⭐ Sabse Important Points · एक नज़र में
- Slope-Intercept Form: y = mx + c, where m is slope and c is y-intercept.
- Standard Equation: (x - h)² + (y - k)² = r², center (h, k), radius r.
- Point-Slope Form: y - y1 = m(x - x1), for a line through (x1, y1) with slope m.
- General Equation: x² + y² + 2gx + 2fy + c = 0, center (-g, -f), radius = √(g² + f² - c).
- Distance between parallel lines: For ax + by + c1 = 0 and ax + by + c2 = 0, distance = |c1 - c2| / √(a² + b²).
- Condition for tangency: A line y = mx + c is tangent to circle x² + y² = r² if c² = r²(1 + m²).
Introduction · परिचय
Analytical Geometry (Coordinate Geometry) is a key topic in NDA Mathematics. It involves representing geometric figures using coordinates and algebraic equations. This revision note covers straight lines, circles, and conic sections, focusing on standard forms, properties, and problem-solving techniques essential for the exam.
Straight Lines
A straight line is defined by its slope and intercepts. Key forms include slope-intercept, point-slope, two-point, and general form. The distance between parallel lines and angle between two lines are frequently tested.
📐 Formula / Rule Box · सूत्र याद रखो
Slope of line through two points
m = (y2 - y1) / (x2 - x1)
Distance from point to line
d = |ax1 + by1 + c| / √(a² + b²)
⭐ Key Points · याद रखने वाली बातें
- Slope-Intercept Form: y = mx + c, where m is slope and c is y-intercept.
- Point-Slope Form: y - y1 = m(x - x1), for a line through (x1, y1) with slope m.
- Distance between parallel lines: For ax + by + c1 = 0 and ax + by + c2 = 0, distance = |c1 - c2| / √(a² + b²).
- Angle between lines: tan θ = |(m1 - m2) / (1 + m1m2)|, where m1, m2 are slopes.
✍️ Solved Example 1 · हल किया हुआ उदाहरण
❓ Question: Find the equation of the line passing through (2, 3) and perpendicular to the line 3x + 4y = 7.
✅ Solution: 4x - 3y + 1 = 0
💡 Explanation: Slope of given line = -3/4. Perpendicular slope = 4/3. Using point-slope: y - 3 = (4/3)(x - 2) → 3y - 9 = 4x - 8 → 4x - 3y + 1 = 0.
Circles
A circle is the locus of points equidistant from a fixed center. Standard equation: (x - h)² + (y - k)² = r². General equation: x² + y² + 2gx + 2fy + c = 0, with center (-g, -f) and radius √(g² + f² - c).
📐 Formula / Rule Box · सूत्र याद रखो
Circle with center (h,k) and radius r
(x - h)² + (y - k)² = r²
Length of tangent from point (x1,y1) to circle x²+y²+2gx+2fy
√(x1² + y1² + 2gx1 + 2fy1 + c)
Standard Equation
(x - h)² + (y - k)² = r², center (h, k), radius r.
General Equation
x² + y² + 2gx + 2fy + c = 0, center (-g, -f), radius = √(g² + f² - c).
⭐ Key Points · याद रखने वाली बातें
- Condition for tangency: A line y = mx + c is tangent to circle x² + y² = r² if c² = r²(1 + m²).
✍️ Solved Example 1 · हल किया हुआ उदाहरण
❓ Question: Find the center and radius of the circle x² + y² - 4x + 6y - 12 = 0.
✅ Solution: Center (2, -3), radius = 5
💡 Explanation: Compare with x² + y² + 2gx + 2fy + c = 0: 2g = -4 → g = -2; 2f = 6 → f = 3; c = -12. Center = (-g, -f) = (2, -3). Radius = √(g² + f² - c) = √(4 + 9 + 12) = √25 = 5.
Practice Questions · खुद हल करके देखो
पहले खुद try karo, phir “Answer dekho” dabakar check karo. 👇
Q1Find the distance between the parallel lines 3x + 4y = 12 and 3x + 4y = 5.
👀 Answer dekho
✅ Answer: 7/5
💡 Explanation: Rewrite as 3x + 4y - 12 = 0 and 3x + 4y - 5 = 0. Distance = |(-12) - (-5)| / √(3² + 4²) = 7/5.
Q2Find the equation of the circle with center (1, -2) and passing through (4, 2).
👀 Answer dekho
✅ Answer: (x - 1)² + (y + 2)² = 25
💡 Explanation: Radius = distance between (1, -2) and (4, 2) = √((4-1)² + (2+2)²) = √(9 + 16) = 5. Equation: (x - 1)² + (y + 2)² = 25.
Q3Find the angle between the lines y = 2x + 3 and y = -x + 1.
👀 Answer dekho
✅ Answer: 45°
💡 Explanation: Slopes m1 = 2, m2 = -1. tan θ = |(2 - (-1)) / (1 + 2*(-1))| = |3 / (1 - 2)| = |3 / -1| = 3. Wait, tan θ = 3 → θ = arctan(3) ≈ 71.565°, not 45°. Correction: tan θ = |(m1 - m2)/(1 + m1m2)| = |(2 - (-1))/(1 + 2*(-1))| = |3/(1-2)| = |3/-1| = 3. So θ = tan⁻¹(3) ≈ 71.565°. The answer is not 45°. Let me recalc: Actually, if m1=2, m2=-1, then tan θ = |(2 - (-1))/(1 + 2*(-1))| = |3/(1-2)| = |3/-1| = 3. So θ = tan⁻¹(3) ≈ 71.565°. So the answer should be tan⁻¹(3). I'll correct: answer = tan⁻¹(3). Explanation: tan θ = |(m1 - m2)/(1 + m1m2)| = |(2 - (-1))/(1 + 2*(-1))| = |3/(1-2)| = 3, so θ = tan⁻¹(3).
⚠️ Common Mistakes · ये गलतियाँ बिल्कुल मत करना
- Forgetting to take absolute value in distance formula: Distance is always positive, so use |c1 - c2|.
- Mixing up center coordinates in general circle equation: Center is (-g, -f), not (g, f).
- In angle between lines, using tan θ = (m1 - m2)/(1 + m1m2) without absolute value: Always take absolute value to get acute angle.
⚡ Quick Revision · Exam से पहले 2 मिनट में दोहराओ
- Slope of line: m = (y2 - y1)/(x2 - x1); for perpendicular lines m1*m2 = -1.
- Circle general equation: x² + y² + 2gx + 2fy + c = 0 has center (-g, -f) and radius √(g² + f² - c).
- Distance between parallel lines ax+by+c1=0 and ax+by+c2=0 is |c1 - c2|/√(a²+b²).