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Coordinate geometry

Lesson 28 of 113

1. Ye topic hai kya?

Har point ko do numbers se lock karo: (x, y). Phir bina figure scale ke distance, beech ka point, slope, teen points ka triangle area. Paper: do points kitni door, m:n mein kaun sa point, x-axis join ko kahan kaat-ti, teen points collinear ya nahi.

Yahan lock

Quadrants, origin se √(x²+y²), distance, section, midpoint, centroid, slope (y₂−y₁)/(x₂−x₁), area ½|…|, collinear ⇒ area 0.

Yahan kya nahi

DE ∥ BC, AA/SAS. a1/a2 vs b1/b2 (do lines). 30° tower. Circle x²+y²=r² Class-XI. Reasoning: 3 km east–north walk. AP 6,10,14.

2. Plane — axes, quadrants

Horizontal x-axis, vertical y-axis, milte origin O(0,0). Point (x, y): pehla number x (right +, left −), dusra y (up +, down −). Order lock: (2, 5) ≠ (5, 2).

y x O I (+,+) II (−,+) III (−,−) IV (+,−) (3, 4) (−4, 4) (3, −3)

Sign se quadrant. Axes pe (4, 0) ya (0, −2) — kisi quadrant mein nahi.

PointxyJagah
(3, 4)++I
(−4, 4)−+II
(−2, −5)−−III
(5, −1)+−IV
(0, 4)0+y-axis
(−7, 0)−0x-axis
(−3, 4) II hai, III nahi. Pehla sign x ka. (0, 5) “up” hai lekin quadrant nahi — axis.

3. Distance — do points

A(x₁, y₁), B(x₂, y₂). Horizontal jump x₂−x₁, vertical y₂−y₁. Yeh do legs of a right triangle (axes ke parallel). Hypotenuse hi AB.

AB = √[(x₂−x₁)² + (y₂−y₁)²]. Origin se: OP = √(x²+y²). Square andar — sign cancel (minus² plus). Order A→B ya B→A same.
A(1, 2) B(5, 5) Δx = 4 Δy = 3

Δx=4, Δy=3 → AB=√(16+9)=5. Geometry wala 8-15-17 ladder yahan nahi — yahan coordinates se √.

Example

A(1, 2), B(5, 5). AB?

Solution

  1. Δ

    5−1=4, 5−2=3.

  2. √

    √(16+9)=√25=5.

Answer5

Origin se (−6, 8): √(36+64)=√100=10. (−2, −3) se (4, 5): Δx=6, Δy=8, √(36+64)=10.

Formula Pythagoras se aati hai, lekin sawal points deta hai, “sihi 13 m” nahi. Right triangle check: teen distances nikaalo, dekho p²+q²=r² (sabse badi r).

4. Section — m : n internally

P, join AB ko internally m : n mein kaat-ta hai matlab AP : PB = m : n. P beech mein hai, A aur B ke darmiyan. m bada → P, B ke kareeb (zyada hissa A se already chal chuka).

P(x, y) divides A(x₁,y₁), B(x₂,y₂) in m : n:
x = (m x₂ + n x₁) / (m+n),   y = (m y₂ + n y₁) / (m+n).
Weighted average: B ko m weight, A ko n weight. Denominator m+n.
Example

P divides A(2, 3) aur B(8, 15) internally 1 : 2 mein. P?

Solution

  1. m=1, n=2

    x = (1·8 + 2·2)/3 = (8+4)/3 = 4.

  2. y

    y = (1·15 + 2·3)/3 = (15+6)/3 = 7. P(4, 7).

  3. Check

    1:2 → P, A ke kareeb (chhota m). 4, 2 se 8 ke beech pehle-tihai: 2+2=4. Theek.

Answer(4, 7)

External (kabhi paper pe): P, AB ke extend pe, AP:PB = m:n lekin P segment ke bahar. x = (m x₂ − n x₁)/(m−n) jab m ≠ n. Sign minus, denominator m−n. Pehle internal lock karo; external alag formula.

(m x₂ + n x₁) — pehla m dusre point B se. Ulta (m x₁ + n x₂) tabhi jab ratio n:m likha ho. Ratio “1:2” padho: m=1, n=2, A first point.

5. Midpoint aur centroid

Midpoint = section 1 : 1. Formula short:

Midpoint of A, B: ((x₁+x₂)/2, (y₁+y₂)/2). Average. Origin se symmetric: (a, b) ka opposite (−a, −b).

(−4, 6) aur (8, −2): x=(−4+8)/2=2, y=(6−2)/2=2 → (2, 2).

Centroid G of ΔABC = medians ka milna. Har median ko vertex se 2 : 1 mein kaat-ta (vertex 2, midpoint 1). Shortcut:

G = ( (x₁+x₂+x₃)/3 , (y₁+y₂+y₃)/3 ). Average of three vertices.

(1, 2), (4, 6), (7, 4): G = (12/3, 12/3) = (4, 4).

Slope of line through A(x₁,y₁), B(x₂,y₂):

m = (y₂ − y₁) / (x₂ − x₁). Horizontal (y same) → m=0. Vertical (x same) → undefined. Pair-linear ki “k” yahan nahi — wahan a₁/a₂.

(1, 1) se (2, 3): m=(3−1)/(2−1)=2. (2, 3) se (3, 5): (5−3)/(3−2)=2. Same slope ⇒ collinear (area 0, neeche).

6. Area of triangle — collinear

Vertices A(x₁,y₁), B(x₂,y₂), C(x₃,y₃). Figure scale pe mat socho — formula:

Area = ½ | x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂) |.
Modulus — area positive. Collinear (ek line pe) ⇔ area = 0 ⇔ woh expression 0.
Example

(2, 3), (6, 3), (6, 8). Area?

Solution

  1. Plug

    ½ |2(3−8) + 6(8−3) + 6(3−3)| = ½ |2(−5) + 6(5) + 0| = ½ |−10+30| = 10.

  2. Picture

    Base horizontal 4, height 5 — right triangle, ½×4×5=10. Same. (Geometry ½bh yahan coordinates se verify, similar-triangles nahi.)

Answer10

Collinear: (1, 1), (2, 3), (3, 5). ½|1(3−5)+2(5−1)+3(1−3)| = ½|−2+8−6|=0. Ek line, slope 2. Triangle nahi.

Modulus bhool → negative “area” option mein trap. Order of vertices change kare to sign flip, | | same. Do points same → area 0, degenerate.

7. Word — axis, equidistant

x-axis pe point: (x, 0). y-axis: (0, y). “X-axis join AB ko kis ratio mein kaat-ti” = section formula mein y=0 se m:n.

Example

x-axis, A(1, 4) aur B(5, −6) ke join ko kis ratio mein kaat-ti?

Solution

  1. y=0

    (m·(−6) + n·4)/(m+n) = 0 → −6m + 4n = 0 → 4n = 6m → m:n = 2:3.

  2. Point

    x = (2·5 + 3·1)/5 = 13/5. Kaatan (13/5, 0). y-axis kaat-ti to x=0 same tarah.

Answer2 : 3 internally
Example

x-axis pe P, A(2, 3) aur B(4, 1) se equal door. P?

Solution

  1. P(x,0)

    (x−2)²+9 = (x−4)²+1.

  2. Expand

    (x−2)² − (x−4)² = −8. (a−b)(a+b)=2·(2x−6)=−8 → 4x−12=−8 → x=1. P(1, 0).

Answer(1, 0)

8. Sawal — basic se pro

Q1 · Basic

(−3, 4) kahan?

  1. I
  2. II
  3. III
  4. y-axis

Solution

  1. Signs

    x−, y+ → II. III dono minus. D: x=0 hota.

AnswerB
Q2 · Basic

A(1, 2), B(5, 5). AB?

  1. 5
  2. 7
  3. √7
  4. 25

Solution

  1. √(16+9)

    5. B: 4+3. D: without √.

Answer5
Q3 · Basic

Origin se (−6, 8). Distance?

  1. 10
  2. 14
  3. 2
  4. √14

Solution

  1. √(36+64)

    10. B: 6+8. Minus sign distance nahi ghatata.

Answer10
Q4 · Basic

Midpoint of (−4, 6) aur (8, −2)?

  1. (2, 2)
  2. (12, 4)
  3. (2, 4)
  4. (4, 2)

Solution

  1. Average

    (4/2, 4/2)=(2, 2). B: sum without /2. C: y mein 6+2.

Answer(2, 2)
Q5 · Medium

P divides (2, 3) aur (8, 15) in 1 : 2 internally. P?

  1. (4, 7)
  2. (6, 11)
  3. (5, 9)
  4. (10, 18)

Solution

  1. Formula

    (8+4)/3=4, (15+6)/3=7. B: ratio ulta 2:1. D: bina divide.

Answer(4, 7)
Q6 · Medium

Area: (2, 3), (6, 3), (6, 8)?

  1. 10
  2. 20
  3. 5
  4. 0

Solution

  1. ½|−10+30|

    10. B: ½ bhool. D: collinear nahi — teesra y alag.

Answer10
Q7 · Medium

(1, 1), (2, 3), (3, 5) collinear?

  1. Haan, area 0
  2. Nahi, area 2
  3. Nahi, quadrant alag
  4. Sirf midpoint se pata chale

Solution

  1. Area

    ½|−2+8−6|=0. C: saare I mein, collinear alag baat.

AnswerA
Q8 · Medium

x-axis, (1, 4) aur (5, −6) ke join ko ratio?

  1. 2 : 3
  2. 3 : 2
  3. 1 : 1
  4. 1 : 2 (signs ulta)

Solution

  1. y=0

    −6m+4n=0 → m:n=2:3. B: n:m ulta. D: 4:6 ko 1:2 bana diya (signs/weights galat).

Answer2 : 3
Q9 · Pro

Centroid of (1, 2), (4, 6), (7, 4)?

  1. (4, 4)
  2. (12, 12)
  3. (4, 6)
  4. (3, 4)

Solution

  1. /3

    (12/3, 12/3)=(4, 4). B: sum. D: do points ka mid.

Answer(4, 4)
Q10 · Pro

x-axis pe P, (2, 3) aur (4, 1) se equal door. P?

  1. (1, 0)
  2. (3, 0)
  3. (0, 0)
  4. (2, 0)

Solution

  1. PA=PB

    (x−2)²+9=(x−4)²+1 → x=1. B: x ka midpoint 3 (y ignore nahi kar sakte — y alag 3 aur 1).

Answer(1, 0)

Practice

Pehle khud, phir neeche kholo.

12345 678910

P1

(0, −5) kahan?

  1. IV
  2. III
  3. y-axis, origin ke neeche
  4. x-axis

Solution

  1. x=0

    y-axis. Quadrant nahi.

AnswerC
P2

(−2, −3) se (4, 5). Distance?

  1. 10
  2. 14
  3. 8
  4. √10

Solution

  1. Δ 6, 8

    √(36+64)=10.

Answer10
P3

Midpoint of (0, 0) aur (8, −6)?

  1. (4, −3)
  2. (8, −6)
  3. (4, 3)
  4. (−4, 3)

Solution

  1. Half

    (4, −3). C: y ka sign.

Answer(4, −3)
P4

P divides (0, 0) aur (9, 6) in 2 : 1. P?

  1. (6, 4)
  2. (3, 2)
  3. (9, 6)
  4. (4.5, 3)

Solution

  1. 2:1

    x=(2·9+1·0)/3=6, y=12/3=4. B: 1:2. D: midpoint.

Answer(6, 4)
P5

Area (0, 0), (4, 0), (0, 6)?

  1. 12
  2. 24
  3. 10
  4. 6

Solution

  1. ½×4×6

    12. B bina ½.

Answer12
P6

(1, 1), (4, 1), (4, 5). Right triangle?

  1. Haan, sides 3, 4, 5
  2. Nahi, distances 3, 4, 7
  3. Sirf similar se pata chale
  4. Equilateral

Solution

  1. Distances

    3 (horizontal), 4 (vertical), √(9+16)=5. 9+16=25. Right at (4,1).

AnswerA
P7

y-axis pe point ka x?

  1. 0
  2. 1
  3. y ke barabar
  4. Undefined

Solution

  1. x=0

    (0, y). D nahi.

Answer0
P8

(2, 3), (6, 3), (4, 3) area?

  1. 0, collinear (horizontal)
  2. 6
  3. 8
  4. 12

Solution

  1. Same y

    Sab y=3. Line y=3. Area 0.

AnswerA
P9

Centroid (0, 0), (6, 0), (0, 9)?

  1. (2, 3)
  2. (2, 0)
  3. (3, 4.5)
  4. (6, 9)

Solution

  1. /3

    (6/3, 9/3)=(2, 3). C: midpoint of hypotenuse (circumcentre yahan coincidentally alag — mid of (6,0)(0,9)=(3, 4.5)).

Answer(2, 3)
P10

P divides (1, 2) aur (7, 8) externally 3 : 1. P?

  1. (10, 11)
  2. (4, 5)
  3. (5.5, 6.5)
  4. (8, 9)

Solution

  1. External

    x=(3·7 − 1·1)/(3−1)=(21−1)/2=10. y=(24−2)/2=11. B: internal 1:1-ish. C: midpoint.

Answer(10, 11)

← Index · Triangles

Agle topic: Introduction to trigonometry.