(1 − 2/3)|
        = |(−5/3)/(1/3)|
        = |−5|
        = 5

  θ = tan⁻¹(5)

The angle between the lines is tan⁻¹(5) ≈ 78.69°.

Q4. A rod of length 10 units slides with its ends on the coordinate axes. Find the locus of the midpoint of the rod and identify the curve.

Let the ends of the rod be A(a, 0) on the x-axis and
B(0, b) on the y-axis.

Given: AB = 10
  √(a² + b²) = 10
  a² + b² = 100          ...(1)

Let P(h, k) be the midpoint of AB.
  h = (a + 0)/2 = a/2  →  a = 2h
  k = (0 + b)/2 = b/2   →  b = 2k

Substituting in (1):
  (2h)² + (2k)² = 100
  4h² + 4k² = 100
  h² + k² = 25

Replacing h by x and k by y:
  x² + y² = 25

The locus is a circle with centre at the origin
and radius 5 units.

The locus of the midpoint is the circle x² + y² = 25.

Q5. Find the equation of the parabola with focus (0, −3) and directrix y = 3. Also find the latus rectum, vertex, and axis.

Given: Focus S = (0, −3), Directrix: y = 3

Let P(x, y) be any point on the parabola.
By definition: distance from P to focus = distance from P to directrix

  √((x − 0)² + (y + 3)²) = |y − 3|

Squaring both sides:
  x² + (y + 3)² = (y − 3)²
  x² + y² + 6y + 9 = y² − 6y + 9
  x² + 12y = 0
  x² = −12y

This is of the form x² = −4ay, where 4a = 12, so a = 3.

Vertex: midpoint of focus and directrix on the axis
  = (0, (−3 + 3)/2) = (0, 0)

Axis: x = 0 (the y-axis)

Latus rectum = |4a| = 12

Summary:
  Equation of parabola: x² = −12y
  Vertex: (0, 0)
  Focus: (0, −3)
  Directrix: y = 3
  Axis: x = 0 (y-axis)
  Latus rectum: 12 units

9. Common Mistakes to Avoid

1. Sign error in slope formula — m = (y₂ − y₁)/(x₂ − x₁), NOT (x₂ − x₁)/(y₂ − y₁). The numerator is always the difference of y-coordinates.

2. Forgetting absolute value in distance formula — The distance from point (x₁, y₁) to line ax + by + c = 0 is **	ax₁ + by₁ + c	** / √(a² + b²). The absolute value is essential; distance is always non-negative.

3. Confusing parallel and perpendicular conditions — Parallel: m₁ = m₂. Perpendicular: m₁ × m₂ = −1. Students often mix up the product and equality.

4. Wrong centre from general circle equation — For x² + y² + 2gx + 2fy + c = 0, the centre is (−g, −f), NOT (g, f). The signs must be reversed.

5. Mixing up parabola standard forms — y² = 4ax opens right (towards positive x); y² = −4ax opens left. x² = 4ay opens up; x² = −4ay opens down. Do not confuse which variable is squared with the direction of opening.

6. Normal form division error — To convert ax + by + c = 0 to normal form x cos α + y sin α = p, you must divide the entire equation (including c) by √(a² + b²), and ensure p > 0. If p comes out negative, multiply the entire equation by −1.

7. Distance between parallel lines requires same coefficients — The formula d =	c₁ − c₂	/√(a² + b²) works only when the coefficients of x and y are identical in both equations. If not, first normalize one equation to match before applying.

8. Omitting the absolute value in the angle formula — tan θ = **	(m₁ − m₂)/(1 + m₁m₂)	**. Without the absolute value, you may get a negative angle.

9. Family of lines misinterpretation — L₁ + λL₂ = 0 represents ALL lines through the intersection of L₁ = 0 and L₂ = 0 except the line L₂ = 0 itself (which corresponds to λ → ∞). To include L₂, mention it separately.

10. Locus equation must be simplified — After eliminating the parameter, always simplify the equation. Leaving it in an unsimplified form costs marks.

10. Exam Checklist

Before entering the exam hall, make sure you can answer Yes to every item below:

[ ] I can find the slope from two points, from a general-form equation, and from an angle with the x-axis.
[ ] I can write the equation of a line in all five forms (slope-intercept, point-slope, two-point, intercept, normal) and convert between them.
[ ] I know the conditions for parallel, perpendicular, and coincident lines — in both slope form and general form.
[ ] I can compute the angle between two lines using tan θ = |(m₁ − m₂)/(1 + m₁m₂)|.
[ ] I can find the distance from a point to a line and the distance between two parallel lines.
[ ] I can use the family of lines (L₁ + λL₂ = 0) to find a line through an intersection point with a given additional condition.
[ ] I can find the centre and radius of a circle from both the standard form and the general form.
[ ] I can write the equation of a circle passing through three non-collinear points.
[ ] I can identify the type, vertex, focus, directrix, axis, and latus rectum for all four standard parabolas.
[ ] I can derive the locus of a point satisfying a geometric condition and simplify the resulting equation.
[ ] I can convert a general-form line equation to normal form correctly (ensuring p > 0).
[ ] I remember to use absolute values in the distance formula and the angle formula.
[ ] I know the sign conventions: centre of circle is (−g, −f), focus of y² = 4ax is (a, 0), and focus of x² = 4ay is (0, a).

<FAQJsonLd faqs={[ { q: "What is the slope-intercept form of a line?", a: "y = mx + c, where m is the slope and c is the y-intercept." },

{ q: "What is the condition for two lines to be parallel?", a: "Two lines are parallel if their slopes are equal: m₁ = m₂. In general form: a₁/a₂ = b₁/b₂ ≠ c₁/c₂." },

]} />

{ q: "How do you find the distance from a point to a line?", a: "For line ax + by + c = 0 and point (x₁, y₁), distance =	ax₁ + by₁ + c	/ √(a² + b²)." },
{ q: "What is the angle between two lines with slopes m₁ and m₂?", a: "tan θ =	(m₁ − m₂)/(1 + m₁m₂)	. If m₁m₂ = −1, the lines are perpendicular (θ = 90°)." },

</ExerciseLayout>

Frequently Asked Questions

y = mx + c, where m is the slope and c is the y-intercept.

Analytical Geometry Solutions | KalviGavani