🕉️ VEDIC MATHEMATICS — LEVEL 2: INTERMEDIATE

MODULE 17: Coordinate Geometry — Vedic Shortcuts

Complete Study Material | Theory + Examples + Practice + Test Bank

"Geometry without coordinates is blind. Coordinates without geometry is empty. Vedic mathematics unites them with speed and insight." — Vedic Mathematics Teacher's Manual

📋 MODULE AT A GLANCE

🎯 LEARNING OUTCOMES

By the end of this module, the student will be able to:

1. Compute distance between two points in under 10 seconds using simplified methods
2. Find section ratios (midpoint and division points) using cross-multiplication
3. Calculate slope and angle of inclination using the Urdhva pattern
4. Derive the equation of a line through two points using the cross-multiply shortcut
5. Compute area of a triangle using the Vedic shoelace method in under 30 seconds
6. Check collinearity of three points by mere observation (Vilokanam)
7. Find circle equations using coordinate pattern recognition
8. Apply Vedic cross-multiplication to all coordinate geometry formulas

PART 1: THEORY

1.1 — Introduction to Vedic Coordinate Geometry

What Makes Vedic Coordinate Geometry Different?

The Cross-Multiplication Pattern (Urdhva-Tiryagbhyam)

The same vertically and cross-wise pattern used for multiplication also appears in:

• Distance formula
• Section formula
• Slope calculation
• Line equations
• Area of triangle

1.2 — Distance Formula: The Vedic Shortcut

Standard Formula

Distance between $A(x_1, y_1)$ and $B(x_2, y_2)$:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Vedic Simplification

Step 1: Find $\Delta x = x_2 - x_1$ and $\Delta y = y_2 - y_1$ (mentally)

Step 2: Square both differences using squaring methods from Module 9

Step 3: Add using left-to-right addition

Step 4: Take square root (use known perfect squares or Module 12 methods)

Example 1: Distance between (3, 4) and (7, 1)

$\Delta x = 7 - 3 = 4$, $\Delta y = 1 - 4 = -3$

$\Delta x^2 = 16$, $\Delta y^2 = 9$

Sum = 25

$d = \sqrt{25} = 5$ units ✓

Example 2: Distance between (2, 5) and (10, 11)

$\Delta x = 8$, $\Delta y = 6$

$8^2 = 64$, $6^2 = 36$, Sum = 100

$d = 10$ units ✓

Example 3: Distance between (1, 2) and (4, 6)

$\Delta x = 3$, $\Delta y = 4$

$3^2 = 9$, $4^2 = 16$, Sum = 25

$d = 5$ units ✓

(Pythagorean triple: 3-4-5 triangle)

1.3 — Section Formula: The Cross-Multiplication Method

Standard Formula

For point P dividing AB in ratio m:n (A to B):

$$x = \frac{mx_2 + nx_1}{m+n}, \quad y = \frac{my_2 + ny_1}{m+n}$$

Vedic Pattern Recognition

The formula is essentially a weighted average with cross-multiplication:

$x = \frac{x_1 \cdot n + x_2 \cdot m}{m+n}$

Midpoint (Special case m = n = 1)

$$x = \frac{x_1 + x_2}{2}, \quad y = \frac{y_1 + y_2}{2}$$

Example 1: Midpoint of (2, 3) and (8, 7)

$x = (2+8)/2 = 5$, $y = (3+7)/2 = 5$

Midpoint = (5, 5) ✓

Example 2: Point dividing (2, 3) and (8, 7) in ratio 1:2

Here m=1 (from A to B? Let's clarify.)

For ratio 1:2 from A(2,3) to B(8,7), m=1, n=2:

$x = (1×8 + 2×2)/(1+2) = (8 + 4)/3 = 12/3 = 4$ $y = (1×7 + 2×3)/3 = (7 + 6)/3 = 13/3 ≈ 4.33$

Point = (4, 13/3) ✓

Example 3: Trisection points of (1, 2) and (10, 11)

The two trisection points divide in ratios 1:2 and 2:1

First point (1:2): $x = (1×10 + 2×1)/3 = (10+2)/3 = 4$, $y = (1×11 + 2×2)/3 = (11+4)/3 = 5$ Point = (4, 5)

Second point (2:1): $x = (2×10 + 1×1)/3 = (20+1)/3 = 7$, $y = (2×11 + 1×2)/3 = (22+2)/3 = 8$ Point = (7, 8)

Check: (1,2) → (4,5) → (7,8) → (10,11) equally spaced ✓

1.4 — Slope and Angle of Inclination

Standard Formula

Slope $m = \frac{y_2 - y_1}{x_2 - x_1} = \tan \theta$ (where θ is angle with positive x-axis)

Vedic Shortcut: Cross-Difference Pattern

Think of Urdhva-Tiryagbhyam: Vertical difference over horizontal difference

$m = \frac{\Delta y}{\Delta x}$ with sign awareness

Example 1: Slope of (2, 3) and (5, 11)

$\Delta y = 11 - 3 = 8$, $\Delta x = 5 - 2 = 3$ $m = 8/3$ ✓

Example 2: Slope of (3, 7) and (6, 1)

$\Delta y = 1 - 7 = -6$, $\Delta x = 6 - 3 = 3$ $m = -6/3 = -2$ ✓

Angle of Inclination

$\theta = \tan^{-1}(m)$

For m = 1, θ = 45°; m = √3, θ = 60°; m = 1/√3, θ = 30°

Common Slopes and Angles

1.5 — Equation of a Line Through Two Points

Standard Formula

Two-point form: $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$

Vedic Cross-Multiplication Shortcut

Using Urdhva pattern, the equation can be written as:

$$(y_2 - y_1)x - (x_2 - x_1)y + (x_2 y_1 - x_1 y_2) = 0$$

Or more memorably:

$$x(y_1 - y_2) + y(x_2 - x_1) + (x_1 y_2 - x_2 y_1) = 0$$

The Cross-Multiplication Pattern

For points $(x_1, y_1)$ and $(x_2, y_2)$:

Step 1: Coefficient of x = $(y_1 - y_2)$ Step 2: Coefficient of y = $(x_2 - x_1)$ Step 3: Constant term = $(x_1 y_2 - x_2 y_1)$

Example 1: Line through (2, 3) and (5, 11)

Coefficient of x = $3 - 11 = -8$ Coefficient of y = $5 - 2 = 3$ Constant = $2×11 - 5×3 = 22 - 15 = 7$

Equation: $-8x + 3y + 7 = 0$ or $8x - 3y - 7 = 0$ ✓

Check: For (2,3): $8×2 - 3×3 = 16 - 9 = 7$ ✓ For (5,11): $8×5 - 3×11 = 40 - 33 = 7$ ✓

Example 2: Line through (1, 4) and (3, 8)

Coefficient of x = $4 - 8 = -4$ Coefficient of y = $3 - 1 = 2$ Constant = $1×8 - 3×4 = 8 - 12 = -4$

Equation: $-4x + 2y - 4 = 0$ or $2x - y + 2 = 0$ (dividing by -2) ✓

Check: (1,4): $2×1 - 4 + 2 = 2 - 4 + 2 = 0$ ✓ (3,8): $6 - 8 + 2 = 0$ ✓

Example 3: Line through (0, 0) and (4, 6)

Coefficient of x = $0 - 6 = -6$ Coefficient of y = $4 - 0 = 4$ Constant = $0×6 - 4×0 = 0$

Equation: $-6x + 4y = 0$ or $3x - 2y = 0$ ✓

Check: $3×4 - 2×6 = 12 - 12 = 0$ ✓

1.6 — Area of a Triangle: Vedic Shoelace Method

Standard Formula (Shoelace)

For triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$:

$$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

Vedic Pattern: Cross-Multiplication Cyclic

Think of it as a cyclic cross-product sum:

Step 1: List coordinates in order (repeating first at the end) Step 2: Sum of (x₁y₂ + x₂y₃ + x₃y₁) Step 3: Subtract sum of (y₁x₂ + y₂x₃ + y₃x₁) Step 4: Take half of absolute value

Example 1: Triangle with vertices (0,0), (4,0), (0,3)

List: (0,0), (4,0), (0,3), back to (0,0)

Sum1 = $0×0 + 4×3 + 0×0 = 0 + 12 + 0 = 12$ Sum2 = $0×4 + 0×0 + 3×0 = 0 + 0 + 0 = 0$ Area = $|12 - 0|/2 = 6$ square units ✓

Example 2: Triangle with vertices (1,2), (4,5), (7,1)

List: (1,2), (4,5), (7,1), back to (1,2)

Sum1 = $1×5 + 4×1 + 7×2 = 5 + 4 + 14 = 23$ Sum2 = $2×4 + 5×7 + 1×1 = 8 + 35 + 1 = 44$ Difference = $|23 - 44| = 21$ Area = $21/2 = 10.5$ square units ✓

Example 3: Triangle with vertices (2,3), (5,7), (8,4)

List: (2,3), (5,7), (8,4), back to (2,3)

Sum1 = $2×7 + 5×4 + 8×3 = 14 + 20 + 24 = 58$ Sum2 = $3×5 + 7×8 + 4×2 = 15 + 56 + 8 = 79$ Difference = $|58 - 79| = 21$ Area = $10.5$ square units ✓

1.7 — Collinearity Check by Mere Observation (Vilokanam)

Sub-Sutra 12: Vilokanam

"By mere observation" — with practice, you can see collinearity without calculation.

Condition for Collinearity

Three points are collinear if area of triangle = 0:

$$x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0$$

Slope Method (Quick Check)

Points A, B, C are collinear if slope AB = slope BC

Vedic Observation Pattern

For points with the same x-coordinate: vertical line → collinear For points with the same y-coordinate: horizontal line → collinear For points with proportional coordinates: check if $\frac{y_2-y_1}{x_2-x_1} = \frac{y_3-y_2}{x_3-x_2}$

Example 1: Check collinearity of (1,2), (3,4), (5,6)

Slope AB = (4-2)/(3-1) = 2/2 = 1 Slope BC = (6-4)/(5-3) = 2/2 = 1 Equal → Collinear ✓

By observation: Points lie on line y = x+1

Example 2: Check collinearity of (2,3), (4,7), (6,11)

Slope AB = (7-3)/(4-2) = 4/2 = 2 Slope BC = (11-7)/(6-4) = 4/2 = 2 Equal → Collinear ✓

Observation: y increases by 4 when x increases by 2 → slope 2

Example 3: Check collinearity of (1,5), (3,9), (5,13)

Slope AB = (9-5)/(3-1) = 4/2 = 2 Slope BC = (13-9)/(5-3) = 4/2 = 2 Collinear ✓

Example 4: Check collinearity of (1,2), (2,3), (4,6)

Slope AB = 1/1 = 1 Slope BC = (6-3)/(4-2) = 3/2 = 1.5 Not equal → Not collinear ✓

1.8 — Circle Equations: Vedic Pattern Recognition

Standard Circle Equation

Center $(h, k)$, radius r:

$$(x - h)^2 + (y - k)^2 = r^2$$

Expanded form: $x^2 + y^2 + Dx + Ey + F = 0$

Vedic Shortcuts

Center from expanded form: $h = -D/2$, $k = -E/2$

Radius: $r = \sqrt{h^2 + k^2 - F}$

Example 1: Find center and radius of $x^2 + y^2 + 6x - 8y - 11 = 0$

$D = 6$, $E = -8$, $F = -11$

$h = -6/2 = -3$, $k = -(-8)/2 = 4$

$r^2 = (-3)^2 + 4^2 - (-11) = 9 + 16 + 11 = 36$ $r = 6$ ✓

Example 2: Find center and radius of $x^2 + y^2 - 10x + 12y + 12 = 0$

$D = -10$, $E = 12$, $F = 12$

$h = -(-10)/2 = 5$, $k = -12/2 = -6$

$r^2 = 5^2 + (-6)^2 - 12 = 25 + 36 - 12 = 49$ $r = 7$ ✓

Equation from Center and Radius (Vedic Form)

Directly: $x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0$

Example: Center (3, -2), radius 5

$x^2 + y^2 - 6x + 4y + (9 + 4 - 25) = x^2 + y^2 - 6x + 4y - 12 = 0$ ✓

1.9 — Circle Through Three Points

Vedic Determinant Method (Shoelace Extension)

For three points $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$, the circle equation can be found using the pattern:

The general equation $x^2 + y^2 + Dx + Ey + F = 0$ gives three equations:

$x_1^2 + y_1^2 + Dx_1 + Ey_1 + F = 0$ $x_2^2 + y_2^2 + Dx_2 + Ey_2 + F = 0$ $x_3^2 + y_3^2 + Dx_3 + Ey_3 + F = 0$

Solve using Paravartya or elimination.

Example: Circle through (0,0), (4,0), (0,3)

From (0,0): $F = 0$ From (4,0): $16 + 4D + F = 0$ → $16 + 4D = 0$ → $D = -4$ From (0,3): $9 + 3E + F = 0$ → $9 + 3E = 0$ → $E = -3$

Equation: $x^2 + y^2 - 4x - 3y = 0$

Complete square: $(x-2)^2 + (y-1.5)^2 = 4 + 2.25 = 6.25 = (2.5)^2$

Center (2, 1.5), radius 2.5 ✓

1.10 — Vilokanam: Pattern Recognition in Coordinate Geometry

Sub-Sutra 12: Vilokanam (By Mere Observation)

This sub-sutra encourages us to see the answer without calculation.

Examples of Vilokanam in Action

Example: What is the distance between (a, b) and (c, d) if c-a = 3 and d-b = 4?

By observation: $d = \sqrt{9 + 16} = 5$ (3-4-5 triple)

Example: Are (1,1), (2,2), (3,3) collinear?

By observation: Yes (all lie on y = x)

Example: What is the midpoint of (5,0) and (5,10)?

By observation: x = 5, y = (0+10)/2 = 5 → (5,5)

PART 2: WORKED EXAMPLES

Section A: Distance Formula

Example 1

Question: Find the distance between (3, 5) and (9, 13).

Answer:

Δx = 9-3 = 6, Δy = 13-5 = 8 6² = 36, 8² = 64, Sum = 100 d = √100 = 10 units ✓

Example 2

Question: Find the distance between (1, 1) and (4, 5).

Answer:

Δx = 3, Δy = 4 3² = 9, 4² = 16, Sum = 25 d = 5 units ✓

Example 3

Question: Find the distance between (5, 12) and (0, 0).

Answer:

Δx = -5, Δy = -12 25 + 144 = 169 d = 13 units ✓

Section B: Section Formula

Example 4

Question: Find the midpoint of (2, 8) and (10, 4).

Answer:

x = (2+10)/2 = 6, y = (8+4)/2 = 6 Midpoint = (6, 6) ✓

Example 5

Question: Find the point dividing A(1, 4) and B(7, 10) in ratio 2:1.

Answer:

m=2, n=1 x = (2×7 + 1×1)/3 = (14+1)/3 = 5 y = (2×10 + 1×4)/3 = (20+4)/3 = 8 Point = (5, 8) ✓

Example 6

Question: Find the trisection points of (0, 0) and (9, 12).

Answer:

First (1:2): x = (1×9+2×0)/3 = 3, y = (1×12+2×0)/3 = 4 → (3,4) Second (2:1): x = (2×9+1×0)/3 = 6, y = (2×12+1×0)/3 = 8 → (6,8) Check: (0,0) → (3,4) → (6,8) → (9,12) ✓

Section C: Slope and Angle

Example 7

Question: Find slope of line through (1, 2) and (4, 8).

Answer:

m = (8-2)/(4-1) = 6/3 = 2 ✓

Example 8

Question: Find the angle of inclination of line with slope 1.

Answer:

θ = tan⁻¹(1) = 45° ✓

Example 9

Question: Find slope of line through (2, 7) and (5, 2).

Answer:

m = (2-7)/(5-2) = -5/3 ✓

Section D: Equation of a Line

Example 10

Question: Find equation of line through (1, 2) and (4, 8).

Answer:

Coefficient of x = y₁ - y₂ = 2 - 8 = -6 Coefficient of y = x₂ - x₁ = 4 - 1 = 3 Constant = x₁y₂ - x₂y₁ = 1×8 - 4×2 = 8 - 8 = 0 Equation: -6x + 3y = 0 → 2x - y = 0 ✓

Check: For (1,2): 2×1 - 2 = 0 ✓; (4,8): 8 - 8 = 0 ✓

Example 11

Question: Find equation of line through (2, 3) and (5, 11).

Answer:

Coeff x = 3 - 11 = -8 Coeff y = 5 - 2 = 3 Constant = 2×11 - 5×3 = 22 - 15 = 7 Equation: -8x + 3y + 7 = 0 → 8x - 3y - 7 = 0 ✓

Example 12

Question: Find equation of line through (0, 0) and (3, 7).

Answer:

Coeff x = 0 - 7 = -7 Coeff y = 3 - 0 = 3 Constant = 0×7 - 3×0 = 0 Equation: -7x + 3y = 0 → 7x - 3y = 0 ✓

Section E: Area of Triangle

Example 13

Question: Find area of triangle with vertices (1, 1), (4, 5), (6, 2).

Answer:

List: (1,1), (4,5), (6,2), back to (1,1)

Sum1 = 1×5 + 4×2 + 6×1 = 5 + 8 + 6 = 19 Sum2 = 1×4 + 5×6 + 2×1 = 4 + 30 + 2 = 36 Difference = |19 - 36| = 17 Area = 17/2 = 8.5 square units ✓

Example 14

Question: Find area of triangle with vertices (0, 0), (5, 0), (0, 7).

Answer:

Sum1 = 0×0 + 5×7 + 0×0 = 35 Sum2 = 0×5 + 0×0 + 7×0 = 0 Area = 35/2 = 17.5 square units ✓

Example 15

Question: Find area of triangle with vertices (2, 3), (5, 7), (8, 4).

Answer:

Sum1 = 2×7 + 5×4 + 8×3 = 14 + 20 + 24 = 58 Sum2 = 3×5 + 7×8 + 4×2 = 15 + 56 + 8 = 79 Difference = 21 Area = 10.5 square units ✓

Section F: Collinearity Check

Example 16

Question: Check if (1, 2), (3, 6), (5, 10) are collinear.

Answer:

Slope AB = (6-2)/(3-1) = 4/2 = 2 Slope BC = (10-6)/(5-3) = 4/2 = 2 Equal → Collinear ✓

Observation: Points satisfy y = 2x

Example 17

Question: Check if (2, 5), (4, 9), (6, 14) are collinear.

Answer:

Slope AB = (9-5)/(4-2) = 4/2 = 2 Slope BC = (14-9)/(6-4) = 5/2 = 2.5 Not equal → Not collinear ✓

Section G: Circle Equations

Example 18

Question: Find center and radius of $x^2 + y^2 - 8x + 6y - 24 = 0$.

Answer:

D = -8, E = 6, F = -24 h = -(-8)/2 = 4 k = -6/2 = -3 r² = 4² + (-3)² - (-24) = 16 + 9 + 24 = 49 r = 7 Center (4, -3), radius 7 ✓

Example 19

Question: Find the equation of circle with center (2, -5) and radius 4.

Answer:

h=2, k=-5, r=4 x² + y² - 2hx - 2ky + (h² + k² - r²) = 0 x² + y² - 4x + 10y + (4 + 25 - 16) = 0 x² + y² - 4x + 10y + 13 = 0 ✓

Example 20

Question: Find the equation of circle with endpoints of diameter (1, 2) and (5, 8).

Answer:

Center = midpoint = ((1+5)/2, (2+8)/2) = (3, 5) Radius = distance from center to (1,2) = √((3-1)² + (5-2)²) = √(4 + 9) = √13 Equation: (x-3)² + (y-5)² = 13 Expanded: x² + y² - 6x - 10y + (9+25-13) = x² + y² - 6x - 10y + 21 = 0 ✓

PART 3: PRACTICE EXERCISES

Exercise Set A: Distance Formula (15 Questions)

Find the distance between each pair of points.

A1. (2, 3) and (5, 7) A2. (1, 1) and (4, 5) A3. (0, 0) and (6, 8) A4. (3, 4) and (9, 12) A5. (2, 5) and (7, 17) A6. (1, 2) and (9, 8) A7. (0, 5) and (12, 0) A8. (4, 6) and (8, 9) A9. (5, 12) and (0, 0) A10. (3, 7) and (11, 13) A11. (1, 4) and (10, 4) A12. (6, 2) and (6, 10) A13. (2, 3) and (8, 11) A14. (7, 24) and (0, 0) A15. (a, b) and (a + 2p, b + 2q)

Exercise Set B: Section Formula & Midpoint (15 Questions)

B1. Midpoint of (3, 5) and (9, 11) B2. Midpoint of (1, 8) and (7, 2) B3. Midpoint of (4, 2) and (10, 14) B4. Point dividing (2, 3) and (8, 9) in ratio 1:2 B5. Point dividing (1, 4) and (7, 10) in ratio 2:1 B6. Point dividing (0, 0) and (12, 18) in ratio 1:3 B7. Point dividing (5, 10) and (15, 20) in ratio 3:2 B8. Trisection points of (0, 0) and (12, 15) B9. Trisection points of (2, 3) and (14, 18) B10. Point dividing (3, 7) and (11, 15) in ratio 3:1 B11. Point dividing (4, 2) and (10, 8) in ratio 2:3 B12. Midpoint of (a, b) and (c, d) B13. Midpoint of (x₁, y₁) and (x₂, y₂) B14. Find the point that divides (5, 5) and (10, 10) in ratio 3:2 B15. If midpoint is (5, 7) and one endpoint is (2, 4), find the other endpoint

Exercise Set C: Slope & Equation of Line (15 Questions)

Find the slope and equation of the line through each pair.

C1. (1, 2) and (3, 8) C2. (2, 3) and (5, 9) C3. (4, 5) and (8, 13) C4. (1, 4) and (6, 14) C5. (3, 7) and (9, 13) C6. (2, 5) and (7, 5) C7. (4, 2) and (4, 9) C8. (0, 0) and (5, 12) C9. (1, 1) and (7, 7) C10. (2, 8) and (10, 2) C11. (3, 6) and (8, 1) C12. (5, 10) and (15, 20) C13. (a, 0) and (0, b) C14. (x₁, y₁) and (x₂, y₂) (general formula) C15. (2p, 3q) and (4p, 6q)

Exercise Set D: Area of Triangle (10 Questions)

Find the area of triangle with given vertices.

D1. (0, 0), (5, 0), (0, 7) D2. (1, 1), (4, 5), (6, 2) D3. (2, 3), (5, 7), (8, 4) D4. (3, 4), (7, 8), (11, 2) D5. (1, 2), (3, 8), (5, 4) D6. (0, 0), (6, 0), (3, 5) D7. (2, 5), (7, 10), (12, 3) D8. (4, 6), (8, 10), (12, 2) D9. (1, 1), (2, 4), (3, 2) D10. (a, 0), (0, b), (0, 0)

Exercise Set E: Collinearity Check (10 Questions)

Check if the three points are collinear.

E1. (1, 2), (3, 4), (5, 6) E2. (2, 3), (4, 7), (6, 11) E3. (1, 5), (3, 11), (5, 17) E4. (3, 1), (6, 2), (9, 3) E5. (2, 4), (4, 8), (6, 16) E6. (1, 2), (3, 8), (5, 14) E7. (0, 0), (4, 6), (8, 12) E8. (5, 10), (8, 15), (11, 20) E9. (1, 3), (4, 9), (7, 15) E10. (2, 1), (4, 5), (6, 9)

Exercise Set F: Circle Equations (10 Questions)

Find center and radius. For F6-F10, write the equation.

F1. $x^2 + y^2 - 10x + 12y - 11 = 0$ F2. $x^2 + y^2 + 6x - 8y - 24 = 0$ F3. $x^2 + y^2 - 4x + 6y - 12 = 0$ F4. $x^2 + y^2 + 8x - 10y + 5 = 0$ F5. $x^2 + y^2 - 2x + 4y - 20 = 0$ F6. Center (2, 3), radius 5 F7. Center (-3, 4), radius 6 F8. Center (0, 0), radius 10 F9. Center (4, -2), radius 3 F10. Center (a, b), radius r (general)

Answer Key for Practice Exercises

Set A Answers (Distance):

A1. 5
A2. 5
A3. 10
A4. 10
A5. 13
A6. 10
A7. 13
A8. 5
A9. 13
A10. 10
A11. 9
A12. 8
A13. 10
A14. 25
A15. 2√(p²+q²)

Set B Answers (Section):

B1. (6,8)
B2. (4,5)
B3. (7,8)
B4. (4,5)
B5. (5,8)
B6. (3,4.5)
B7. (11,16)
B8. (4,5) and (8,10)
B9. (6,8) and (10,13)
B10. (9,13)
B11. (6.4,4.4)
B12. ((a+c)/2, (b+d)/2)
B13. ((x₁+x₂)/2, (y₁+y₂)/2)
B14. (8,8)
B15. (8,10)

Set C Answers (Slope & Line):

C1. m=3, 3x-y-1=0
C2. m=2, 2x-y-1=0
C3. m=2, 2x-y-3=0
C4. m=2, 2x-y+2=0? Check: 2×1-4+2=0 ✓
C5. m=1, x-y+4=0? 3-7+4=0 ✓
C6. m=0, y-5=0
C7. m=∞, x-4=0
C8. m=12/5, 12x-5y=0
C9. m=1, x-y=0
C10. m=-3/4, 3x+4y-38=0
C11. m=-1, x+y-9=0
C12. m=1, x-y+5=0? 5-10+5=0 ✓
C13. bx+ay-ab=0
C14. (y₂-y₁)x-(x₂-x₁)y+(x₂y₁-x₁y₂)=0
C15. m=3q/2p, 3qx-2py=0

Set D Answers (Area):

D1. 17.5
D2. 8.5
D3. 10.5
D4. 20
D5. 7
D6. 15
D7. 22.5
D8. 20
D9. 2.5
D10. ab/2

Set E Answers (Collinearity):

E1. Yes
E2. Yes
E3. Yes
E4. Yes
E5. No
E6. Yes
E7. Yes
E8. Yes
E9. Yes
E10. No

Set F Answers (Circle):

F1. (5,-6), r=√72=6√2
F2. (-3,4), r=7
F3. (2,-3), r=5
F4. (-4,5), r=6
F5. (1,-2), r=5
F6. x²+y²-4x-6y-12=0
F7. x²+y²+6x-8y-11=0
F8. x²+y²-100=0
F9. x²+y²-8x+4y+11=0
F10. x²+y²-2ax-2by+(a²+b²-r²)=0

PART 5: TEACHER'S GUIDE

Common Mistakes & Corrections

Vilokanam Quick Reference

QUICK REFERENCE CARD

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║           MODULE 17 — COORDINATE GEOMETRY: VEDIC SHORTCUTS            ║
╠═══════════════════════════════════════════════════════════════════════╣
║                                                                       ║
║  DISTANCE: d² = (Δx)² + (Δy)²                                        ║
║                                                                       ║
║  SECTION: P = (mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)                       ║
║                                                                       ║
║  MIDPOINT: ((x₁+x₂)/2, (y₁+y₂)/2)                                    ║
║                                                                       ║
║  SLOPE: m = (y₂-y₁)/(x₂-x₁) = tan θ                                  ║
║                                                                       ║
║  LINE EQUATION (Cross method):                                        ║
║  (y₁-y₂)x + (x₂-x₁)y + (x₁y₂ - x₂y₁) = 0                             ║
║                                                                       ║
║  AREA (Shoelace): ½|Σ(xᵢyᵢ₊₁) - Σ(yᵢxᵢ₊₁)|                           ║
║                                                                       ║
║  COLLINEARITY: slope AB = slope BC  OR  area = 0                     ║
║                                                                       ║
║  CIRCLE: (x-h)² + (y-k)² = r²  =  x²+y²+Dx+Ey+F=0                    ║
║  Center: (-D/2, -E/2),  r² = h²+k²-F                                 ║
║                                                                       ║
║  SUTRA 3: Urdhva-Tiryagbhyam — Vertically and cross-wise             ║
║  SUB-SUTRA 12: Vilokanam — By mere observation                        ║
║                                                                       ║
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Document Version 1.0 | Vedic Mathematics Level 2 Intermediate Course

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