πŸ“ Coordinate Geometry

Slope, line equations, distance formula aur section formula β€” sab ek jagah!

πŸ“ 1. Slope aur Inclination

Ek line ka slope batata hai ki woh kitni tezi se upar/neeche jaati hai. Ramp, road, graph β€” sab jagah slope use hota hai.

m > 0 A(x₁,y₁) B(xβ‚‚,yβ‚‚) Run = xβ‚‚βˆ’x₁ Rise = yβ‚‚βˆ’y₁ m < 0 m = 0 Slope Types
Slope Formula: m = (yβ‚‚ βˆ’ y₁) / (xβ‚‚ βˆ’ x₁) = Rise / Run = tan ΞΈ Jahan ΞΈ = angle line aur x-axis ke beech (inclination angle) m > 0 β†’ line upar jaati hai (left se right) m < 0 β†’ line neeche jaati hai m = 0 β†’ horizontal line m undefined β†’ vertical line (x = constant)
Example: Slope Nikaalte Hain

Do points A(2, 3) aur B(6, 11) ke beech slope nikalo. Inclination angle bhi batao.

1
m = (yβ‚‚βˆ’y₁)/(xβ‚‚βˆ’x₁) = (11βˆ’3)/(6βˆ’2) = 8/4 = 2
2
m = tan ΞΈ = 2 β†’ ΞΈ = tan⁻¹(2) β‰ˆ 63.43Β°

Parallel aur Perpendicular Lines

Parallel lines: m₁ = mβ‚‚ (same slope, different y-intercept) Perpendicular lines: m₁ Γ— mβ‚‚ = βˆ’1 (slopes ka product = βˆ’1) ya: mβ‚‚ = βˆ’1/m₁ (negative reciprocal) Example: Line 1 ka slope = 3/4 β†’ Parallel line ka slope = 3/4 β†’ Perpendicular line ka slope = βˆ’4/3
πŸ’‘ Angle between two lines: tan Ξ± = |(m₁ βˆ’ mβ‚‚)/(1 + m₁mβ‚‚)| β€” agar answer = 0, lines parallel; agar denominator = 0, lines perpendicular!

πŸ“ 2. Straight Line ke Equations

Ek line ko describe karne ke 4 main forms hain. Sab equivalent hain β€” situation ke hisaab se sahi form use karo.

1. Slope-Intercept Form

y = mx + c

m = slope, c = y-intercept (jahan line y-axis ko cut karti hai). Sabse zyada use hone wali form!

2. Point-Slope Form

y βˆ’ y₁ = m(x βˆ’ x₁)

Jab ek point aur slope diya ho. (x₁,y₁) us point ke coordinates hain.

3. Two-Point Form

(yβˆ’y₁)/(yβ‚‚βˆ’y₁) = (xβˆ’x₁)/(xβ‚‚βˆ’x₁)

Jab do points diye hoon. Slope nikaalte hain phir point-slope use karte hain.

4. Intercept Form

x/a + y/b = 1

a = x-intercept, b = y-intercept. Jab dono intercepts diye hon.

General Form

General Form: ax + by + c = 0 Slope = βˆ’a/b x-intercept = βˆ’c/a (y=0 rakho) y-intercept = βˆ’c/b (x=0 rakho) Kisi bhi form ko general form mein convert kar sakte hain!
Example 1: Slope-Intercept se General Form

Line y = 3x βˆ’ 5 ko general form mein likho. Dono intercepts nikalo.

1
General form: 3x βˆ’ y βˆ’ 5 = 0 (a=3, b=βˆ’1, c=βˆ’5)
2
x-intercept: y=0 β†’ 3x=5 β†’ x = 5/3
3
y-intercept: x=0 β†’ βˆ’y=5 β†’ y = βˆ’5 (yaani c = βˆ’5)
Example 2: Two Points se Line Equation

Points P(1, 2) aur Q(4, 8) se guzarne wali line ki equation nikalo.

1
m = (8βˆ’2)/(4βˆ’1) = 6/3 = 2
2
Point-slope: y βˆ’ 2 = 2(x βˆ’ 1)
3
y βˆ’ 2 = 2x βˆ’ 2 β†’ y = 2x (line origin se guzarti hai!)

Do Lines ka Intersection

Do lines ka intersection: simultaneous equations solve karo L1: a₁x + b₁y + c₁ = 0 L2: aβ‚‚x + bβ‚‚y + cβ‚‚ = 0 Cramer's rule ya substitution se solve karo. Example: x + y = 5 aur 2x βˆ’ y = 1 β†’ Add: 3x = 6 β†’ x = 2, y = 3 β†’ Intersection point: (2, 3)

πŸ“ 3. Distance aur Section Formula

Distance Formula

Do points P(x₁,y₁) aur Q(xβ‚‚,yβ‚‚) ke beech: d = √[(xβ‚‚βˆ’x₁)Β² + (yβ‚‚βˆ’y₁)Β²] (Pythagoras theorem ka use! Horizontal aur vertical distances ka hypotenuse)
P(x₁,y₁) Q(xβ‚‚,yβ‚‚) xβ‚‚βˆ’x₁ yβ‚‚βˆ’y₁ d = √[(Ξ”x)Β²+(Ξ”y)Β²]
Example: Distance Nikaalte Hain

A(3, βˆ’4) aur B(βˆ’1, 2) ke beech distance nikalo.

1
d = √[(-1-3)² + (2-(-4))²]
2
= √[(-4)² + (6)²] = √[16 + 36] = √52
3
= 2√13 β‰ˆ 7.21 units

Section Formula

Point P(x,y) jo A(x₁,y₁) aur B(xβ‚‚,yβ‚‚) ko m:n mein divide karta hai: Internal Division: x = (mxβ‚‚ + nx₁)/(m+n) y = (myβ‚‚ + ny₁)/(m+n) External Division: x = (mxβ‚‚ βˆ’ nx₁)/(mβˆ’n) y = (myβ‚‚ βˆ’ ny₁)/(mβˆ’n) Midpoint (m=n=1): x = (x₁+xβ‚‚)/2, y = (y₁+yβ‚‚)/2
Example: Section Formula

A(2,5) aur B(8,βˆ’1) ko join karne wali line ko point P, 2:1 mein internally divide karta hai. P nikalo.

1
m=2, n=1, (x₁,y₁)=(2,5), (xβ‚‚,yβ‚‚)=(8,βˆ’1)
2
x = (2Γ—8 + 1Γ—2)/(2+1) = (16+2)/3 = 18/3 = 6
3
y = (2Γ—(βˆ’1) + 1Γ—5)/(2+1) = (βˆ’2+5)/3 = 3/3 = 1
4
P = (6, 1)

Point se Line ki Distance

Point P(xβ‚€,yβ‚€) se line ax + by + c = 0 ki perpendicular distance: d = |axβ‚€ + byβ‚€ + c| / √(aΒ² + bΒ²) Example: Point (3,4) se line 3x βˆ’ 4y + 5 = 0 ki distance: d = |3Γ—3 βˆ’ 4Γ—4 + 5| / √(9+16) = |9βˆ’16+5| / 5 = |βˆ’2| / 5 = 2/5 = 0.4 units
FormulaUse KabYaad karne ki trick
Distance formulaDo points ke beechPythagoras theorem
Section formula (internal)Point ratio mein divide karemΓ—far + nΓ—near
MidpointSection 1:1Average of coordinates
Point-to-line distancePoint aur line ke beech|Plug in + c| / √(a²+b²)
πŸ”¬ Coordinate Geometry Lab Open Karo β†’