Class 10 Maths · Chapter 7
Short answer:
Is chapter me sirf do cheezein cover hoti hain — Distance Formula (Exercise 7.1, 7 questions) aur Section Formula (Exercise 7.2, 7 questions). Distance formula do points ke beech ki seedhi-line distance nikalta hai; section formula ek point nikalta hai jo do points ko diye gaye ratio me baantta hai — midpoint formula iska hi special case hai jab ratio 1:1 ho. Total 14 questions, sab calculation-based, koi proof nahi.
Coordinate geometry algebra aur geometry ko jodta hai — shapes ko numbers se solve karte hain. Har point ek (x, y) pair hai, aur do points ke beech distance ya kisi ratio me baantne wala point — dono ko seedha formula se nikal sakte hain, bina graph banaye. Yeh chapter sirf do formulas pe based hai: distance formula aur section formula.
Chapter 7 Summary — 5 Minute Revision
7.1 Distance Formula
Do points P(x₁, y₁) aur Q(x₂, y₂) ke beech distance nikalne ke liye Pythagoras theorem use hota hai. Horizontal difference (x₂−x₁) aur vertical difference (y₂−y₁) lekar ek right triangle banta hai, aur PQ uska hypotenuse hai.
| Case | Formula |
|---|---|
| General points P(x₁,y₁), Q(x₂,y₂) | d = √[(x₂−x₁)² + (y₂−y₁)²] |
| Origin O(0,0) se point P(x,y) | OP = √(x² + y²) |
↔ Table ko side me swipe karein
Use hota hai: collinearity check karne ke liye (teen points ek line par hain ya nahi — teeno distances me se do ka sum teesre ke barabar hona chahiye), aur triangle/quadrilateral ka type identify karne ke liye (sides + diagonals compare karke — scalene, isosceles, equilateral, square, rectangle, rhombus).
7.2 Section Formula
Point P(x, y) jo A(x₁,y₁) aur B(x₂,y₂) ko join karne wale line segment ko ratio m₁:m₂ me (A se B ki taraf) baantta hai:
| Case | Formula |
|---|---|
| Ratio m₁ : m₂ me | x = (m₁x₂ + m₂x₁)/(m₁+m₂), y = (m₁y₂ + m₂y₁)/(m₁+m₂) |
| Midpoint (ratio 1:1) | x = (x₁+x₂)/2, y = (y₁+y₂)/2 |
↔ Table ko side me swipe karein
Important applications:
- Trisection points — ek line segment ko 3 barabar hisso me baantne wale do points, ratio 1:2 aur 2:1 se nikalte hain.
- Ratio find karna — jab point diya ho aur ratio nikalna ho, section formula ko unknown ratio k:1 maan kar solve karte hain.
- Parallelogram vertices — parallelogram ke diagonals ek dusre ko bisect karte hain, isliye dono diagonals ka midpoint barabar hota hai (yahi property missing vertex nikalne me use hoti hai).
- Circle ka diameter — agar center given ho aur ek end point given ho, to center hi diameter ka midpoint hota hai — is se doosra end point nikalta hai.
Note: Area of a triangle by coordinates (determinant formula) is chapter ke current rationalised version me nahi hai — sirf distance aur section formula hi board syllabus me hai.
Poore Class 10 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q14)
Q1. Find the distance between the points (2, 3) and (4, 1).
Distance formula: d = √[(x₂−x₁)² + (y₂−y₁)²]
d = √[(4−2)² + (1−3)²] = √[2² + (−2)²] = √[4 + 4] = √8 = 2√2 units
Q2. Find the distance between the points (0, 0) and (36, 15). Also, find the distance between two towns A and B if town B is located at 36 km east and 15 km north of town A.
Yahan A(0,0) aur B(36,15) maan sakte hain — east ko x-axis, north ko y-axis.
d = √[(36−0)² + (15−0)²] = √[1296 + 225] = √1521 = 39
Distance = 39 units, aur town A aur town B ke beech distance bhi 39 km hai.
Q3. Determine if the points (1, 5), (2, 3) and (−2, −11) are collinear.
Let A(1,5), B(2,3), C(−2,−11). Teeno sides ki distance nikalte hain.
AB = √[(2−1)² + (3−5)²] = √[1+4] = √5
BC = √[(−2−2)² + (−11−3)²] = √[16+196] = √212 = 2√53
AC = √[(−2−1)² + (−11−5)²] = √[9+256] = √265
Agar points collinear hote, to AB + BC = AC hona chahiye tha. Check: √5 + 2√53 ≈ 16.79 jabki √265 ≈ 16.28 — dono barabar nahi hain (algebraically bhi √5+2√53 = √265 tabhi possible hota jab √265=12, jo galat hai). Isliye points collinear nahi hain.
Q4. Check whether (5, −2), (6, 4) and (7, −2) are the vertices of an isosceles triangle.
Let A(5,−2), B(6,4), C(7,−2).
AB = √[(6−5)² + (4−(−2))²] = √[1+36] = √37
BC = √[(7−6)² + (−2−4)²] = √[1+36] = √37
AC = √[(7−5)² + (−2−(−2))²] = √[4+0] = 2
AB = BC = √37, aur AC alag hai. Do sides barabar hain, isliye ABC ek isosceles triangle hai.
Q5. Four friends are seated at points A(3, 4), B(6, 7), C(9, 4) and D(6, 1) in a classroom. Champa says ABCD is a square, Chameli disagrees. Who is correct? Give reasons using the distance formula.
Chaaro sides aur dono diagonals ki length nikalte hain.
AB = √[(6−3)² + (7−4)²] = √[9+9] = 3√2
BC = √[(9−6)² + (4−7)²] = √[9+9] = 3√2
CD = √[(6−9)² + (1−4)²] = √[9+9] = 3√2
DA = √[(3−6)² + (4−1)²] = √[9+9] = 3√2
AC = √[(9−3)² + (4−4)²] = √36 = 6
BD = √[(6−6)² + (1−7)²] = √36 = 6
Chaaro sides equal (3√2) hain aur dono diagonals bhi equal (6) hain. Isliye Champa sahi hai — ABCD ek square hai.
Q6. Find a point on the x-axis which is equidistant from the points (2, −5) and (−2, 9).
x-axis par point (x, 0) maan lo. Distance formula se donon distances barabar rakhte hain.
(x−2)² + (0−(−5))² = (x−(−2))² + (0−9)²
(x−2)² + 25 = (x+2)² + 81
x² −4x +4 +25 = x² +4x +4 +81
−4x + 29 = 4x + 85 ⟹ −8x = 56 ⟹ x = −7
Required point = (−7, 0).
Q7. If the distance between the points P(2, −3) and Q(10, y) is 10 units, find the value(s) of y.
Distance formula:
(10−2)² + (y−(−3))² = 10²
64 + (y+3)² = 100
(y+3)² = 36 ⟹ y + 3 = ±6
Isliye y = 3 ya y = −9.
Q8. Find the coordinates of the point which divides the line segment joining (−1, 7) and (4, −3) in the ratio 2 : 3.
Section formula: x = (m₁x₂+m₂x₁)/(m₁+m₂), y = (m₁y₂+m₂y₁)/(m₁+m₂), ratio 2:3.
x = (2×4 + 3×(−1))/(2+3) = (8−3)/5 = 1
y = (2×(−3) + 3×7)/(2+3) = (−6+21)/5 = 3
Required point = (1, 3).
Q9. Find the coordinates of the points of trisection of the line segment joining (4, −1) and (−2, −3).
Trisection points P aur Q line segment ko 1:2 aur 2:1 ratio me baantte hain.
P (1:2): x = (1×(−2)+2×4)/3 = 6/3 = 2, y = (1×(−3)+2×(−1))/3 = −5/3
Q (2:1): x = (2×(−2)+1×4)/3 = 0, y = (2×(−3)+1×(−1))/3 = −7/3
P = (2, −5/3), Q = (0, −7/3).
Q10. Find the ratio in which the line segment joining (−3, 10) and (6, −8) is divided by the point (−1, 6).
Let ratio k : 1 (from (−3,10) towards (6,−8)). x-coordinate se k nikalte hain.
−1 = (k×6 + 1×(−3))/(k+1)
−(k+1) = 6k − 3 ⟹ −k−1 = 6k−3 ⟹ 2 = 7k ⟹ k = 2/7
Ratio = 2/7 : 1 = 2 : 7. (y-coordinate se check: y = ((2/7)(−8)+10)/(9/7) = 6 ✓, matches diye gaye point se.)
Q11. Find the ratio in which the y-axis divides the line segment joining the points A(1, −5) and B(−4, 5). Also find the point of intersection.
y-axis par x = 0 hota hai. Let ratio k : 1 (A se B ki taraf).
0 = (k×(−4) + 1×1)/(k+1) ⟹ −4k + 1 = 0 ⟹ k = 1/4
Ratio = 1/4 : 1 = 1 : 4.
y = ((1/4)×5 + 1×(−5))/(1/4+1) = (5/4 − 5)/(5/4) = (−15/4)/(5/4) = −3
Point of intersection = (0, −3).
Q12. If the points (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram ABCD, taken in order, find x and y.
Parallelogram ke diagonals ek dusre ko bisect karte hain — isliye AC aur BD ke midpoints barabar honge (section formula, ratio 1:1).
Midpoint of AC = ((1+x)/2, (2+6)/2) = ((1+x)/2, 4)
Midpoint of BD = ((4+3)/2, (y+5)/2) = (3.5, (y+5)/2)
(1+x)/2 = 3.5 ⟹ x = 6
4 = (y+5)/2 ⟹ y = 3
Isliye x = 6, y = 3.
Q13. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, −3) and B is (1, 4).
Circle ka centre diameter AB ka midpoint hota hai — yeh section formula ka 1:1 (midpoint) case hai. Let A = (x, y).
(x+1)/2 = 2 ⟹ x = 3
(y+4)/2 = −3 ⟹ y = −10
A = (3, −10).
Q14. Find the coordinates of the points P and Q that trisect the line segment joining A(2, 2) and B(−7, 4).
P divides AB in ratio 1:2, Q divides AB in ratio 2:1.
P: x = (1×(−7)+2×2)/3 = −1, y = (1×4+2×2)/3 = 8/3
Q: x = (2×(−7)+1×2)/3 = −4, y = (2×4+1×2)/3 = 10/3
P = (−1, 8/3), Q = (−4, 10/3).
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| Distance between P(x₁,y₁) and Q(x₂,y₂) | d = √[(x₂−x₁)² + (y₂−y₁)²] |
| Distance from origin to P(x,y) | OP = √(x² + y²) |
| Section formula (ratio m₁:m₂ from A to B) | x = (m₁x₂+m₂x₁)/(m₁+m₂), y = (m₁y₂+m₂y₁)/(m₁+m₂) |
| Midpoint formula (special case, ratio 1:1) | x = (x₁+x₂)/2, y = (y₁+y₂)/2 |
| Collinearity test (A, B, C) | Teen distances me se koi do ka sum teesre ke barabar ho, e.g. AB + BC = AC |
| Parallelogram diagonal property | Diagonals ek dusre ko bisect karte hain ⟹ midpoint of one diagonal = midpoint of other |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Section formula me x₁, x₂ ka order ulta karna. Ratio m₁:m₂ 'A se B ki taraf' hota hai — formula me x₂ (doosra point) ko m₁ ke saath aur x₁ (pehla point) ko m₂ ke saath multiply karna hai: x=(m₁x₂+m₂x₁)/(m₁+m₂). Bahut students ulta likh dete hain aur galat point mil jaata hai.
- Negative coordinates ke sign me galti. (x₂−x₁) ya (y₂−y₁) nikalte waqt jab koi coordinate negative ho, minus ka minus plus ho jaata hai — jaise (−2−9) ko log galti se (−2+9) likh dete hain. Har substitution ke baad bracket dobara check karo.
- Distance formula me square root lena bhool jaana. Kai students (x₂−x₁)²+(y₂−y₁)² tak calculate karke wahi final answer likh dete hain — yaad rakho distance = √[sum], sirf sum nahi.
- Midpoint ko section formula se alag treat karna. Midpoint sirf section formula ka special case hai jab ratio 1:1 ho. Isko alag se yaad karne ki zaroorat nahi — section formula me m₁=m₂=1 daal do.
- Unknown ratio wale question me ratio ko k:1 ki jagah 1:k maan lena. Jab ratio nikalna ho aur point diya ho, direction clear rakho — kis point se kis point ki taraf baant rahe ho, warna answer ka reciprocal aa jaata hai.
- Collinearity check me galat pair compare karna. Teen distances AB, BC, AC nikalne ke baad check karna hai ki sabse badi distance = doosri do ka sum hai ya nahi — sirf koi bhi do random distances add karke teesri se compare mat karo, order match nahi karega.
Board-Style Important Questions
- 1 mark: Find the distance between the points (a, 0) and (0, b).
- 2 marks: Find the value of k, if the point P(2, 4) is equidistant from the points A(5, k) and B(k, 7).
- 2 marks: Find the coordinates of the point which divides the line segment joining (2, −5) and (−2, 9) in the ratio 3:1 internally.
- 3 marks: Find the ratio in which the point (−4, 6) divides the line segment joining the points A(−6, 10) and B(3, −8).
- 4 marks: If the points A(6, 1), B(8, 2), C(9, 4) and D(x, y) are the vertices of a parallelogram taken in order, find the value of x and y.
Quick Quiz — Score Check Karein
Q1. Class 10 Maths Chapter 7 (Coordinate Geometry) me kaunse do formulas cover hote hain?
Q2. Distance formula kis theorem se derive hoti hai?
Q3. Points (2, 3) aur (4, 1) ke beech distance kitni hai?
Q4. Origin O(0,0) se point P(x, y) ki distance ka formula kya hai?
Q5. Teen points A, B, C ko collinear check karne ke liye distance formula se kya condition dekhi jaati hai?
Q6. Section formula me point P jo A(x₁,y₁) aur B(x₂,y₂) ko ratio m₁:m₂ me (A se B ki taraf) divide karta hai, uska x-coordinate kya hoga?
Q7. Midpoint formula section formula ka special case kab banta hai?
Q8. Line segment joining (−1, 7) aur (4, −3) ko ratio 2:3 me divide karne wale point ke coordinates kya hain?
Q9. Parallelogram ke missing vertex nikalne ke liye section formula ka kaunsa property use hoti hai?
Q10. Circle ka diameter AB hai, jisme centre (2, −3) aur B(1, 4) diya hai. A ke coordinates nikalne ke liye kya use hota hai?
Aksar Poochhe Jaane Wale Sawaal
Distance formula Pythagoras theorem se kaise related hai?
Do points ke beech ka horizontal difference (x₂−x₁) aur vertical difference (y₂−y₁) ek right-angled triangle ke do legs bante hain, aur unke beech ki seedhi-line distance us triangle ka hypotenuse hoti hai — isliye d = √[(x₂−x₁)²+(y₂−y₁)²], jo Pythagoras theorem (hypotenuse² = base² + height²) ka hi form hai.
Section formula me m1 aur m2 kaise decide karte hain?
Ratio m1:m2 hamesha 'pehle point se doosre point ki taraf' hota hai. Agar point P line segment AB ko m1:m2 me divide karta hai, to AP:PB = m1:m2, aur formula me x2 (B ka x-coordinate) ko m1 ke saath aur x1 (A ka x-coordinate) ko m2 ke saath multiply karte hain.
Midpoint formula alag se yaad karna zaroori hai kya?
Nahi. Midpoint sirf section formula ka special case hai jaha ratio 1:1 hota hai (m1=m2=1). Section formula yaad ho to midpoint apne aap nikal jaata hai: x=(x1+x2)/2, y=(y1+y2)/2.
Area of triangle wala determinant formula is chapter me kyun nahi hai?
Current rationalised NCERT syllabus me area-of-triangle-by-coordinates section hata diya gaya hai. Ab is chapter me sirf distance formula aur section formula hi cover hote hain.
Teen points collinear hain ya nahi, distance formula se kaise pata karein?
Teeno points ke beech ke teen distances nikalo (AB, BC, AC). Agar sabse badi distance baaki do ke sum ke barabar hai (jaise AB+BC=AC), to teeno points ek hi seedhi line par hain, yaani collinear hain.
Parallelogram ke missing vertex nikalne me section formula kaise use hoti hai?
Parallelogram ke dono diagonals ek dusre ko bisect karte hain, matlab dono diagonals ka midpoint same hota hai. Yeh midpoint formula (section formula ka 1:1 case) use karke unknown vertex ke coordinates nikal sakte ho.
Class 10 Maths — Saare Chapters
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