Maths Week 2 - Coordinate Geometry and Straight Lines

Mathematics for Data Science I - Week 2: Coordinate Geometry and Straight Lines

📚 Table of Contents

1. Fundamental Concepts
2. Question Pattern Analysis
3. Detailed Solutions by Pattern
4. Practice Exercises
5. Visual Learning
6. Common Pitfalls & Traps
7. Quick Refresher Handbook

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Fundamental Concepts

🎯 Rectangular Coordinate System

The rectangular coordinate system (also called Cartesian coordinate system) is a fundamental tool for representing geometric relationships algebraically.

Key Components:

• X-axis: Horizontal axis (represents independent variable)
• Y-axis: Vertical axis (represents dependent variable)
• Origin: Point (0,0) where axes intersect
• Quadrants: Four regions created by the axes
• Ordered Pair: (x,y) representing a point’s position

Distance Formula:

For two points $A(x_1,y_1)$ and $B(x_2,y_2)$: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

Section Formula:

Point dividing line segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ in ratio $m:n$:

• Internal Division: $\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$
• External Division: $\left(\frac{m{x}_2-n{x}_1}{m-n},\frac{m{y}_2-n{y}_1}{m-n}\right)$

📈 Straight Lines

1. Slope of a Line

For line passing through points $(x_1,y_1)$ and $(x_2,y_2)$: $m=\frac{y_2-y_1}{x_2-x_1}=\tan \theta$

where $\theta$ is the angle the line makes with positive x-axis.

2. Forms of Linear Equations

Form	Equation	When to Use	Key Features
Slope-Intercept	$y=mx+c$	When slope and y-intercept are known	$m$ = slope, $c$ = y-intercept
Point-Slope	$y-y_1=m(x-x_1)$	When slope and one point are known	Most versatile form
Two-Point	$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$	When two points are known	Direct application of slope formula
General/Standard	$Ax+By+C=0$	For theoretical analysis	$A,B,C$ are constants
Intercept Form	$\frac{x}{a}+\frac{y}{b}=1$	When intercepts are known	$a$ = x-intercept, $b$ = y-intercept

3. Special Cases of Lines

Line Type	Slope	Equation	Characteristics
Horizontal	$m=0$	$y=c$	Parallel to x-axis
Vertical	Undefined	$x=k$	Parallel to y-axis
Line through origin	$m$	$y=mx$	Passes through (0,0)

4. Parallel and Perpendicular Lines

For lines with slopes $m_1$ and $m_2$:

• Parallel: $m_1=m_2$
• Perpendicular: $m_1\times m_2=-1$

5. Distance from Point to Line

Distance from point $(x_0,y_0)$ to line $Ax+By+C=0$: $d=\frac{|A{x}_0+B{y}_0+C|}{\sqrt{A^2+B^2}}$

6. Angle Between Two Lines

For lines with slopes $m_1$ and $m_2$: $\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$

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Question Pattern Analysis

Based on analysis of previous year questions, Week 2 follows these distinct patterns:

🎯 Pattern 1: Line Intersection & Collision Detection

Frequency: High | Difficulty: Easy-Medium

• Tests understanding of solving systems of linear equations
• Often framed as real-world scenarios (bird/plane collision, paths)

🎯 Pattern 2: Distance & Rate Problems

Frequency: Medium | Difficulty: Medium

• Combines geometry with calculus concepts (rates of change)
• Circular ripples, expanding objects

🎯 Pattern 3: Reflection & Optics

Frequency: Medium | Difficulty: Medium-Hard

• Uses laws of reflection in coordinate geometry
• Light rays, mirror problems

🎯 Pattern 4: Geometric Properties (Parallelograms, Triangles)

Frequency: High | Difficulty: Medium

• Tests coordinate geometry properties of quadrilaterals
• Section formula, area calculations

🎯 Pattern 5: Linear Models & Cost Analysis

Frequency: Medium | Difficulty: Easy-Medium

• Real-world applications (cost optimization, decision making)
• Comparing linear equations

🎯 Pattern 6: Shortest Distance Problems

Frequency: High | Difficulty: Medium

• Point-to-line distance formula
• Optimization scenarios

🎯 Pattern 7: Linear Regression & Data Fitting

Frequency: High | Difficulty: Medium-Hard

• Sum of Squared Errors (SSE)
• Best fit lines, data interpretation

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Detailed Solutions by Pattern

🎯 Pattern 1: Line Intersection & Collision Detection

Example from PYQ:

A bird is flying along the straight line $2y-6x=6$. In the same plane, an aeroplane starts to fly in a straight line from the origin and passes through the point $(4,12)$. If the bird and plane collide, enter the answer as 1, and if not, then 0.

Step-by-Step Solution:

Step 1: Convert both equations to standard form

• Bird’s line: $2y-6x=6$

  • Rearrange: $2y=6x+6$
  • Simplify: $y=3x+3$ ← Slope = 3, y-intercept = 3

• Plane’s line: Passes through $(0,0)$ and $(4,12)$

  • Slope: $m=\frac{12-0}{4-0}=\frac{12}{4}=3$
  • Equation: $y=3x+0$ ← Slope = 3, y-intercept = 0

Step 2: Check for collision

• Two lines collide if they intersect
• For collision: Lines must have same slope AND same y-intercept
• Here: Both have slope = 3 ✓
• But: y-intercepts are 3 and 0 ✗

Step 3: Conclusion Since lines are parallel but distinct, they never intersect.

Answer: 0

💡 Key Insight:

Parallel lines with different intercepts never meet. This is a fundamental property used in collision detection problems.

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🎯 Pattern 2: Distance & Rate Problems

Example from PYQ:

A rock is thrown in a pond and creates circular ripples whose radius increases at a rate of 0.2 meters per second. What will be the value of $\frac{A}{\pi }$, where $A$ is the area (in square meters) of the circle after 5 seconds?

Step-by-Step Solution:

Step 1: Understand the relationship

• Radius increases at constant rate: $\frac{dr}{dt}=0.2$ m/s
• Time: $t=5$ seconds
• Area of circle: $A=\pi r^2$

Step 2: Find radius after 5 seconds

• $r=r_0+\frac{dr}{dt}\times t$
• $r=0+0.2\times 5=1$ meter

Step 3: Calculate area

• $A=\pi r^2=\pi \times 1^2=\pi$ square meters

Step 4: Find $\frac{A}{\pi }$

• $\frac{A}{\pi }=\frac{\pi }{\pi }=1$

Answer: 1.0

💡 Key Insight:

When asked for $\frac{A}{\pi }$ in circle problems, they’re simplifying to find $r^2$, avoiding $\pi$ calculations.

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🎯 Pattern 3: Reflection & Optics

Example from PYQ:

A ray of light passing through the point $A(1,2)$ is reflected at a point $B$ on the X-axis and then passes through the point $(5,3)$. What is the equation of the straight line segment $AB$?

Step-by-Step Solution:

Step 1: Understand reflection law

• Angle of incidence = Angle of reflection
• For reflection on x-axis: y-coordinate changes sign, x-coordinate remains same
• If $B=(x,0)$ is reflection point, then reflected point of $A(1,2)$ is $A^{\prime }(1,-2)$

Step 2: Use collinearity

• Points $A^{\prime }(1,-2)$, $B(x,0)$, and $(5,3)$ are collinear
• Slope between $A^{\prime }$ and $(5,3)$ = Slope between $A^{\prime }$ and $B$

Step 3: Calculate slopes

• Slope $A^{\prime }$ to $(5,3)$: $m_1=\frac{3-(-2)}{5-1}=\frac{5}{4}$
• Slope $A^{\prime }$ to $B$: $m_2=\frac{0-(-2)}{x-1}=\frac{2}{x-1}$

Step 4: Equate slopes

• $\frac{2}{x-1}=\frac{5}{4}$
• $8=5(x-1)$
• $8=5x-5$
• $5x=13$
• $x=\frac{13}{5}$

Step 5: Find equation of AB

• Points: $A(1,2)$ and $B(\frac{13}{5},0)$
• Slope: $m=\frac{0-2}{\frac{13}{5}-1}=\frac{-2}{\frac{8}{5}}=-\frac{5}{4}$

Using point-slope form with point $A(1,2)$:

• $y-2=-\frac{5}{4}(x-1)$
• $4(y-2)=-5(x-1)$
• $4y-8=-5x+5$
• $5x+4y=13$

Answer: $5x+4y=13$

💡 Key Insight:

Reflection problems use the property that the incident ray, reflected ray, and normal make equal angles. For coordinate axes reflections, simply change the sign of the perpendicular coordinate.

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🎯 Pattern 4: Geometric Properties

Example from PYQ:

Let $ABCD$ be a parallelogram with vertices $A(x_1,y_1)$, $B(x_2,y_2)$, and $C(x_3,y_3)$. Which of the following always denotes the coordinate of the fourth vertex $D$?

Step-by-Step Solution:

Step 1: Use parallelogram property

• In a parallelogram, diagonals bisect each other
• Midpoint of $AC$ = Midpoint of $BD$

Step 2: Apply midpoint formula

• Midpoint of $AC$: $\left(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2}\right)$
• Let $D=(x_4,y_4)$
• Midpoint of $BD$: $\left(\frac{x_2+x_4}{2},\frac{y_2+y_4}{2}\right)$

Step 3: Equate midpoints

• $\frac{x_1+x_3}{2}=\frac{x_2+x_4}{2}\Rightarrow x_1+x_3=x_2+x_4$
• $\frac{y_1+y_3}{2}=\frac{y_2+y_4}{2}\Rightarrow y_1+y_3=y_2+y_4$

Step 4: Solve for $D(x_4,y_4)$

• $x_4=x_1+x_3-x_2$
• $y_4=y_1+y_3-y_2$

Answer: $(x_1-x_2+x_3,y_1-y_2+y_3)$

💡 Key Insight:

The fourth vertex formula comes from vector addition: $\overrightarrow{AD}=\overrightarrow{BC}$, so $D=A+C-B$.

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🎯 Pattern 5: Linear Models & Cost Analysis

Example from PYQ:

Lalith wants to buy a new mobile and needs 200 minutes of calls per month. He has two options: Company Offer: Mobile and a 1-year Astron Network plan (unlimited calls) for ₹34000. Separate Purchase: Buy only the mobile for ₹22000 and choose a network provider…

Step-by-Step Solution:

Step 1: Calculate costs for Company Offer

• Total cost = ₹34,000

Step 2: Calculate costs for Separate Purchase

• Mobile cost = ₹22,000
• Network costs for 200 minutes/month:
  • Astron: ₹100/month + ₹2/min × 200 = ₹100 + ₹400 = ₹500/month
  • Proton: ₹200/month + ₹0.5/min × 200 = ₹200 + ₹100 = ₹300/month
• Annual network costs:
  • Astron: ₹500 × 12 = ₹6,000
  • Proton: ₹300 × 12 = ₹3,600
• Total separate costs:
  • With Astron: ₹22,000 + ₹6,000 = ₹28,000
  • With Proton: ₹22,000 + ₹3,600 = ₹25,600

Step 3: Find best option and savings

• Best separate option: ₹25,600 (with Proton)
• Savings: ₹34,000 - ₹25,600 = ₹8,400

Answer: 8400

💡 Key Insight:

Always compare all options systematically. The “cheaper” per-minute rate may not always give the lowest total cost due to fixed charges.

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🎯 Pattern 6: Shortest Distance Problems

Example from PYQ:

The government wants to connect a town to a national highway. The national highway is a straight line connecting points $(2,1)$ and $(10,7)$. There are 3 possible locations in the town to build the connecting road from: A(3,8), B(5,7), and C(6,9)…

Step-by-Step Solution:

Step 1: Find equation of highway line

• Points: $(2,1)$ and $(10,7)$
• Slope: $m=\frac{7-1}{10-2}=\frac{6}{8}=\frac{3}{4}$
• Equation: $y-1=\frac{3}{4}(x-2)$
• $4(y-1)=3(x-2)$
• $4y-4=3x-6$
• $3x-4y-2=0$

Step 2: Use distance formula from point to line For point $(x_0,y_0)$ to line $Ax+By+C=0$: $d=\frac{|A{x}_0+B{y}_0+C|}{\sqrt{A^2+B^2}}$

Here: $A=3$, $B=-4$, $C=-2$

Step 3: Calculate distances

• For $A(3,8)$: $d_A=\frac{|3(3)-4(8)-2|}{\sqrt{3^2+(-4{)}^2}}=\frac{|9-32-2|}{5}=\frac{25}{5}=5$
• For $B(5,7)$: $d_B=\frac{|3(5)-4(7)-2|}{5}=\frac{|15-28-2|}{5}=\frac{15}{5}=3$
• For $C(6,9)$: $d_C=\frac{|3(6)-4(9)-2|}{5}=\frac{|18-36-2|}{5}=\frac{20}{5}=4$

Step 4: Convert to meters and select minimum

• Minimum distance = 3 units = 3 × 100 = 300 meters
• Best location: Point B

Answers: Location B, 300 meters

💡 Key Insight:

The shortest distance from a point to a line is always along the perpendicular. This is why we use the point-to-line distance formula.

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🎯 Pattern 7: Linear Regression & Data Fitting

Example from PYQ:

A fitness trainer models a client’s weight loss with the equation $W=-8t+98$, where $W$ is weight in Kg and $t$ is time in months. The actual data is in the table below… An equation is a “good fit” if its Sum of Squared Errors (SSE) is less than 5. Is the trainer’s equation a good fit?

Step-by-Step Solution:

Step 1: Understand SSE $SSE={\sum }_{i=1}^n(y_i-{\hat{y}}_i{)}^2$ where $y_i$ = actual value, ${\hat{y}}_i$ = predicted value

Step 2: Calculate predicted values Using $W=-8t+98$:

• $t=0$: $\hat{W}=-8(0)+98=98$
• $t=1$: $\hat{W}=-8(1)+98=90$
• $t=2$: $\hat{W}=-8(2)+98=82$
• $t=3$: $\hat{W}=-8(3)+98=74$
• $t=4$: $\hat{W}=-8(4)+98=66$
• $t=5$: $\hat{W}=-8(5)+98=58$
• $t=6$: $\hat{W}=-8(6)+98=50$

Step 3: Calculate squared errors

$t$	Actual $W$	Predicted $\hat{W}$	Error	Error²
0	98	98	0	0
1	90	90	0	0
2	82	82	0	0
3	74	74	0	0
4	66	66	0	0
5	57	58	-1	1
6	49	50	-1	1

Step 4: Calculate SSE $SSE=0+0+0+0+0+1+1=2$

Step 5: Compare with threshold

• SSE = 2 < 5 ✓

Answer: True (it is a good fit)

💡 Key Insight:

Perfect matches contribute 0 to SSE. Even small deviations can make SSE exceed the threshold, so precision matters.

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4. Handbook / Quick Refresher

Concept	Formula
Slope	$m=\frac{y_2-y_1}{x_2-x_1}$
Slope-Intercept Form	$y=mx+c$
Point-Slope Form	$y-y_1=m(x-x_1)$
General Form	$Ax+By+C=0$
Parallel Lines	$m_1=m_2$
Perpendicular Lines	$m_1\cdot m_2=-1$
Distance (Point-Point)	$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
Distance (Point-Line)	$d=\frac{A{x}_0+B{y}_0+C}{\sqrt{A^2+B^2}}$
Midpoint	$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
Section Formula (Internal)	$(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$
Area of Triangle	$\frac{1}{2}{x}_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$
Sum of Squared Errors	SSE = $\sum (y_{actual}-y_{predicted}{)}^2$

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Practice Exercises

🎯 Exercise Set 1: Fundamental Concepts

Exercise 1.1: Find the equation of the line passing through $(2,5)$ and $(-1,1)$.

Hint

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

Solution

Step 1: Calculate slope $m=\frac{1-5}{-1-2}=\frac{-4}{-3}=\frac{4}{3}$

Step 2: Use point-slope form with $(2,5)$ $y-5=\frac{4}{3}(x-2)$

Step 3: Convert to standard form $3(y-5)=4(x-2)$ $3y-15=4x-8$ $4x-3y+7=0$

Answer: $4x-3y+7=0$ or $y=\frac{4}{3}x+\frac{7}{3}$

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Exercise 1.2: Find the distance between parallel lines $2x+3y=6$ and $4x+6y=20$.

Hint

First check if lines are parallel, then find distance from any point on one line to the other line.

Solution

Step 1: Check if parallel

• Line 1: $2x+3y=6$
• Line 2: $4x+6y=20$ or $2x+3y=10$
• Both have same coefficients for $x$ and $y$ → Parallel ✓

Step 2: Find a point on Line 1 When $x=0$: $3y=6\Rightarrow y=2$ Point: $(0,2)$

Step 3: Use distance formula Line 2 in standard form: $2x+3y-10=0$ Distance from $(0,2)$ to Line 2: $d=\frac{|2(0)+3(2)-10|}{\sqrt{2^2+3^2}}=\frac{|6-10|}{\sqrt{13}}=\frac{4}{\sqrt{13}}$

Answer: $\frac{4}{\sqrt{13}}$ units

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🎯 Exercise Set 2: Application Problems

Exercise 2.1: A laser beam is aimed from point $(2,4)$ to a mirror placed along the line $y=2$. After reflection, the beam passes through $(6,0)$. Find the point of incidence on the mirror.

Hint

For reflection on a horizontal line $y = k$, the y-coordinate becomes $2k - y$ while x-coordinate remains the same.

Solution

Step 1: Find reflected point

• Mirror: $y=2$
• Original point: $(2,4)$
• Reflected point: $(2,2\times 2-4)=(2,0)$

Step 2: Use collinearity Points $(2,0)$, incidence point $(x,2)$, and $(6,0)$ are collinear.

Step 3: Calculate using section formula Since $(2,0)$ and $(6,0)$ have same y-coordinate, the incidence point must have:

• $x=\frac{2+6}{2}=4$ (midpoint)
• $y=2$ (on mirror)

Answer: $(4,2)$

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Exercise 2.2: A company produces two products. Product A costs ₹50 per unit plus ₹2000 setup cost. Product B costs ₹80 per unit plus ₹1500 setup cost. Find the break-even point where total costs are equal.

Hint

Set up linear cost equations for both products and solve for equality.

Solution

Step 1: Set up cost equations

• Product A: $C_A=50x+2000$
• Product B: $C_B=80x+1500$

Step 2: Find break-even point $50x+2000=80x+1500$ $500=30x$ $x=\frac{500}{30}=\frac{50}{3}\approx 16.67$

Step 3: Calculate cost at break-even $C=50\times \frac{50}{3}+2000=\frac{2500}{3}+2000=\frac{8500}{3}\approx ₹2833.33$

Answer: At approximately 16.67 units, costing ₹2833.33

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Visual Learning

📊 Coordinate System Visualization

graph TD
    A[Origin 0,0] --> B[Quadrant I +,+]
    A --> C[Quadrant II -,+]
    A --> D[Quadrant III -,-]
    A --> E[Quadrant IV +,-]

    F[Point x₁,y₁] --> G[Distance Formula]
    G --> H["√x₂-x₁² + y₂-y₁²"]

    I[Line Properties] --> J[Slope m = Δy/Δx]
    I --> K[Parallel m₁ = m₂]
    I --> L[Perpendicular m₁ × m₂ = -1]

📈 Line Relationships

graph LR
    subgraph Parallel Lines
        direction LR
        L1[Line 1: y = 2x + 3]
        L2[Line 2: y = 2x - 1]
        L1 -.-> L2
    end

    subgraph Perpendicular Lines
        direction TB
        L3[Line 3: y = -1/2x + 4]
        L4[Line 4: y = 2x + 1]
        L3 -- 90° --> L4
    end

    subgraph Distance Problem
        direction RL
        P[Point x₀,y₀]
        L[Line: Ax + By + C = 0]
        P -.->|Shortest Distance| L
        D["d = |Ax₀ + By₀ + C|/√A² + B²"]
    end

🔍 Reflection Geometry

sequenceDiagram
    participant Light as Light Ray
    participant Mirror as Mirror Line
    participant Point as Original Point
    participant Reflected as Reflected Point

    Point->>Mirror: Incident Ray
    Note over Mirror: Angle of Incidence
    Mirror->>Reflected: Reflected Ray
    Note over Mirror: Angle of Reflection

    Note right of Point: For horizontal mirror:<br/>y-coordinate changes sign<br/>x-coordinate stays same

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Common Pitfalls & Traps

🚨 Pitfall 1: Sign Errors in Distance Formula

Problem: Forgetting absolute value in distance formula $d=\frac{|A{x}_0+B{y}_0+C|}{\sqrt{A^2+B^2}}$

Trap: Using $d=\frac{A{x}_0+B{y}_0+C}{\sqrt{A^2+B^2}}$ (without absolute value)

Example: Distance from $(1,1)$ to line $x+y-1=0$

• ❌ Wrong: $\frac{1+1-1}{\sqrt{2}}=\frac{1}{\sqrt{2}}$
• ✅ Correct: $\frac{|1+1-1|}{\sqrt{2}}=\frac{1}{\sqrt{2}}$

Prevention: Always use absolute value for distance calculations.

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🚨 Pitfall 2: Confusing Parallel and Perpendicular Conditions

Problem: Mixing up the conditions for parallel and perpendicular lines

Common Mistakes:

• Thinking parallel means $m_1=-m_2$ ❌
• Thinking perpendicular means $m_1=m_2$ ❌

Correct Conditions:

• Parallel: $m_1=m_2$ ✅
• Perpendicular: $m_1\times m_2=-1$ ✅

Memory Trick:

• Parallel = “Same direction” = same slope
• Perpendicular = “Opposite direction” = negative reciprocal

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🚨 Pitfall 3: Section Formula Direction

Problem: Not distinguishing between internal and external division

Internal Division (point between the two points): $\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

External Division (point outside the segment): $\left(\frac{m{x}_2-n{x}_1}{m-n},\frac{m{y}_2-n{y}_1}{m-n}\right)$

Trap: Using internal division formula for external division problems

Prevention: Check if the dividing point lies between the given points.

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🚨 Pitfall 4: Rate Problems Unit Conversion

Problem: Not converting units properly in rate problems

Example: Radius increases at 0.2 m/s for 5 seconds

• ❌ Wrong: $r=0.2\times 5=1$ (forgetting units)
• ❌ Wrong: $r=20\times 5=100$ (wrong conversion)
• ✅ Correct: $r=0.2$ m/s × $5$ s = $1$ m

Prevention: Always track units throughout the calculation.

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🚨 Pitfall 5: SSE Calculation Errors

Problem: Order of operations in Sum of Squared Errors

Common Error: $(y_i-{\hat{y}}_i^2)$ instead of $(y_i-{\hat{y}}_i{)}^2$

Example: Actual = 10, Predicted = 8

• ❌ Wrong: $10-8^2=10-64=-54$
• ✅ Correct: $(10-8{)}^2=2^2=4$

Prevention: Calculate the difference first, then square the result.

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Quick Refresher Handbook

📋 Essential Formulas (Week 2)

| Concept | Formula | When to Use | | ------------------------------ | ------------------------------------------------ | ----------------------------------- | ------------------ | ------------------------------------ | | Distance between points | $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$ | Finding distance between two points | | Slope | $\frac{y_2-y_1}{x_2-x_1}$ | Finding inclination of line | | Point-slope form | $y-y_1=m(x-x_1)$ | When slope and one point known | | Point to line distance | $\frac{|A{x}_0+B{y}_0+C|}{\sqrt{A^2+B^2}}$ | Shortest distance from point to line | | Section formula (internal) | $(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$ | Point dividing segment internally | | Angle between lines | $\tan \theta =|\frac{m_2-m_1}{1+m_1{m}_2}|$ | Finding angle between two lines | | SSE | $\sum (y_i-{\hat{y}}_i{)}^2$ | Measuring fit quality |

🎯 Problem-Solving Strategy

1. Read Carefully: Identify what’s given and what’s asked
2. Choose Right Formula: Match the problem type to the correct formula
3. Draw Diagram: Visual representation helps avoid errors
4. Check Units: Ensure consistency throughout calculations
5. Verify Answer: Plug back into original conditions

🔍 Quick Check Techniques

For Line Equations:

• Does the point satisfy the equation?
• Is the slope correct for given points?

For Distance Problems:

• Is the result positive?
• Does it make sense in context?

For Parallel/Perpendicular:

• Check the condition: $m_1=m_2$ or $m_1\times m_2=-1$

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💡 Memory Tricks

• SLOPE: “Rise over Run” = $\frac{\Delta y}{\Delta x}$
• Parallel Lines: “Same family” = same slope
• Perpendicular: “Negative reciprocal” = flip and change sign
• Distance Formula: “ABC over root A²+B²” = $\frac{|A{x}_0+B{y}_0+C|}{\sqrt{A^2+B^2}}$

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📝 Consolidated Question Patterns

Pattern Summary with Solution Approaches

Pattern	Key Concept	Solution Strategy	Frequency
Line Intersection	System of equations	Solve simultaneously or compare slopes	High
Distance & Rate	Distance formula + rates	Find relationship, substitute time	Medium
Reflection	Coordinate transformation	Reflect point, use collinearity	Medium
Geometric Properties	Section formula, area	Apply coordinate geometry theorems	High
Linear Models	Cost comparison	Set up equations, compare	Medium
Shortest Distance	Point-to-line distance	Use formula directly	High
Data Fitting	SSE calculation	Compute predicted values, sum squares	High

🎯 Practice Recommendations

1. Master the Basics: Ensure you can quickly apply all formulas
2. Pattern Recognition: Learn to identify question types quickly
3. Time Management: Allocate time based on question frequency
4. Verification: Always check your answers
5. Real-world Applications: Practice converting word problems to mathematical form

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This comprehensive guide covers all essential concepts, patterns, and strategies for Mathematics Week 2. Practice regularly with these approaches to build strong foundational skills in coordinate geometry and straight lines.