📍 COORDINATE GEOMETRY

Class 10 Mathematics • Chapter 7

Distance Formula • Section Formula • Coordinate Applications

THE GOAL • HANDWRITTEN EXAM NOTES
01. Cartesian Plane
The Cartesian plane is formed by two perpendicular number lines:

• Horizontal line → x-axis
• Vertical line → y-axis
• Their intersection → Origin O(0,0)
O(0,0) x-axis y-axis I II III IV
💡 Easy Hindi: Coordinate plane ko ek map samjho. Kisi bhi point ki exact position batane ke liye hum uske coordinates `(x,y)` use karte hain.

Pehle x-coordinate, phir y-coordinate likha jata hai.
02. Coordinates of a Point
A point is represented as:
P(x,y)
Here:
x → abscissa
y → ordinate
P(3,2) x = 3 y = 2 x y
✏ Example
Find the abscissa and ordinate of P(−4, 7).

Abscissa = x-coordinate = −4
Ordinate = y-coordinate = 7
✓ Answer: Abscissa = −4, Ordinate = 7
03. Signs in Four Quadrants
I II III IV (+,+) (−,+) (−,−) (+,-) X Y
Quadrant x y
I + +
II +
III
IV +
✏ Example
Point A(−3, 5) lies in which quadrant?

x is negative and y is positive.
✓ A lies in Quadrant II.
04. Distance Between Two Points ⭐⭐⭐
For two points:
A(x₁,y₁) and B(x₂,y₂)

the distance AB is given by the Distance Formula.
A(x₁,y₁) B(x₂,y₂) x₂ − x₁ y₂ − y₁ AB
AB = √[ (x₂ − x₁)² + (y₂ − y₁)² ]
Why?
This comes from the Pythagoras theorem.
💡 Hindi: Do points ke beech straight-line distance nikalne ke liye x ka difference aur y ka difference lekar Pythagoras theorem apply karte hain.
✏ Example 1 — Direct Formula
Find the distance between A(2,3) and B(6,6).

x₁ = 2, y₁ = 3
x₂ = 6, y₂ = 6

AB = √[(6−2)² + (6−3)²]
= √[4² + 3²]
= √25
= 5 units
✓ Distance = 5 units
05. Distance from Origin ⭐⭐
O(0,0) P(x,y) OP
For P(x,y):
OP = √(x² + y²)
✏ Example
Find the distance of P(3,4) from origin.

OP = √(3² + 4²)
= √25
= 5 units
06. Points on Same Vertical / Horizontal Line
Same x-coordinate

A(x,y₁), B(x,y₂)

Distance:
|y₂ − y₁|
Same y-coordinate

A(x₁,y), B(x₂,y)

Distance:
|x₂ − x₁|
A(x,y₂) B(x,y₁) |y₂−y₁|
✏ Example
Find the distance between A(5,2) and B(5,9).

Both have the same x-coordinate.
Distance = |9−2| = 7 units
✓ Answer = 7 units
07. Finding an Unknown Coordinate Using Distance
✏ Exam-Type Example
Find x if the distance between A(x,2) and B(3,6) is 5 units.
A(x,2) B(3,6) 5 units
5 = √[(3−x)² + (6−2)²]
25 = (3−x)² + 16
(3−x)² = 9
3−x = ±3

Therefore:
x = 0 or x = 6
💡 Equation ko solve karte time square root remove karne ke baad ± possibility ko ignore mat karo.
08. Section Formula ⭐⭐⭐
Suppose a point P divides the line segment joining
A(x₁,y₁) and B(x₂,y₂)
internally in the ratio:
AP : PB = m : n

Then the coordinates of P are given by the Section Formula.
A(x₁,y₁) P B(x₂,y₂) m n AP : PB = m : n
P = (
mx₂ + nx₁ m+n
,
my₂ + ny₁ m+n
)
💡 Hindi trick: Jis point ko find karna hai, uske coordinates mein opposite ratio ka multiplication aata hai.

For P dividing AB in m:n:
x-coordinate = `(m × x₂ + n × x₁)/(m+n)`
y-coordinate = `(m × y₂ + n × y₁)/(m+n)`
09. Section Formula — Direct Example ⭐⭐⭐
✏ Example
Find the coordinates of P which divides A(2,3) and B(8,9) internally in the ratio 1:2.
A(2,3) P B(8,9) 1 2
Here:
x₁ = 2, y₁ = 3
x₂ = 8, y₂ = 9
m = 1, n = 2

x-coordinate:
x = (1×8 + 2×2)/(1+2)
= 12/3
= 4
y-coordinate:
y = (1×9 + 2×3)/(1+2)
= 15/3
= 5
✓ P = (4,5)
10. Midpoint Formula ⭐⭐⭐
Midpoint is the point which divides a line segment into two equal parts. So:
AP : PB = 1 : 1
Therefore midpoint formula is obtained from section formula.
A(x₁,y₁) M B(x₂,y₂) equal equal
M = (
x₁+x₂ 2
,
y₁+y₂ 2
)
✏ Example
Find the midpoint of A(2,4) and B(8,10).

M = ( (2+8)/2 , (4+10)/2 )
= (5,7)
✓ Midpoint = (5,7)
11. Finding a Missing Coordinate Using Midpoint
✏ Exam-Type Example
The midpoint of A(x,4) and B(6,8) is (4,6). Find x.
A(x,4) M(4,6) B(6,8)
Midpoint x-coordinate:
4 = (x + 6)/2
8 = x + 6
x = 2

Check y:
(4+8)/2 = 6 ✓
✓ A = (2,4)
12. Finding Ratio When Point is Given
✏ Exam-Type Example
Point P(4,5) divides A(2,3) and B(8,9). Find AP:PB.
A(2,3) P(4,5) B(8,9)
Let:
AP : PB = m : n

Using x-coordinate:
4 = (8m + 2n)/(m+n)
4m + 4n = 8m + 2n
2n = 4m
n = 2m

Therefore:
m:n = 1:2
✓ AP : PB = 1 : 2
13. Internal Division — Figure Based Question
✏ Example
A point P divides the line joining A(1,2) and B(7,8) internally in the ratio 2:1. Find P.
A(1,2) P B(7,8) 2 1
x = (2×7 + 1×1)/(2+1) = 15/3 = 5
y = (2×8 + 1×2)/(2+1) = 18/3 = 6
✓ P = (5,6)
14. Proving a Triangle is Isosceles
Coordinate geometry can be used to prove geometric properties by calculating lengths. For an isosceles triangle:

Two sides must be equal.
✏ Example
Show that A(0,0), B(4,0), C(2,3) form an isosceles triangle.
A(0,0) B(4,0) C(2,3)
AB:
AB = √[(4−0)²+(0−0)²] = 4
AC:
AC = √[(2−0)²+(3−0)²] = √13
BC:
BC = √[(2−4)²+(3−0)²] = √13
Therefore:
AC = BC
✓ Triangle ABC is isosceles.
15. Proving a Triangle is Right-Angled
✏ Exam-Type Example
Check whether A(0,0), B(3,0), C(3,4) form a right-angled triangle.
A(0,0) B(3,0) C(3,4)
AB = 3
BC = 4
AC = 5

Since:
AB² + BC² = AC²
3² + 4² = 5²
9 + 16 = 25 ✓
✓ Triangle ABC is right-angled at B.
💡 Teen points se triangle right-angled prove karne ke liye sabse badi distance ko hypotenuse maan kar Pythagoras check karo.
16. Important Exam Pattern — Equidistant Point
✏ Example
Find the point on the x-axis which is equidistant from A(2,3) and B(4,5).
A(2,3) B(4,5) P(x,0)
Let P(x,0). Since P is equidistant:
PA = PB
Squaring both sides:
(x−2)² + (0−3)² = (x−4)² + (0−5)²
x² − 4x + 13 = x² − 8x + 41
4x = 28
x = 7
✓ Required point = (7,0)
17. Important Formula Sheet
Distance Formula
d = √[(x₂−x₁)² + (y₂−y₁)²]

Distance from Origin
√(x²+y²)

Section Formula
P = (
mx₂+nx₁ m+n ,
my₂+ny₁ m+n
)

Midpoint Formula
M = (
x₁+x₂ 2
,
y₁+y₂ 2
)
18. Common Exam Mistakes ⚠️
1. Coordinate `(x,y)` mein x aur y interchange karna.

2. Negative number ka square incorrectly calculate karna.

3. Distance formula mein `(x₂−x₁)` ya `(y₂−y₁)` ka square bhool jaana.

4. Section formula mein ratio ko ulta use karna.

5. Final answer mein units na likhna when distance is asked.

6. Midpoint ko section formula se connect na karna.

7. Square root ko incorrectly simplify karna.
19. Exam Question Selection Guide
Question mein kya diya hai? Method
Two points + distance Distance Formula
Point dividing line in m:n Section Formula
Line divided equally Midpoint Formula
Unknown coordinate + distance Distance Formula + Equation
Equidistant from two points Equal Distances
Check right triangle Pythagoras using distances
Check isosceles triangle Compare side lengths
20. Final Concept Check 🎯
✓ Can you identify x-coordinate and y-coordinate?

✓ Can you identify the quadrant?

✓ Can you use the distance formula correctly?

✓ Can you find distance from origin?

✓ Can you find an unknown coordinate using distance?

✓ Can you apply section formula?

✓ Can you find midpoint?

✓ Can you find a missing coordinate using midpoint?

✓ Can you find the ratio in which a point divides a line?

✓ Can you use coordinates to verify geometric properties?
💡 Chapter ka main idea:

Coordinates hume geometry ko numbers mein convert karne dete hain.

Point → Coordinates → Formula → Calculation → Geometric Result