Board Buddy

Page 1: Introduction to Coordinate Geometry

Coordinate Geometry (also called Analytical Geometry) uses numbers and algebra to study geometry.

It was developed by René Descartes — that’s why the plane is called the Cartesian plane.

We assign coordinates (ordered pairs) to points and use formulas to find distances, midpoints, areas, etc.

This chapter introduces the Cartesian system, plotting points, and basic formulas.

Page 2: The Cartesian Plane

Two perpendicular lines: horizontal X-axis and vertical Y-axis.

Intersection point O → Origin (0,0).

Any point P is represented as (x, y) where:

x = horizontal distance from Y-axis (abscissa)
y = vertical distance from X-axis (ordinate)

Ordered pair (x, y) — x always first.

Page 3: Quadrants and Sign Convention

I Quadrant: (+, +) → e.g., (3, 4)
II Quadrant: (-, +) → e.g., (-2, 5)
III Quadrant: (-, -) → e.g., (-1, -3)
IV Quadrant: (+, -) → e.g., (4, -2)

Points on axes: (x, 0) on X-axis, (0, y) on Y-axis.

Origin: (0, 0)

Page 4: Plotting Points in the Plane

Steps:

Draw X and Y axes.
Mark units equally.
From origin, move x units along X-axis (right +, left -).
From there, move y units parallel to Y-axis (up +, down -).

Example: Plot A(3, 2), B(-4, 5), C(-1, -3), D(0, 4)

Page 5: Points on Axes and Special Cases

(5, 0) → 5 units right on X-axis

(0, -3) → 3 units down on Y-axis

(0, 0) → Origin

All points with same x → vertical line

All points with same y → horizontal line

Page 6: Distance Formula

Distance between two points (x₁, y₁) and (x₂, y₂):

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

Derived from Pythagoras theorem.

Example: Distance between (1, 2) and (4, 6)
= √[(4-1)² + (6-2)²] = √[9 + 16] = √25 = 5 units

Page 7: Distance Formula Applications

Find length of sides of triangle, verify equilateral/isosceles/right triangle.

Example: Show points A(0,0), B(3,0), C(0,4) form right triangle.
AB=3, AC=4, BC=5 → 3²+4²=5² → right-angled at A.

Page 8: Special Cases in Distance

Points on same horizontal line (y same): distance = |x₂ - x₁|

Points on same vertical line (x same): distance = |y₂ - y₁|

Distance from origin (0,0) to (x,y): √(x² + y²)

Page 9: Section Formula (Internal Division)

Point dividing join of (x₁, y₁) and (x₂, y₂) in ratio m:n internally:

( (m x₂ + n x₁)/(m + n) , (m y₂ + n y₁)/(m + n) )

Special case m:n = 1:1 → Midpoint: ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )

Page 10: Section Formula Examples

Example 1: Midpoint of (2,3) and (4,7)
= ((2+4)/2, (3+7)/2) = (3, 5)

Example 2: Point dividing (1,2) and (7,8) in 2:3
= ((2·7 + 3·1)/(5), (2·8 + 3·2)/5) = (17/5, 22/5)

Page 11: External Section Formula

Point dividing externally in ratio m:n:

( (m x₂ - n x₁)/(m - n) , (m y₂ - n y₁)/(m - n) )

Used when point is outside the segment.

Page 12: Area of Triangle

Area with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃):

(1/2) | x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) |

If area = 0 → points are collinear.

Page 13: Area Formula Examples

Example: Area of (0,0), (4,0), (0,3)
= (1/2)| 0(0-3) + 4(3-0) + 0(0-0) | = (1/2)|12| = 6 sq units

Example: Check collinearity of (1,1), (2,2), (3,3)
Area = 0 → collinear.

Page 14: Key Formulas Summary

Distance: √[(x₂-x₁)² + (y₂-y₁)²]
Midpoint: ((x₁+x₂)/2, (y₁+y₂)/2)
Section (m:n): ((m x₂ + n x₁)/(m+n), ...)
Area: (1/2)|x₁(y₂-y₃)+...|
Collinear if area=0

Page 15: Practice Questions - Easy (1-10)

Plot (3,-2), (-4,0), (0,5).
Name quadrant of (-3,-5).
Find distance between (0,0) and (5,12).
Midpoint of (1,2) and (3,4).
Abscissa of (4,-7).
Points on X-axis have y=?
Distance between (2,3) and (2,7).
Is (0,0) in any quadrant?
Ordinate of point on X-axis.
Plot origin and label.

Page 16: Practice Questions - Medium (11-20)

Find distance (1,1) to (5,4).
Show (0,0),(3,0),(0,4) right triangle.
Point dividing (2,3) and (8,7) in 1:1.
Section formula 3:2 for (1,0) and (6,0).
Area of (1,2),(3,4),(5,6).
Are (1,1),(2,2),(4,4) collinear?
Find k if (k,3) is 5 units from (2,7).
Midpoint (0,0) and (a,b).
External division 1:1.
Verify equilateral triangle.

Page 17: Practice Questions - Hard (21-30)

Centroid of triangle (coordinates).
Point dividing externally 3:2.
Prove collinearity using area.
Find ratio if point divides in k:1.
Coordinates of point equidistant.
Area zero proof.
Section formula derivation hint.
Word problem: vertices given.
Distance from origin condition.
Combined distance + section.

Page 18: NCERT Exercise 4.1 Type

Basic plotting and quadrant identification.

Page 19: NCERT Exercise 4.2 Type

Distance formula applications.

Page 20: NCERT Exercise 4.3 Type

Section formula (internal and external).

Page 21: NCERT Exercise 4.4 Type

Area of triangle and collinearity.

Page 22: Common Mistakes to Avoid

Swapping x and y in ordered pair
Forgetting square root in distance
Wrong sign in section formula
Missing absolute value in area
Wrong ratio order (m:n vs n:m)

Page 23: Previous Year Board Questions

Typical: Find distance (2 marks)

Section formula (3 marks)

Area + collinearity (4 marks)

Plotting + verification (3 marks)

Page 24: Exam Strategy Tips

Draw neat axes with labels
Show all steps in formulas
Check signs carefully
Memorise all formulas exactly
Practice plotting quickly

Page 25: Quick Revision Formula Sheet

Distance: √[(Δx)² + (Δy)²]
Midpoint: average of coordinates
Section internal: weighted average
Area: (1/2)|sum|
Collinear → area=0

Master these → full marks!

Page 26: Final Motivation

You've completed the 27-page Coordinate Geometry guide!

This chapter connects algebra and geometry perfectly.

It's the base for Class 10 circles and conic sections.

Keep practising formulas and plotting.

You're unstoppable now 🦖

Class 9 Mathematics