Maths Week 2: Coordinate Geometry and Straight Lines

This week, we build the bridge between algebra and geometry. Coordinate geometry allows us to represent geometric shapes using algebraic equations, providing a powerful framework for analysis and problem-solving. We will focus on the most fundamental geometric object: the straight line.

📚 Table of Contents

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

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1. Fundamental Concepts

🎯 The Rectangular Coordinate System

Also known as the Cartesian plane, this system uses two perpendicular axes to define the position of any point.

• Axes: The horizontal x-axis and the vertical y-axis.
• Origin: The intersection point of the axes, $(0,0)$.
• Coordinates: A point is represented by an ordered pair $(x,y)$.
• Key Formulas:
  • Distance between 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 (Internal Division): Coordinates of a point $P$ that divides the segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ in the ratio $m:n$. $P(x,y)=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$
  • Midpoint Formula (a special case of section formula where m=n=1): $M(x,y)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

📈 The Straight Line

Slope (m)

The slope measures the “steepness” or inclination of a line.

• Formula: Given two points $(x_1,y_1)$ and $(x_2,y_2)$: $m=\frac{\text{rise}}{\text{run}}=\frac{y_2-y_1}{x_2-x_1}$
• Special Cases:
  • Horizontal line: $m=0$.
  • Vertical line: $m$ is undefined.

Forms of Linear Equations

Form Name	Equation	When to Use It
Slope-Intercept	$y=mx+c$	Given slope ($m$) and y-intercept ($c$).
Point-Slope	$y-y_1=m(x-x_1)$	Given slope ($m$) and one point $(x_1,y_1)$.
General Form	$Ax+By+C=0$	Often used for final answers and formulas.
Two-Point Form	$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$	Given two points $(x_1,y_1),(x_2,y_2)$.

Parallel and Perpendicular Lines

Given two lines with slopes $m_1$ and $m_2$:

• Parallel Lines: Their slopes are equal. $m_1=m_2$
• Perpendicular Lines: The product of their slopes is -1. $m_1\times m_2=-1$

Distance from a Point to a Line

The shortest distance from a point $(x_0,y_0)$ to a line given in the general form $Ax+By+C=0$ is: $d=\frac{|A{x}_0+B{y}_0+C|}{\sqrt{A^2+B^2}}$

📊 Straight-Line Fit (Sum of Squared Errors)

When we try to model a set of data points with a line, the Sum of Squared Errors (SSE) measures how well the line fits the data.

• Error (Residual): For a data point $(x_i,y_i)$ and a line $y=f(x)$, the error is the vertical distance between the actual point and the line: ${\text{Error}}_i=y_i-f(x_i)$
• SSE Formula: $\text{SSE}={\sum }_{i=1}^n({\text{Error}}_i{)}^2={\sum }_{i=1}^n(y_i-f(x_i){)}^2$ A smaller SSE indicates a better fit.

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2. Question Pattern Analysis

From the provided Week_2_Graded_Assignment, several distinct problem patterns emerge.

🎯 Pattern 1: Line Intersection & Collision

• Frequency: High
• Description: Determining if and where two lines intersect. Often framed as a “collision” problem between objects moving in straight lines.
• Example: Questions 1, 12, 14.

🎯 Pattern 2: Geometric Properties & Formulas

• Frequency: High
• Description: Using coordinate geometry formulas (section, midpoint, area) to solve problems involving geometric shapes like parallelograms and triangles.
• Example: Questions 4, 13, 16.

🎯 Pattern 3: Data Fitting & Sum of Squared Errors (SSE)

• Frequency: High
• Description: Calculating the SSE for a given linear model and a set of data points to evaluate the “goodness of fit”.
• Example: Questions 9, 18.

🎯 Pattern 4: Shortest Distance Problems

• Frequency: Medium
• Description: Finding the shortest distance from a point to a line, often in a real-world context like building a road to a highway.
• Example: Questions 7, 8.

🎯 Pattern 5: Reflection Geometry

• Frequency: Medium
• Description: Problems involving the reflection of light rays off an axis.
• Example: Question 3.

🎯 Pattern 6: Rate & Application Problems

• Frequency: Medium
• Description: Simple real-world scenarios that combine basic geometry with rates of change or cost analysis.
• Example: Questions 2, 6.

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3. Detailed Solutions by Pattern

🎯 Pattern 1: Line Intersection & Collision

Example (from Graded Assignment Q12):

A bird is flying along the straight line $2y-6x=6$. An aeroplane follows a path with a slope of 2 and passes through the point $(4,8)$. Let $(\alpha ,\beta )$ be the collision point. Find $\alpha +\beta$.

Step-by-Step Solution:

1. Find the Equation for the Bird’s Path: Rearrange the given equation into the familiar $y=mx+c$ form. $2y-6x=6$ $2y=6x+6$ $y=3x+3(\text{Line 1})$

2. Find the Equation for the Aeroplane’s Path: Use the point-slope form $y-y_1=m(x-x_1)$ with slope $m=2$ and point $(x_1,y_1)=(4,8)$. $y-8=2(x-4)$ $y-8=2x-8$ $y=2x(\text{Line 2})$

3. Find the Intersection Point: The collision occurs where the paths intersect. At this point, the $(x,y)$ coordinates are the same for both lines. So, we set the expressions for $y$ equal to each other. $3x+3=2x$ $x=-3$ Now, substitute $x=-3$ into either equation to find $y$. Using the simpler Line 2: $y=2(-3)=-6$ The collision point $(\alpha ,\beta )$ is $(-3,-6)$.

4. Calculate the Final Value: $\alpha +\beta =-3+(-6)=-9$

💡 Key Insight: A “collision” is just a geometric intersection. The problem is solved by finding the equations for both paths and solving them as a system of linear equations.

🎯 Pattern 3: Data Fitting & Sum of Squared Errors (SSE)

Example (from Graded Assignment Q18):

Radhika fits a line $y=4x+2$ to her data of expenses ($y$) vs. outings ($x$). What is the SSE for this fit? Data: (1, 6), (3, 14), (5, 24), (7, 29), (9, 39), (11, 45).

Step-by-Step Solution:

1. Understand the Goal: We need to calculate SSE = $\sum (y_{actual}-y_{predicted}{)}^2$.

2. Create a Table to organize the calculations. For each data point, calculate the predicted $y$ using the formula $y=4x+2$, then find the error, and finally square the error.

x (outings)	y (actual)	y (predicted) = 4x+2	Error (actual - predicted)	Error²
1	6	$4(1)+2=6$	$6-6=0$	0
3	14	$4(3)+2=14$	$14-14=0$	0
5	24	$4(5)+2=22$	$24-22=2$	4
7	29	$4(7)+2=30$	$29-30=-1$	1
9	39	$4(9)+2=38$	$39-38=1$	1
11	45	$4(11)+2=46$	$45-46=-1$	1

3. Sum the Squared Errors: Add the values in the last column: $\text{SSE}=0+0+4+1+1+1=7$

💡 Key Insight: SSE is a measure of total error. Each deviation from the line adds to the SSE, and squaring ensures that both positive and negative errors contribute positively.

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4. Practice Exercises

Exercise 1: Parallel and Perpendicular Lines

Find the equation of a line that passes through the point $(2,-3)$ and is perpendicular to the line $4x-2y+5=0$.

Hint

First, find the slope of the given line. Then, use the perpendicular condition ($m_1 \times m_2 = -1$) to find the slope of the new line. Finally, use the point-slope form.

Solution

1. **Find the slope of the given line:** Rearrange $4x - 2y + 5 = 0$ to the form $y=mx+c$. $2y = 4x + 5 \implies y = 2x + 2.5$. The slope of this line, $m_1$, is 2. 2. **Find the slope of the perpendicular line:** Let the new slope be $m_2$. We have $m_1 \times m_2 = -1$. $2 \times m_2 = -1 \implies m_2 = -1/2$. 3. **Find the equation of the new line:** Use the point-slope form with $m = -1/2$ and point $(2, -3)$. $y - (-3) = -\frac{1}{2}(x - 2)$ $y + 3 = -\frac{1}{2}x + 1$ $y = -\frac{1}{2}x - 2$ **Answer:** $y = -\frac{1}{2}x - 2$ or $x + 2y + 4 = 0$.

Exercise 2: Section Formula

Find the coordinates of the point that divides the line segment connecting $A(-1,7)$ and $B(4,-3)$ in the ratio $2:3$.

Hint

Apply the section formula directly with $(x_1, y_1) = (-1, 7)$, $(x_2, y_2) = (4, -3)$, $m=2$, and $n=3$.

Solution

Using the section formula $P(x,y) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)$: * **x-coordinate:** $x = \frac{2(4) + 3(-1)}{2+3} = \frac{8 - 3}{5} = \frac{5}{5} = 1$ * **y-coordinate:** $y = \frac{2(-3) + 3(7)}{2+3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3$ **Answer:** The coordinates of the point are $(1, 3)$.

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5. Visual Learning: Mermaid Diagrams

Relationship Between Line Forms

graph TD
    A[Given: 2 Points] --> B(Calculate Slope);
    A --> C[Two-Point Form];
    B --> D[Point-Slope Form];
    D --> E[Slope-Intercept Form];
    E --> F[General Form];
    C --> F;

Geometry of Reflection

sequenceDiagram
    participant A as Original Point A(x,y)
    participant B as Reflection Point on x-axis
    participant C as Final Point
    participant A_prime as Reflected Point A'(x,-y)

    A->>B: Incident Ray
    B->>C: Reflected Ray
    Note over A_prime, B, C: These three points are collinear!

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6. Common Pitfalls & Traps

• Slope Signs: Be careful with signs when calculating slope. A common mistake is $\frac{y_1-y_2}{x_2-x_1}$ instead of $\frac{y_2-y_1}{x_2-x_1}$. Keep the order consistent.
• Perpendicular Slope: The perpendicular slope is the negative reciprocal (e.g., if $m=2$, $m_{\perp }=-1/2$), not just the negative ($m=-2$) or the reciprocal ($m=1/2$).
• Distance Formula Absolute Value: The numerator in the point-to-line distance formula, $|A{x}_0+B{y}_0+C|$, uses an absolute value because distance must be non-negative. Forgetting this can lead to incorrect negative distances.
• SSE Calculation: Always remember to square the error $(y_{actual}-y_{predicted})$. A common slip-up is to sum the errors directly, which can lead to cancellation and an incorrect, smaller result.

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7. Quick Refresher Handbook

Concept	Formula
Slope (m)	$m=\frac{y_2-y_1}{x_2-x_1}$
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{
Midpoint Formula	$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
Section Formula (Internal)	$(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$
Sum of Squared Errors	SSE = $\sum (y_{actual}-y_{predicted}{)}^2$