Coordinate Geometry — Class 10 Mathematics
1. The Distance Formula
The distance between two points A(x₁, y₁) and B(x₂, y₂): AB = √[(x₂−x₁)² + (y₂−y₁)²]
Derivation Using Pythagoras Theorem
Draw perpendiculars from A and B to the axes. The horizontal distance = |x₂−x₁|. The vertical distance = |y₂−y₁|. These form the legs of a right triangle. The hypotenuse AB = √[(horizontal)² + (vertical)²].
Worked Examples
Example 1: Find distance between (2,3) and (6,7). AB = √[(6−2)²+(7−3)²] = √[16+16] = √32 = 4√2.
Example 2: Which point is farther from origin — (3,4) or (5,1)? Distance from origin to (3,4): √(9+16) = 5. To (5,1): √(25+1) = √26 ≈ 5.1. So (5,1) is slightly farther.
Example 3 — Proving a Shape: Show that A(1,2), B(4,6), C(7,2) form an isosceles triangle. AB = √(9+16)=5. BC = √(9+16)=5. CA = √(36+0)=6. Two sides equal → isosceles ✓.
Example 4 — Proving a Right Triangle: A(1,2), B(4,6), C(−1,4). AB² = (4−1)²+(6−2)² = 9+16 = 25. BC² = (−1−4)²+(4−6)² = 25+4 = 29. CA² = (1+1)²+(2−4)² = 4+4 = 8. Check: 25+8 = 33 ≠ 29. Not right. Actually: AB²=25, BC²=29, CA²=8. None = sum of other two. Not right-angled.
2. Section Formula (Internal Division)
The coordinates of point P that divides the segment joining A(x₁, y₁) and B(x₂, y₂) INTERNALLY in the ratio m:n:
P(x, y) = [(mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)]
Midpoint Formula (Special Case, m=n=1)
M(x, y) = [(x₁+x₂)/2, (y₁+y₂)/2]
Worked Examples — Section Formula
Example 1 — Midpoint: Find midpoint of A(3,−5) and B(−1,7). M = [(3−1)/2, (−5+7)/2] = (1, 1).
Example 2 — Section: Find the point dividing AB (where A=(2,3), B=(8,9)) internally in ratio 2:1. P = [(2×8+1×2)/(2+1), (2×9+1×3)/(2+1)] = [(16+2)/3, (18+3)/3] = (6, 7).
Example 3 — Ratio from Given Point: If P(4,−1) lies on the line joining A(2,−3) and B(6,1), find the ratio AP:PB. Let ratio = k:1. x-coordinate: 4 = (k×6+1×2)/(k+1) → 4(k+1) = 6k+2 → 4k+4 = 6k+2 → 2k = 2 → k = 1. Ratio = 1:1 → P is the MIDPOINT. Check y: (−1) = (1×1+1×(−3))/2 = −2/2 = −1 ✓.
3. Area of a Triangle
For vertices A(x₁,y₁), B(x₂,y₂), C(x₃,y₃): Area = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|. The ABSOLUTE VALUE ensures area is positive.
Derivation
The three vertices, along with their projections on the x-axis, form three trapeziums. The area of the triangle = algebraic sum of the areas of these trapeziums. The formula is compact: it's ½ times the absolute value of the determinant-like expression.
Worked Examples
Example 1: Area of triangle with vertices (1,2), (4,6), (5,3). = ½|1(6−3)+4(3−2)+5(2−6)| = ½|1(3)+4(1)+5(−4)| = ½|3+4−20| = ½|−13| = 6.5 sq units.
Example 2: Area with one vertex at origin. (0,0), (3,0), (0,4). = ½|0+3(0−0)+0(0−0)| → Actually: ½|0(0−4)+3(4−0)+0(0−0)| = ½|0+12+0| = 6 sq units. 'When one vertex is the origin, the formula simplifies. But it's safer to use the full formula.'
4. Checking Collinearity — Three Points
Three points A, B, C are COLLINEAR (lie on a straight line) if and only if the AREA of the triangle formed by them is ZERO. Test: ½|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)| = 0 → points ARE collinear.
Worked Example
Check if (1,2), (3,4), (5,6) are collinear. = ½|1(4−6)+3(6−2)+5(2−4)| = ½|−2+12−10| = ½|0| = 0 → Collinear ✓. 'Points form an arithmetic progression in both x and y: difference of 2 in x and 2 in y each time. This is a visual clue of collinearity.'
Alternative Method — Slope Equality: Points are collinear if slope AB = slope BC. Between (1,2) and (3,4): slope = 2/2 = 1. Between (3,4) and (5,6): slope = 2/2 = 1. Equal → collinear.
5. Common Mistakes
- Forgetting square root in distance formula: Distance = √(Δx²+Δy²), NOT just Δx²+Δy².
- Order of subtraction matters for section formula: The ratio m:n corresponds to AP:PB, where P divides AB. m is for PB, n is for AP... wait, actually m corresponds to the segment FROM B (since mx₂+nx₁). Be VERY careful.
- Forgetting absolute value for area: Area must be POSITIVE. The determinant can be negative.
- Section formula for external division: The formula DIFFERS — the sign becomes MINUS instead of PLUS in the denominator for the weighted terms.
6. AP Exam Focus
| Topic | Marks |
|---|---|
| Distance formula application | 2-3 |
| Section formula / Midpoint | 3-4 |
| Area of triangle | 3-4 |
| Collinearity test | 2-3 |
Quick Self-Test
- Distance between (−3,4) and (1,−2)? (Answer: √[(4)²+(−6)²] = √(16+36) = √52 = 2√13.)
- Midpoint of (−4,6) and (2,−8)? (Answer: M = (−1, −1).)
- Point dividing (2,3) and (8,9) in ratio 1:2? (Answer: [(2×8+1×2)/3, (2×9+1×3)/3] = (6, 7).)
- Area of Δ with vertices (0,0), (4,0), (0,6)? (Answer: ½|0+0+24| = 12 sq units.)
- Are (2,3), (4,5), (6,7) collinear? (Answer: Area = ½|2(5−7)+4(7−3)+6(3−5)| = ½|−4+16−12| = 0. Yes.)
More Section Formula Examples
Example 4 — Trisection Point: Find the points of trisection of segment joining A(2, −2) and B(−7, 4). The points divide AB in ratios 1:2 and 2:1. Point P (ratio 1:2): P = [(1×(−7)+2×2)/3, (1×4+2×(−2))/3] = [(−7+4)/3, (4−4)/3] = (−1, 0). Point Q (ratio 2:1): Q = [(2×(−7)+1×2)/3, (2×4+1×(−2))/3] = [(−14+2)/3, (8−2)/3] = (−4, 2).
Example 5 — Finding Ratio from Given Point: If C(−1, 6) divides AB where A(−3, 10) and B(6, −8), find the ratio. Let ratio = k:1. x: −1 = (k×6+1×(−3))/(k+1) → −1(k+1) = 6k−3 → −k−1 = 6k−3 → −7k = −2 → k = 2/7. Ratio = 2:7.
Distance Formula — Advanced Applications
Example — Proving a Quadrilateral is a Square: Show that A(4, 6), B(5, 3), C(2, 2), D(1, 5) form a square. AB = √(1+9) = √10. BC = √(9+1) = √10. CD = √(1+9) = √10. DA = √(9+1) = √10. All sides equal. Diagonal AC = √(4+16) = √20 = 2√5. Diagonal BD = √(16+4) = √20 = 2√5. Diagonals equal. Since all sides equal AND diagonals equal → SQUARE.
Example — Point on x-axis Equidistant from Two Points: Find the point on the x-axis equidistant from A(−2, 5) and B(2, −3). Point on x-axis: (x, 0). PA² = (x+2)² + (0−5)² = x²+4x+4+25. PB² = (x−2)² + (0+3)² = x²−4x+4+9. Equating: x²+4x+29 = x²−4x+13 → 8x = −16 → x = −2. Point: (−2, 0).
Coordinate Geometry in Real Life
- GPS navigation uses coordinate geometry to calculate distances continuously.
- Surveyors use the area formula to calculate land areas from corner coordinates.
- Video game graphics render objects using 3D coordinate geometry — every pixel on screen is a coordinate pair.
AP Exam Strategies
- Always draw a rough diagram with coordinates marked. This prevents sign errors.
- Distance from origin to (x,y): √(x²+y²) — simplify by dropping the (0−x)² terms.
- For 'prove this is an isosceles/right/equilateral triangle': find all three side lengths. Compare.
- Section formula: carefully note whether ratio is m:n (from A) or n:m (from B). Many students swap this.
Key Formulas — Quick Reference Card
| Formula | Expression | Used For |
|---|---|---|
| Distance | √[(x₂−x₁)²+(y₂−y₁)²] | Length between two points |
| Midpoint | ((x₁+x₂)/2, (y₁+y₂)/2) | Centre of a segment |
| Section (m:n) | ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)) | Point dividing segment |
| Triangle Area | ½ | x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂) |
| Collinearity | Area = 0 | Three points on same line |
| Distance from origin | √(x²+y²) | Point to (0,0) |
'Memorise these five formulas. They are the FOUNDATION of Class 10 coordinate geometry and will be used again in Class 11 and 12 for more advanced analytic geometry.' 5. Are (2,3), (4,5), (6,7) collinear? (Answer: Area = ½|2(5−7)+4(7−3)+6(3−5)| = ½|−4+16−12| = 0. Yes.)
