Coordinate Geometry 📍

Distance formula, Section formula, Midpoint — plane geometry ko algebra se solve karo.

🌐 Cartesian Plane — Quick Revision

🗺️ Map ki tarah socho:

Google Maps mein har jagah ka ek location (latitude, longitude) hota hai. Coordinate geometry mein bhi — har point ka ek address (x, y) hota hai. René Descartes ne 17th century mein yeh system banaya tha — isliye ise Cartesian System kehte hain!

x y I (+,+) II (−,+) III (−,−) IV (+,−) A(4,3) B(−2,2) C(−2,−2) 14 −2−4 2−2

📏 Distance Formula

Do points A(x₁,y₁) aur B(x₂,y₂) ke beech ki distance Pythagoras theorem se nikaali jaati hai.
A(x₁,y₁) B(x₂,y₂) C(x₂,y₁) AC = x₂−x₁ (horizontal) BC = y₂−y₁ AB = distance
AB = √[(x₂−x₁)² + (y₂−y₁)²]
Example 1: Distance between A(4,−3) aur B(−2,5)
AB = √[(−2−4)² + (5−(−3))²]
= √[36 + 64] = √100 = 10 units
Example 2: Check karo ki A(−1, 4), B(−5, 1), C(−3, 4) ek isosceles triangle banate hain ya nahi
AB = √[16+9] = 5
BC = √[4+9] = √13
CA = √[4+0] = 2
AB ≠ BC ≠ CA — no two sides equal → Scalene triangle (no isosceles)
Note: for isosceles, exactly 2 sides equal honi chahiye
Example 3: Point P(x, y) point A(1,3) se equidistant hai aur B(−2, 5) se. Relation nikalo.
PA = PB → PA² = PB²
(x−1)² + (y−3)² = (x+2)² + (y−5)²
x²−2x+1+y²−6y+9 = x²+4x+4+y²−10y+25
−2x−6y+10 = 4x−10y+29
6x − 4y + 19 = 0

✂️ Section Formula

Point P jo line segment AB ko m:n mein internally divide karta hai:
A(x₁,y₁) B(x₂,y₂) P(x,y) m n
P = ( (mx₂+nx₁)/(m+n) , (my₂+ny₁)/(m+n) )
Midpoint (Special case — m:n = 1:1):
M = ( (x₁+x₂)/2 , (y₁+y₂)/2 )
Example 1: A(2,−3) aur B(5,6) ko 1:2 mein divide karne wala point
P = ( (1×5+2×2)/(1+2), (1×6+2×(−3))/(1+2) )
= ( (5+4)/3, (6−6)/3 ) = (3, 0)
Example 2: Midpoint of (3, 4) aur (7, −2)
M = ((3+7)/2, (4+(−2))/2) = (10/2, 2/2) = (5, 1)
Example 3: A(2,3) aur B(10,8) ko y-axis kaata hai kis ratio mein?
y-axis par x=0. Section formula se: 0 = (m×10+n×2)/(m+n) → 10m+2n=0 → 10m=−2n → m/n = −1/5
Negative ratio = external division, 1:5 ratio mein
(ya kehte hain y-axis A(2,3) aur B(10,8) ko 1:5 externally divide karta hai)
Example 4: Triangle ke vertices A(8,−4), B(9,5), C(−4,2). Centroid nikalo.
Centroid G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)
G = ((8+9−4)/3, (−4+5+2)/3) = (13/3, 3/3) = (13/3, 1)

📐 Area of Triangle

Vertices A(x₁,y₁), B(x₂,y₂), C(x₃,y₃)
Area = ½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|

Note: |...| = absolute value (hamesha positive lena)
Collinear points check: Agar A, B, C collinear hain (ek seedhi line par), to Area = 0.
x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂) = 0
Example: Area of triangle A(1,0), B(5,2), C(3,4)
Area = ½ |1(2−4) + 5(4−0) + 3(0−2)|
= ½ |−2 + 20 − 6| = ½ × 12 = 6 sq. units
Example: Check if A(3,5), B(1,1), C(0,−1) are collinear
3(1−(−1)) + 1(−1−5) + 0(5−1) = 3×2 + 1×(−6) + 0 = 6−6 = 0
Yes, collinear!

📝 Practice Questions

Q1. Distance formula se: A(0,0) aur B(3,4) ke beech distance nikalo.

AB = √(9+16) = √25 = 5 units

Q2. Kya A(4,3), B(8,5), C(4,7) ek equilateral triangle banate hain?

AB=√(16+4)=√20; BC=√(16+4)=√20; CA=√(0+16)=4. AB=BC≠CA → Isosceles, not equilateral

Q3. Section formula: A(1,4) aur B(5,−2) ko 3:1 ratio mein divide karta point.

P = ((3×5+1×1)/4, (3×(−2)+1×4)/4) = (16/4, −2/4) = (4, −1/2)

Q4. Triangle A(2,3), B(−1,0), C(4,−5) ka area nikalo.

Area = ½|2(0−(−5)) + (−1)(−5−3) + 4(3−0)| = ½|10+8+12| = ½×30 = 15 sq. units

Q5. Agar A(1,2), B(4,6), C(x,y) ek triangle hai jiska centroid (3,4) hai. C ke coordinates nikalo.

(1+4+x)/3=3 → x=4. (2+6+y)/3=4 → y=4. C = (4, 4)
🎮 Coordinate Lab →