Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Quick Summary
Two-Dimensional Analytical Geometry translates geometric shapes into algebraic equations. Chapter 6 covers straight lines in detail — different forms, angles, distances — then introduces circles and conics. This chapter is fundamental for Class 12 Analytical Geometry (conics).
Straight Lines
Key Formulas
| Formula | Expression |
|---|---|
| Distance between (x₁,y₁) and (x₂,y₂) | √((x₂−x₁)² + (y₂−y₁)²) |
| Midpoint | ((x₁+x₂)/2, (y₁+y₂)/2) |
| Slope (gradient) | m = (y₂−y₁)/(x₂−x₁) = tan θ |
| Slope from ax+by+c=0 | m = −a/b |
Equation of a Line — Forms
| Form | Equation |
|---|---|
| Slope-intercept | y = mx + c |
| Point-slope | y − y₁ = m(x − x₁) |
| Two-point | (y−y₁)/(y₂−y₁) = (x−x₁)/(x₂−x₁) |
| Intercept form | x/a + y/b = 1 |
| Normal form | x cosα + y sinα = p |
| General form | ax + by + c = 0 |
Parallel and Perpendicular Lines
- Parallel: m₁ = m₂
- Perpendicular: m₁ × m₂ = −1
Distances
- Distance from point (x₁,y₁) to line ax+by+c=0: |ax₁+by₁+c| / √(a²+b²)
- Distance between parallel lines ax+by+c₁=0 and ax+by+c₂=0: |c₁−c₂| / √(a²+b²)
Angle Between Two Lines
| tan θ = | (m₁ − m₂) / (1 + m₁m₂) |
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Circles
Standard form: (x−h)² + (y−k)² = r² (centre (h,k), radius r)
General form: x² + y² + 2gx + 2fy + c = 0
- Centre: (−g, −f)
- Radius: r = √(g² + f² − c)
Condition for general equation to represent a circle: coefficient of x² = coefficient of y², no xy term.
Circle Through Three Points
Substitute all three points into x² + y² + Dx + Ey + F = 0, solve the three simultaneous equations.
Conics (Introduction)
A conic section is the intersection of a cone and a plane.
| Conic | Equation (standard) | Condition |
|---|---|---|
| Circle | x² + y² = r² | e = 0 |
| Parabola | y² = 4ax | e = 1 |
| Ellipse | x²/a² + y²/b² = 1 (a>b) | e < 1 |
| Hyperbola | x²/a² − y²/b² = 1 | e > 1 |
Worked Examples
Example 1: Find the equation of the line through (2, 3) with slope −2. y − 3 = −2(x − 2) → y = −2x + 7 → 2x + y − 7 = 0
Example 2: Find the distance from (3, −4) to the line 3x − 4y + 5 = 0.
| d = | 3(3) − 4(−4) + 5 | / √(9 + 16) = | 9 + 16 + 5 | / 5 = 30/5 = 6 units |
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Example 3: Find the centre and radius of x² + y² − 6x + 4y − 12 = 0. 2g = −6 → g = −3; 2f = 4 → f = 2; c = −12. Centre: (3, −2). Radius: √(9 + 4 + 12) = √25 = 5 units.
Example 4: Find the angle between lines y = 2x+1 and y = 3x−2.
| m₁ = 2, m₂ = 3. tan θ = | (2−3)/(1+6) | = 1/7. θ = tan⁻¹(1/7). |
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Important Questions
One Mark
- Find the slope of the line joining (1, 2) and (4, 8).
- Write the equation of the x-axis.
- What is the radius of the circle x² + y² = 25?
- State the condition for two lines to be perpendicular.
Short Answer (3 marks)
- Find the equation of the line passing through (1, −2) and parallel to 3x − 4y + 5 = 0.
- Find the distance between the parallel lines 3x + 4y − 6 = 0 and 3x + 4y + 4 = 0.
- Find the equation of the circle with centre (2, −3) passing through (5, 1).
Long Answer (5 marks)
- Find the equation of a circle passing through (1, 2), (3, −4), and (5, −6).
- A straight line cuts equal intercepts on both axes and passes through (2, 3). Find its equation and the intercepts.
Common Mistakes
- Using m = (x₂−x₁)/(y₂−y₁) instead of (y₂−y₁)/(x₂−x₁) — always rise over run.
- Forgetting the absolute value bars when computing distance from a point to a line.
- Finding the centre of a circle from general form: centre is (−g, −f), not (g, f).
- Perpendicular condition: m₁ × m₂ = −1 does NOT apply when one line is vertical (slope undefined).
Board Exam Tips
- Distance from a point to a line is a very common 3-mark question — memorise the formula.
- Circle problems (find equation given centre and radius, or three points) appear as 5-mark questions.
- The intercept form x/a + y/b = 1 is useful for problems stating "cuts equal intercepts" or "sum of intercepts is k".
- JEE: Locus problems (find the equation of a point satisfying a distance/ratio condition) are tested regularly.